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Evaluate the integral: $int_0^1 fracln(x+1)x^2+1 mathrm dx$
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Compute
$$int_0^1 fracln(x+1)x^2+1 mathrm dx$$
calculus real-analysis integration definite-integrals
edited Aug 29 '14 at 22:39
Ali Caglayan
3,68763161
asked Jun 9 '12 at 7:49
user 23571113
21.7k870218
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show 2 more comments
up vote
72
down vote
favorite
Compute
$$int_0^1 fracln(x+1)x^2+1 mathrm dx$$
calculus real-analysis integration definite-integrals
edited Aug 29 '14 at 22:39
Ali Caglayan
3,68763161
asked Jun 9 '12 at 7:49
user 23571113
21.7k870218
wolfram gives the answer as 0.272198
– user9413
Jun 9 '12 at 7:57
Mathematica gives the answer $fracpi8 log 2$. But I do not know how it computed this number...
– Siminore
Jun 9 '12 at 8:25
@Chris: Even, i thought of that only :)
– user9413
Jun 9 '12 at 8:26
7
Why dont you try and solve these Integrals yourself. Browsing through your most recent questions, you have had this type of question almost exclusively. Other users get downvoted for this. I see no reason not to hint the same to you and at least show some effort. -1
– CBenni
Mar 20 '13 at 17:01
3
Why don't you try to show some of your work....
– user210387
Jun 12 '15 at 10:29
 |Â
show 2 more comments
up vote
72
down vote
favorite
up vote
72
down vote
favorite
Compute
$$int_0^1 fracln(x+1)x^2+1 mathrm dx$$
calculus real-analysis integration definite-integrals
edited Aug 29 '14 at 22:39
Ali Caglayan
3,68763161
asked Jun 9 '12 at 7:49
user 23571113
21.7k870218
Compute
$$int_0^1 fracln(x+1)x^2+1 mathrm dx$$
calculus real-analysis integration definite-integrals
edited Aug 29 '14 at 22:39
Ali Caglayan
3,68763161
asked Jun 9 '12 at 7:49
user 23571113
21.7k870218
edited Aug 29 '14 at 22:39
Ali Caglayan
3,68763161
edited Aug 29 '14 at 22:39
Ali Caglayan
3,68763161
edited Aug 29 '14 at 22:39
Ali Caglayan
3,68763161
3,68763161
asked Jun 9 '12 at 7:49
user 23571113
21.7k870218
asked Jun 9 '12 at 7:49
user 23571113
21.7k870218
asked Jun 9 '12 at 7:49
user 23571113
21.7k870218
21.7k870218
wolfram gives the answer as 0.272198
– user9413
Jun 9 '12 at 7:57
Mathematica gives the answer $fracpi8 log 2$. But I do not know how it computed this number...
– Siminore
Jun 9 '12 at 8:25
@Chris: Even, i thought of that only :)
– user9413
Jun 9 '12 at 8:26
7
Why dont you try and solve these Integrals yourself. Browsing through your most recent questions, you have had this type of question almost exclusively. Other users get downvoted for this. I see no reason not to hint the same to you and at least show some effort. -1
– CBenni
Mar 20 '13 at 17:01
3
Why don't you try to show some of your work....
– user210387
Jun 12 '15 at 10:29
 |Â
show 2 more comments
wolfram gives the answer as 0.272198
– user9413
Jun 9 '12 at 7:57
Mathematica gives the answer $fracpi8 log 2$. But I do not know how it computed this number...
– Siminore
Jun 9 '12 at 8:25
@Chris: Even, i thought of that only :)
– user9413
Jun 9 '12 at 8:26
7
Why dont you try and solve these Integrals yourself. Browsing through your most recent questions, you have had this type of question almost exclusively. Other users get downvoted for this. I see no reason not to hint the same to you and at least show some effort. -1
– CBenni
Mar 20 '13 at 17:01
3
Why don't you try to show some of your work....
– user210387
Jun 12 '15 at 10:29
wolfram gives the answer as 0.272198
– user9413
Jun 9 '12 at 7:57
wolfram gives the answer as 0.272198
– user9413
Jun 9 '12 at 7:57
Mathematica gives the answer $fracpi8 log 2$. But I do not know how it computed this number...
– Siminore
Jun 9 '12 at 8:25
Mathematica gives the answer $fracpi8 log 2$. But I do not know how it computed this number...
– Siminore
Jun 9 '12 at 8:25
@Chris: Even, i thought of that only :)
– user9413
Jun 9 '12 at 8:26
@Chris: Even, i thought of that only :)
– user9413
Jun 9 '12 at 8:26
7
7
Why dont you try and solve these Integrals yourself. Browsing through your most recent questions, you have had this type of question almost exclusively. Other users get downvoted for this. I see no reason not to hint the same to you and at least show some effort. -1
– CBenni
Mar 20 '13 at 17:01
Why dont you try and solve these Integrals yourself. Browsing through your most recent questions, you have had this type of question almost exclusively. Other users get downvoted for this. I see no reason not to hint the same to you and at least show some effort. -1
– CBenni
Mar 20 '13 at 17:01
3
3
Why don't you try to show some of your work....
– user210387
Jun 12 '15 at 10:29
Why don't you try to show some of your work....
– user210387
Jun 12 '15 at 10:29
 |Â
show 2 more comments
7 Answers
7
active
oldest
votes
up vote
92
down vote
accepted
Put $x = tantheta$, then your integral transforms to $$I= int_0^pi/4 log(1+tantheta) dtheta tag1$$
Now using the property that $$int_0^a f(x) dx = int_0^a f(a-x) dx$$ we have $$I = int_0^pi/4 logbiggl(1+tanBigl(fracpi4-thetaBigr)biggr) dtheta = int_0^pi/4 logbiggl(frac21+tantheta biggr) dthetatag2$$
Adding $(1)$ and $(2)$ we get $$2I = int_0^pi/4 log(2) dthetaRightarrow I= log(2) cdot fracpi8$$
add a comment |Â
up vote
50
down vote
Consider:
$$I(a) = int_0^1 fracln (1+ax)1+x^2 , dx$$
than, the derivative $I'$ is equal:
$$I'(a) = int_0^1 fracx(1+ax)(1+x^2) , dx = frac2 a arctan x - 2ln (1+a x) + ln (1+x^2)2(1+a^2) Big|_0^1\
= fracpi a + 2 ln 24(1+a^2) - fracln (1+a)1+a^2$$
Hence:
$$I(1) = int_0^1 left( fracpi a + 2 ln 24(1+a^2) - fracln (1+a)1+a^2 right) , da \
2 I(1) = int_0^1 fracpi a + 2 ln 24(1+a^2) , da = fracpi4 ln 2$$
Divide both sides by $2$ and you're done.
answered Jun 9 '12 at 9:05
qoqosz
2,6771224
Yes. It's very excellent and different method.
– Prasad G
Jun 9 '12 at 9:11
Thanks, but it needs much more calculations than the soulution proposed by @Chandrasekhar :)
– qoqosz
Jun 9 '12 at 9:13
@qoqosz Nice way of seeing how differentiation under the integral sign works. Thanks +1.
– user9413
Jun 9 '12 at 9:25
@Mark Hurd: right.
– user 23571113
Jun 9 '12 at 11:49
1
@Astrobleme in case you are still wondering, it is partial fractions!
– Vincent
Nov 27 '17 at 14:01
 |Â
show 1 more comment
up vote
38
down vote
Good evening,
I've got another method by putting $x=(1-t)/(1+t)$, we obtain
$$int_0^1fracln (x+1)x^2+1dx=int_1^0fraclnfrac21+tleft(frac1-t1+tright)^2+1cdotleft-frac2(1+t)^2rightdt =int_0^1fracln 2-ln (1+t)t^2+1 dt.$$
You can finish easily.
edited Jun 4 '13 at 21:10
mrf
36.6k54484
answered Jun 4 '13 at 21:05
Boulid
38132
4
That substitution is very powerful, and used it when I met ugly integrals. I didn't use it here (unfortunately). Great solution! (+1)
– user 23571113
Jun 4 '13 at 21:09
How is the last expression any easier to computre than what we started with ?
– CivilSigma
Sep 6 '15 at 19:58
1
@CivilSigma, if you look at the first and last expressions, you see you have an equation of the form $A=B-A$, where $A$ is the integral you're trying to evaluate and $B$ is an integral that's easy to evaluate.
– Barry Cipra
Sep 13 '15 at 15:10
Great solution (+1) but how does one even think of that substitution in such situation? I mean, this is a clearly a "let's try it out and hope it works method". The accepted solution is more straightforward.
– Gaurang Tandon
Mar 10 at 9:50
add a comment |Â
up vote
31
down vote
If $(1+x)(1+y)=2$, then
$$beginalign
x&=frac1-y1+y\
1+x^2&=2frac1+y^2(1+y)^2\
frac1+x^21+x&=frac1+y^21+y
endaligntag1
$$
and since $(1+y),mathrmdx+(1+x),mathrmdy=0$ we get
$$
fracmathrmdx1+x^2=-fracmathrmdy1+y^2tag2
$$
Therefore,
$$
beginalign
int_0^1fraclog(1+x)1+x^2mathrmdx
&=int_0^1fraclog(2)-log(1+y)1+y^2mathrmdytag3
endalign
$$
Adding the left side to both sides and dividing by $2$ yields
$$
beginalign
int_0^1fraclog(1+x)1+x^2mathrmdx
&=frac12int_0^1fraclog(2)1+y^2mathrmdy\
&=fracpi8log(2)tag4
endalign
$$
answered Mar 20 '13 at 13:35
robjohn♦
258k26297612
I excuse Mr robjohn. I did'nt see your answer.
– Boulid
Jun 4 '13 at 21:09
add a comment |Â
up vote
22
down vote
Start with $$beginalign*
int_0^pi/4 ln(1+tan x)dx &= int_0^pi/4 ln(sin x+cos x)dx - int_0^pi/4 ln(cos x)dx \
&= int_0^pi/4 lnleft(cos(x-fracpi4)right)dx +int_0^pi/4 ln(sqrt 2)dx - int_0^pi/4 ln(cos x)dx.
endalign*$$ Now change $pi/4-x=t$ in the first integral: $$=int_0^pi/4 ln(cos t) dt +int_0^pi/4 ln(sqrt 2)dx - int_0^pi/4 ln(cos x)dx$$ and the result follows. Changing $x=tan u$ in the first integral yields your integral. As far as I know these are said Bertrand's integrals.
@Chandrasehkar: see here http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf
edited Jun 9 '12 at 12:07
Ben Millwood
11.1k32049
answered Jun 9 '12 at 9:16
Unoqualunque
65144
1
@Unoqualunque: The link was really helpful. thanks
– user9413
Jun 9 '12 at 10:01
add a comment |Â
up vote
4
down vote
Let us consider $$A=iint_[0,1]^2fracx(1+xy)(1+x^2)dx dy$$
By Fubini's theorem, we have : $$A=int_0^1left[frac11+x^2int_0^1fracx,dy1+xyright]dx=int_0^1fracln(1+x)1+x^2dx$$ and$$A=int_0^1left[int_0^1fracx(1+xy)(1+x^2)dxright]dy$$But$$fracx(1+xy)(1+x^2)=frac11+y^2left(frac-y1+xy+fracx+y1+x^2right)$$and therefore$$A=int_0^1frac11+y^2left(-ln(1+y)+fracln(2)2+fracpi yyright)dy=-A+fracpiln(2)4$$Finally :$$boxedint_0^1fracln(1+x)1+x^2dx=fracpiln(2)8$$
answered Jan 12 '17 at 12:43
Adren
5,272319
How did you see this magical decomposition: $fracx(1+xy)(1+x^2) = frac11+y^2 left( frac-y1+xy + fracx+y1+x^2 right)$ ?
– Sandeep Silwal
Mar 17 '17 at 20:37
add a comment |Â
up vote
-2
down vote
you can use the residu theory to calculate the integral of ln|1+x|/(1+x^2) between - inf and + inf by putting f(z)=ln|1+z|/(1+z^2) amd we obtain (Pi/2)*ln(2)
then int(ln|1+x|/(1+x^2),x=-inf..inf)=
int(same function,x=-inf..-1)+int(same function,x=-1..0)+int(same function,x=0..1)+int(same function,x=1..inf) then by putting t=-x in the first and second integrals we obtain:
int(ln|1-x^2|/(1+x^2),x=0..1)+int(ln|1-x^2|/(1+x^2),x=1..inf)=(Pi/2)*ln(2)
then by putting t=1/x in the second integral we obtain:
int(ln|1-x^2|/(1+x^2),x=0..1)-int(ln(x)/(1+x^2),x=0..1)=(Pi/4)*ln(2)
and we put x=(1-t)/(1+t) in the first integral
we obtain after calculus :
int(ln(1+x)/(1+x^2),x=0..1)=(Pi/8)*ln(2)
answered Jul 9 '16 at 17:15
Michel
11
7
It's almost impossible to read this. You should find and read an introduction to MathJax, like this.
– Henrik
Jul 9 '16 at 17:39
In posting an Answer to a (by now four year old) Question having already an Accepted Answer, it is helpful to your Readers to highlight what new information you provide.
– hardmath
Jul 9 '16 at 19:48
why it's impossible ? 😳
– Michel
Jul 30 '16 at 9:15
@Michel I mean, compared to the accepted answer, which one is much easier to read? Also, you spelled "residue" wrong.
– Frank W.
May 20 at 4:18
add a comment |Â
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
92
down vote
accepted
Put $x = tantheta$, then your integral transforms to $$I= int_0^pi/4 log(1+tantheta) dtheta tag1$$
Now using the property that $$int_0^a f(x) dx = int_0^a f(a-x) dx$$ we have $$I = int_0^pi/4 logbiggl(1+tanBigl(fracpi4-thetaBigr)biggr) dtheta = int_0^pi/4 logbiggl(frac21+tantheta biggr) dthetatag2$$
Adding $(1)$ and $(2)$ we get $$2I = int_0^pi/4 log(2) dthetaRightarrow I= log(2) cdot fracpi8$$
add a comment |Â
up vote
92
down vote
accepted
Put $x = tantheta$, then your integral transforms to $$I= int_0^pi/4 log(1+tantheta) dtheta tag1$$
Now using the property that $$int_0^a f(x) dx = int_0^a f(a-x) dx$$ we have $$I = int_0^pi/4 logbiggl(1+tanBigl(fracpi4-thetaBigr)biggr) dtheta = int_0^pi/4 logbiggl(frac21+tantheta biggr) dthetatag2$$
Adding $(1)$ and $(2)$ we get $$2I = int_0^pi/4 log(2) dthetaRightarrow I= log(2) cdot fracpi8$$
add a comment |Â
up vote
92
down vote
accepted
up vote
92
down vote
accepted
Put $x = tantheta$, then your integral transforms to $$I= int_0^pi/4 log(1+tantheta) dtheta tag1$$
Now using the property that $$int_0^a f(x) dx = int_0^a f(a-x) dx$$ we have $$I = int_0^pi/4 logbiggl(1+tanBigl(fracpi4-thetaBigr)biggr) dtheta = int_0^pi/4 logbiggl(frac21+tantheta biggr) dthetatag2$$
Adding $(1)$ and $(2)$ we get $$2I = int_0^pi/4 log(2) dthetaRightarrow I= log(2) cdot fracpi8$$
Put $x = tantheta$, then your integral transforms to $$I= int_0^pi/4 log(1+tantheta) dtheta tag1$$
Now using the property that $$int_0^a f(x) dx = int_0^a f(a-x) dx$$ we have $$I = int_0^pi/4 logbiggl(1+tanBigl(fracpi4-thetaBigr)biggr) dtheta = int_0^pi/4 logbiggl(frac21+tantheta biggr) dthetatag2$$
Adding $(1)$ and $(2)$ we get $$2I = int_0^pi/4 log(2) dthetaRightarrow I= log(2) cdot fracpi8$$
edited Jul 23 '14 at 16:38
answered Jun 9 '12 at 9:04
user9413
add a comment |Â
add a comment |Â
up vote
50
down vote
Consider:
$$I(a) = int_0^1 fracln (1+ax)1+x^2 , dx$$
than, the derivative $I'$ is equal:
$$I'(a) = int_0^1 fracx(1+ax)(1+x^2) , dx = frac2 a arctan x - 2ln (1+a x) + ln (1+x^2)2(1+a^2) Big|_0^1\
= fracpi a + 2 ln 24(1+a^2) - fracln (1+a)1+a^2$$
Hence:
$$I(1) = int_0^1 left( fracpi a + 2 ln 24(1+a^2) - fracln (1+a)1+a^2 right) , da \
2 I(1) = int_0^1 fracpi a + 2 ln 24(1+a^2) , da = fracpi4 ln 2$$
Divide both sides by $2$ and you're done.
answered Jun 9 '12 at 9:05
qoqosz
2,6771224
Yes. It's very excellent and different method.
– Prasad G
Jun 9 '12 at 9:11
Thanks, but it needs much more calculations than the soulution proposed by @Chandrasekhar :)
– qoqosz
Jun 9 '12 at 9:13
@qoqosz Nice way of seeing how differentiation under the integral sign works. Thanks +1.
– user9413
Jun 9 '12 at 9:25
@Mark Hurd: right.
– user 23571113
Jun 9 '12 at 11:49
1
@Astrobleme in case you are still wondering, it is partial fractions!
– Vincent
Nov 27 '17 at 14:01
 |Â
show 1 more comment
up vote
50
down vote
Consider:
$$I(a) = int_0^1 fracln (1+ax)1+x^2 , dx$$
than, the derivative $I'$ is equal:
$$I'(a) = int_0^1 fracx(1+ax)(1+x^2) , dx = frac2 a arctan x - 2ln (1+a x) + ln (1+x^2)2(1+a^2) Big|_0^1\
= fracpi a + 2 ln 24(1+a^2) - fracln (1+a)1+a^2$$
Hence:
$$I(1) = int_0^1 left( fracpi a + 2 ln 24(1+a^2) - fracln (1+a)1+a^2 right) , da \
2 I(1) = int_0^1 fracpi a + 2 ln 24(1+a^2) , da = fracpi4 ln 2$$
Divide both sides by $2$ and you're done.
answered Jun 9 '12 at 9:05
qoqosz
2,6771224
Yes. It's very excellent and different method.
– Prasad G
Jun 9 '12 at 9:11
Thanks, but it needs much more calculations than the soulution proposed by @Chandrasekhar :)
– qoqosz
Jun 9 '12 at 9:13
@qoqosz Nice way of seeing how differentiation under the integral sign works. Thanks +1.
– user9413
Jun 9 '12 at 9:25
@Mark Hurd: right.
– user 23571113
Jun 9 '12 at 11:49
1
@Astrobleme in case you are still wondering, it is partial fractions!
– Vincent
Nov 27 '17 at 14:01
 |Â
show 1 more comment
up vote
50
down vote
up vote
50
down vote
Consider:
$$I(a) = int_0^1 fracln (1+ax)1+x^2 , dx$$
than, the derivative $I'$ is equal:
$$I'(a) = int_0^1 fracx(1+ax)(1+x^2) , dx = frac2 a arctan x - 2ln (1+a x) + ln (1+x^2)2(1+a^2) Big|_0^1\
= fracpi a + 2 ln 24(1+a^2) - fracln (1+a)1+a^2$$
Hence:
$$I(1) = int_0^1 left( fracpi a + 2 ln 24(1+a^2) - fracln (1+a)1+a^2 right) , da \
2 I(1) = int_0^1 fracpi a + 2 ln 24(1+a^2) , da = fracpi4 ln 2$$
Divide both sides by $2$ and you're done.
answered Jun 9 '12 at 9:05
qoqosz
2,6771224
Consider:
$$I(a) = int_0^1 fracln (1+ax)1+x^2 , dx$$
than, the derivative $I'$ is equal:
$$I'(a) = int_0^1 fracx(1+ax)(1+x^2) , dx = frac2 a arctan x - 2ln (1+a x) + ln (1+x^2)2(1+a^2) Big|_0^1\
= fracpi a + 2 ln 24(1+a^2) - fracln (1+a)1+a^2$$
Hence:
$$I(1) = int_0^1 left( fracpi a + 2 ln 24(1+a^2) - fracln (1+a)1+a^2 right) , da \
2 I(1) = int_0^1 fracpi a + 2 ln 24(1+a^2) , da = fracpi4 ln 2$$
Divide both sides by $2$ and you're done.
answered Jun 9 '12 at 9:05
qoqosz
2,6771224
answered Jun 9 '12 at 9:05
qoqosz
2,6771224
answered Jun 9 '12 at 9:05
qoqosz
2,6771224
answered Jun 9 '12 at 9:05
qoqosz
2,6771224
2,6771224
Yes. It's very excellent and different method.
– Prasad G
Jun 9 '12 at 9:11
Thanks, but it needs much more calculations than the soulution proposed by @Chandrasekhar :)
– qoqosz
Jun 9 '12 at 9:13
@qoqosz Nice way of seeing how differentiation under the integral sign works. Thanks +1.
– user9413
Jun 9 '12 at 9:25
@Mark Hurd: right.
– user 23571113
Jun 9 '12 at 11:49
1
@Astrobleme in case you are still wondering, it is partial fractions!
– Vincent
Nov 27 '17 at 14:01
 |Â
show 1 more comment
Yes. It's very excellent and different method.
– Prasad G
Jun 9 '12 at 9:11
Thanks, but it needs much more calculations than the soulution proposed by @Chandrasekhar :)
– qoqosz
Jun 9 '12 at 9:13
@qoqosz Nice way of seeing how differentiation under the integral sign works. Thanks +1.
– user9413
Jun 9 '12 at 9:25
@Mark Hurd: right.
– user 23571113
Jun 9 '12 at 11:49
1
@Astrobleme in case you are still wondering, it is partial fractions!
– Vincent
Nov 27 '17 at 14:01
Yes. It's very excellent and different method.
– Prasad G
Jun 9 '12 at 9:11
Yes. It's very excellent and different method.
– Prasad G
Jun 9 '12 at 9:11
Thanks, but it needs much more calculations than the soulution proposed by @Chandrasekhar :)
– qoqosz
Jun 9 '12 at 9:13
Thanks, but it needs much more calculations than the soulution proposed by @Chandrasekhar :)
– qoqosz
Jun 9 '12 at 9:13
@qoqosz Nice way of seeing how differentiation under the integral sign works. Thanks +1.
– user9413
Jun 9 '12 at 9:25
@qoqosz Nice way of seeing how differentiation under the integral sign works. Thanks +1.
– user9413
Jun 9 '12 at 9:25
@Mark Hurd: right.
– user 23571113
Jun 9 '12 at 11:49
@Mark Hurd: right.
– user 23571113
Jun 9 '12 at 11:49
1
1
@Astrobleme in case you are still wondering, it is partial fractions!
– Vincent
Nov 27 '17 at 14:01
@Astrobleme in case you are still wondering, it is partial fractions!
– Vincent
Nov 27 '17 at 14:01
 |Â
show 1 more comment
up vote
38
down vote
Good evening,
I've got another method by putting $x=(1-t)/(1+t)$, we obtain
$$int_0^1fracln (x+1)x^2+1dx=int_1^0fraclnfrac21+tleft(frac1-t1+tright)^2+1cdotleft-frac2(1+t)^2rightdt =int_0^1fracln 2-ln (1+t)t^2+1 dt.$$
You can finish easily.
edited Jun 4 '13 at 21:10
mrf
36.6k54484
answered Jun 4 '13 at 21:05
Boulid
38132
4
That substitution is very powerful, and used it when I met ugly integrals. I didn't use it here (unfortunately). Great solution! (+1)
– user 23571113
Jun 4 '13 at 21:09
How is the last expression any easier to computre than what we started with ?
– CivilSigma
Sep 6 '15 at 19:58
1
@CivilSigma, if you look at the first and last expressions, you see you have an equation of the form $A=B-A$, where $A$ is the integral you're trying to evaluate and $B$ is an integral that's easy to evaluate.
– Barry Cipra
Sep 13 '15 at 15:10
Great solution (+1) but how does one even think of that substitution in such situation? I mean, this is a clearly a "let's try it out and hope it works method". The accepted solution is more straightforward.
– Gaurang Tandon
Mar 10 at 9:50
add a comment |Â
up vote
38
down vote
Good evening,
I've got another method by putting $x=(1-t)/(1+t)$, we obtain
$$int_0^1fracln (x+1)x^2+1dx=int_1^0fraclnfrac21+tleft(frac1-t1+tright)^2+1cdotleft-frac2(1+t)^2rightdt =int_0^1fracln 2-ln (1+t)t^2+1 dt.$$
You can finish easily.
edited Jun 4 '13 at 21:10
mrf
36.6k54484
answered Jun 4 '13 at 21:05
Boulid
38132
4
That substitution is very powerful, and used it when I met ugly integrals. I didn't use it here (unfortunately). Great solution! (+1)
– user 23571113
Jun 4 '13 at 21:09
How is the last expression any easier to computre than what we started with ?
– CivilSigma
Sep 6 '15 at 19:58
1
@CivilSigma, if you look at the first and last expressions, you see you have an equation of the form $A=B-A$, where $A$ is the integral you're trying to evaluate and $B$ is an integral that's easy to evaluate.
– Barry Cipra
Sep 13 '15 at 15:10
Great solution (+1) but how does one even think of that substitution in such situation? I mean, this is a clearly a "let's try it out and hope it works method". The accepted solution is more straightforward.
– Gaurang Tandon
Mar 10 at 9:50
add a comment |Â
up vote
38
down vote
up vote
38
down vote
Good evening,
I've got another method by putting $x=(1-t)/(1+t)$, we obtain
$$int_0^1fracln (x+1)x^2+1dx=int_1^0fraclnfrac21+tleft(frac1-t1+tright)^2+1cdotleft-frac2(1+t)^2rightdt =int_0^1fracln 2-ln (1+t)t^2+1 dt.$$
You can finish easily.
edited Jun 4 '13 at 21:10
mrf
36.6k54484
answered Jun 4 '13 at 21:05
Boulid
38132
Good evening,
I've got another method by putting $x=(1-t)/(1+t)$, we obtain
$$int_0^1fracln (x+1)x^2+1dx=int_1^0fraclnfrac21+tleft(frac1-t1+tright)^2+1cdotleft-frac2(1+t)^2rightdt =int_0^1fracln 2-ln (1+t)t^2+1 dt.$$
You can finish easily.
edited Jun 4 '13 at 21:10
mrf
36.6k54484
answered Jun 4 '13 at 21:05
Boulid
38132
edited Jun 4 '13 at 21:10
mrf
36.6k54484
edited Jun 4 '13 at 21:10
mrf
36.6k54484
edited Jun 4 '13 at 21:10
mrf
36.6k54484
36.6k54484
answered Jun 4 '13 at 21:05
Boulid
38132
answered Jun 4 '13 at 21:05
Boulid
38132
answered Jun 4 '13 at 21:05
Boulid
38132
38132
4
That substitution is very powerful, and used it when I met ugly integrals. I didn't use it here (unfortunately). Great solution! (+1)
– user 23571113
Jun 4 '13 at 21:09
How is the last expression any easier to computre than what we started with ?
– CivilSigma
Sep 6 '15 at 19:58
1
@CivilSigma, if you look at the first and last expressions, you see you have an equation of the form $A=B-A$, where $A$ is the integral you're trying to evaluate and $B$ is an integral that's easy to evaluate.
– Barry Cipra
Sep 13 '15 at 15:10
Great solution (+1) but how does one even think of that substitution in such situation? I mean, this is a clearly a "let's try it out and hope it works method". The accepted solution is more straightforward.
– Gaurang Tandon
Mar 10 at 9:50
add a comment |Â
4
That substitution is very powerful, and used it when I met ugly integrals. I didn't use it here (unfortunately). Great solution! (+1)
– user 23571113
Jun 4 '13 at 21:09
How is the last expression any easier to computre than what we started with ?
– CivilSigma
Sep 6 '15 at 19:58
1
@CivilSigma, if you look at the first and last expressions, you see you have an equation of the form $A=B-A$, where $A$ is the integral you're trying to evaluate and $B$ is an integral that's easy to evaluate.
– Barry Cipra
Sep 13 '15 at 15:10
Great solution (+1) but how does one even think of that substitution in such situation? I mean, this is a clearly a "let's try it out and hope it works method". The accepted solution is more straightforward.
– Gaurang Tandon
Mar 10 at 9:50
4
4
That substitution is very powerful, and used it when I met ugly integrals. I didn't use it here (unfortunately). Great solution! (+1)
– user 23571113
Jun 4 '13 at 21:09
That substitution is very powerful, and used it when I met ugly integrals. I didn't use it here (unfortunately). Great solution! (+1)
– user 23571113
Jun 4 '13 at 21:09
How is the last expression any easier to computre than what we started with ?
– CivilSigma
Sep 6 '15 at 19:58
How is the last expression any easier to computre than what we started with ?
– CivilSigma
Sep 6 '15 at 19:58
1
1
@CivilSigma, if you look at the first and last expressions, you see you have an equation of the form $A=B-A$, where $A$ is the integral you're trying to evaluate and $B$ is an integral that's easy to evaluate.
– Barry Cipra
Sep 13 '15 at 15:10
@CivilSigma, if you look at the first and last expressions, you see you have an equation of the form $A=B-A$, where $A$ is the integral you're trying to evaluate and $B$ is an integral that's easy to evaluate.
– Barry Cipra
Sep 13 '15 at 15:10
Great solution (+1) but how does one even think of that substitution in such situation? I mean, this is a clearly a "let's try it out and hope it works method". The accepted solution is more straightforward.
– Gaurang Tandon
Mar 10 at 9:50
Great solution (+1) but how does one even think of that substitution in such situation? I mean, this is a clearly a "let's try it out and hope it works method". The accepted solution is more straightforward.
– Gaurang Tandon
Mar 10 at 9:50
add a comment |Â
up vote
31
down vote
If $(1+x)(1+y)=2$, then
$$beginalign
x&=frac1-y1+y\
1+x^2&=2frac1+y^2(1+y)^2\
frac1+x^21+x&=frac1+y^21+y
endaligntag1
$$
and since $(1+y),mathrmdx+(1+x),mathrmdy=0$ we get
$$
fracmathrmdx1+x^2=-fracmathrmdy1+y^2tag2
$$
Therefore,
$$
beginalign
int_0^1fraclog(1+x)1+x^2mathrmdx
&=int_0^1fraclog(2)-log(1+y)1+y^2mathrmdytag3
endalign
$$
Adding the left side to both sides and dividing by $2$ yields
$$
beginalign
int_0^1fraclog(1+x)1+x^2mathrmdx
&=frac12int_0^1fraclog(2)1+y^2mathrmdy\
&=fracpi8log(2)tag4
endalign
$$
answered Mar 20 '13 at 13:35
robjohn♦
258k26297612
I excuse Mr robjohn. I did'nt see your answer.
– Boulid
Jun 4 '13 at 21:09
add a comment |Â
up vote
31
down vote
If $(1+x)(1+y)=2$, then
$$beginalign
x&=frac1-y1+y\
1+x^2&=2frac1+y^2(1+y)^2\
frac1+x^21+x&=frac1+y^21+y
endaligntag1
$$
and since $(1+y),mathrmdx+(1+x),mathrmdy=0$ we get
$$
fracmathrmdx1+x^2=-fracmathrmdy1+y^2tag2
$$
Therefore,
$$
beginalign
int_0^1fraclog(1+x)1+x^2mathrmdx
&=int_0^1fraclog(2)-log(1+y)1+y^2mathrmdytag3
endalign
$$
Adding the left side to both sides and dividing by $2$ yields
$$
beginalign
int_0^1fraclog(1+x)1+x^2mathrmdx
&=frac12int_0^1fraclog(2)1+y^2mathrmdy\
&=fracpi8log(2)tag4
endalign
$$
answered Mar 20 '13 at 13:35
robjohn♦
258k26297612
I excuse Mr robjohn. I did'nt see your answer.
– Boulid
Jun 4 '13 at 21:09
add a comment |Â
up vote
31
down vote
up vote
31
down vote
If $(1+x)(1+y)=2$, then
$$beginalign
x&=frac1-y1+y\
1+x^2&=2frac1+y^2(1+y)^2\
frac1+x^21+x&=frac1+y^21+y
endaligntag1
$$
and since $(1+y),mathrmdx+(1+x),mathrmdy=0$ we get
$$
fracmathrmdx1+x^2=-fracmathrmdy1+y^2tag2
$$
Therefore,
$$
beginalign
int_0^1fraclog(1+x)1+x^2mathrmdx
&=int_0^1fraclog(2)-log(1+y)1+y^2mathrmdytag3
endalign
$$
Adding the left side to both sides and dividing by $2$ yields
$$
beginalign
int_0^1fraclog(1+x)1+x^2mathrmdx
&=frac12int_0^1fraclog(2)1+y^2mathrmdy\
&=fracpi8log(2)tag4
endalign
$$
answered Mar 20 '13 at 13:35
robjohn♦
258k26297612
If $(1+x)(1+y)=2$, then
$$beginalign
x&=frac1-y1+y\
1+x^2&=2frac1+y^2(1+y)^2\
frac1+x^21+x&=frac1+y^21+y
endaligntag1
$$
and since $(1+y),mathrmdx+(1+x),mathrmdy=0$ we get
$$
fracmathrmdx1+x^2=-fracmathrmdy1+y^2tag2
$$
Therefore,
$$
beginalign
int_0^1fraclog(1+x)1+x^2mathrmdx
&=int_0^1fraclog(2)-log(1+y)1+y^2mathrmdytag3
endalign
$$
Adding the left side to both sides and dividing by $2$ yields
$$
beginalign
int_0^1fraclog(1+x)1+x^2mathrmdx
&=frac12int_0^1fraclog(2)1+y^2mathrmdy\
&=fracpi8log(2)tag4
endalign
$$
answered Mar 20 '13 at 13:35
robjohn♦
258k26297612
answered Mar 20 '13 at 13:35
robjohn♦
258k26297612
answered Mar 20 '13 at 13:35
robjohn♦
258k26297612
answered Mar 20 '13 at 13:35
robjohn♦
258k26297612
258k26297612
I excuse Mr robjohn. I did'nt see your answer.
– Boulid
Jun 4 '13 at 21:09
add a comment |Â
I excuse Mr robjohn. I did'nt see your answer.
– Boulid
Jun 4 '13 at 21:09
I excuse Mr robjohn. I did'nt see your answer.
– Boulid
Jun 4 '13 at 21:09
I excuse Mr robjohn. I did'nt see your answer.
– Boulid
Jun 4 '13 at 21:09
add a comment |Â
up vote
22
down vote
Start with $$beginalign*
int_0^pi/4 ln(1+tan x)dx &= int_0^pi/4 ln(sin x+cos x)dx - int_0^pi/4 ln(cos x)dx \
&= int_0^pi/4 lnleft(cos(x-fracpi4)right)dx +int_0^pi/4 ln(sqrt 2)dx - int_0^pi/4 ln(cos x)dx.
endalign*$$ Now change $pi/4-x=t$ in the first integral: $$=int_0^pi/4 ln(cos t) dt +int_0^pi/4 ln(sqrt 2)dx - int_0^pi/4 ln(cos x)dx$$ and the result follows. Changing $x=tan u$ in the first integral yields your integral. As far as I know these are said Bertrand's integrals.
@Chandrasehkar: see here http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf
edited Jun 9 '12 at 12:07
Ben Millwood
11.1k32049
answered Jun 9 '12 at 9:16
Unoqualunque
65144
1
@Unoqualunque: The link was really helpful. thanks
– user9413
Jun 9 '12 at 10:01
add a comment |Â
up vote
22
down vote
Start with $$beginalign*
int_0^pi/4 ln(1+tan x)dx &= int_0^pi/4 ln(sin x+cos x)dx - int_0^pi/4 ln(cos x)dx \
&= int_0^pi/4 lnleft(cos(x-fracpi4)right)dx +int_0^pi/4 ln(sqrt 2)dx - int_0^pi/4 ln(cos x)dx.
endalign*$$ Now change $pi/4-x=t$ in the first integral: $$=int_0^pi/4 ln(cos t) dt +int_0^pi/4 ln(sqrt 2)dx - int_0^pi/4 ln(cos x)dx$$ and the result follows. Changing $x=tan u$ in the first integral yields your integral. As far as I know these are said Bertrand's integrals.
@Chandrasehkar: see here http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf
edited Jun 9 '12 at 12:07
Ben Millwood
11.1k32049
answered Jun 9 '12 at 9:16
Unoqualunque
65144
1
@Unoqualunque: The link was really helpful. thanks
– user9413
Jun 9 '12 at 10:01
add a comment |Â
up vote
22
down vote
up vote
22
down vote
Start with $$beginalign*
int_0^pi/4 ln(1+tan x)dx &= int_0^pi/4 ln(sin x+cos x)dx - int_0^pi/4 ln(cos x)dx \
&= int_0^pi/4 lnleft(cos(x-fracpi4)right)dx +int_0^pi/4 ln(sqrt 2)dx - int_0^pi/4 ln(cos x)dx.
endalign*$$ Now change $pi/4-x=t$ in the first integral: $$=int_0^pi/4 ln(cos t) dt +int_0^pi/4 ln(sqrt 2)dx - int_0^pi/4 ln(cos x)dx$$ and the result follows. Changing $x=tan u$ in the first integral yields your integral. As far as I know these are said Bertrand's integrals.
@Chandrasehkar: see here http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf
edited Jun 9 '12 at 12:07
Ben Millwood
11.1k32049
answered Jun 9 '12 at 9:16
Unoqualunque
65144
Start with $$beginalign*
int_0^pi/4 ln(1+tan x)dx &= int_0^pi/4 ln(sin x+cos x)dx - int_0^pi/4 ln(cos x)dx \
&= int_0^pi/4 lnleft(cos(x-fracpi4)right)dx +int_0^pi/4 ln(sqrt 2)dx - int_0^pi/4 ln(cos x)dx.
endalign*$$ Now change $pi/4-x=t$ in the first integral: $$=int_0^pi/4 ln(cos t) dt +int_0^pi/4 ln(sqrt 2)dx - int_0^pi/4 ln(cos x)dx$$ and the result follows. Changing $x=tan u$ in the first integral yields your integral. As far as I know these are said Bertrand's integrals.
@Chandrasehkar: see here http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf
edited Jun 9 '12 at 12:07
Ben Millwood
11.1k32049
answered Jun 9 '12 at 9:16
Unoqualunque
65144
edited Jun 9 '12 at 12:07
Ben Millwood
11.1k32049
edited Jun 9 '12 at 12:07
Ben Millwood
11.1k32049
edited Jun 9 '12 at 12:07
Ben Millwood
11.1k32049
11.1k32049
answered Jun 9 '12 at 9:16
Unoqualunque
65144
answered Jun 9 '12 at 9:16
Unoqualunque
65144
answered Jun 9 '12 at 9:16
Unoqualunque
65144
65144
1
@Unoqualunque: The link was really helpful. thanks
– user9413
Jun 9 '12 at 10:01
add a comment |Â
1
@Unoqualunque: The link was really helpful. thanks
– user9413
Jun 9 '12 at 10:01
1
1
@Unoqualunque: The link was really helpful. thanks
– user9413
Jun 9 '12 at 10:01
@Unoqualunque: The link was really helpful. thanks
– user9413
Jun 9 '12 at 10:01
add a comment |Â
up vote
4
down vote
Let us consider $$A=iint_[0,1]^2fracx(1+xy)(1+x^2)dx dy$$
By Fubini's theorem, we have : $$A=int_0^1left[frac11+x^2int_0^1fracx,dy1+xyright]dx=int_0^1fracln(1+x)1+x^2dx$$ and$$A=int_0^1left[int_0^1fracx(1+xy)(1+x^2)dxright]dy$$But$$fracx(1+xy)(1+x^2)=frac11+y^2left(frac-y1+xy+fracx+y1+x^2right)$$and therefore$$A=int_0^1frac11+y^2left(-ln(1+y)+fracln(2)2+fracpi yyright)dy=-A+fracpiln(2)4$$Finally :$$boxedint_0^1fracln(1+x)1+x^2dx=fracpiln(2)8$$
answered Jan 12 '17 at 12:43
Adren
5,272319
How did you see this magical decomposition: $fracx(1+xy)(1+x^2) = frac11+y^2 left( frac-y1+xy + fracx+y1+x^2 right)$ ?
– Sandeep Silwal
Mar 17 '17 at 20:37
add a comment |Â
up vote
4
down vote
Let us consider $$A=iint_[0,1]^2fracx(1+xy)(1+x^2)dx dy$$
By Fubini's theorem, we have : $$A=int_0^1left[frac11+x^2int_0^1fracx,dy1+xyright]dx=int_0^1fracln(1+x)1+x^2dx$$ and$$A=int_0^1left[int_0^1fracx(1+xy)(1+x^2)dxright]dy$$But$$fracx(1+xy)(1+x^2)=frac11+y^2left(frac-y1+xy+fracx+y1+x^2right)$$and therefore$$A=int_0^1frac11+y^2left(-ln(1+y)+fracln(2)2+fracpi yyright)dy=-A+fracpiln(2)4$$Finally :$$boxedint_0^1fracln(1+x)1+x^2dx=fracpiln(2)8$$
answered Jan 12 '17 at 12:43
Adren
5,272319
How did you see this magical decomposition: $fracx(1+xy)(1+x^2) = frac11+y^2 left( frac-y1+xy + fracx+y1+x^2 right)$ ?
– Sandeep Silwal
Mar 17 '17 at 20:37
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Let us consider $$A=iint_[0,1]^2fracx(1+xy)(1+x^2)dx dy$$
By Fubini's theorem, we have : $$A=int_0^1left[frac11+x^2int_0^1fracx,dy1+xyright]dx=int_0^1fracln(1+x)1+x^2dx$$ and$$A=int_0^1left[int_0^1fracx(1+xy)(1+x^2)dxright]dy$$But$$fracx(1+xy)(1+x^2)=frac11+y^2left(frac-y1+xy+fracx+y1+x^2right)$$and therefore$$A=int_0^1frac11+y^2left(-ln(1+y)+fracln(2)2+fracpi yyright)dy=-A+fracpiln(2)4$$Finally :$$boxedint_0^1fracln(1+x)1+x^2dx=fracpiln(2)8$$
answered Jan 12 '17 at 12:43
Adren
5,272319
Let us consider $$A=iint_[0,1]^2fracx(1+xy)(1+x^2)dx dy$$
By Fubini's theorem, we have : $$A=int_0^1left[frac11+x^2int_0^1fracx,dy1+xyright]dx=int_0^1fracln(1+x)1+x^2dx$$ and$$A=int_0^1left[int_0^1fracx(1+xy)(1+x^2)dxright]dy$$But$$fracx(1+xy)(1+x^2)=frac11+y^2left(frac-y1+xy+fracx+y1+x^2right)$$and therefore$$A=int_0^1frac11+y^2left(-ln(1+y)+fracln(2)2+fracpi yyright)dy=-A+fracpiln(2)4$$Finally :$$boxedint_0^1fracln(1+x)1+x^2dx=fracpiln(2)8$$
answered Jan 12 '17 at 12:43
Adren
5,272319
answered Jan 12 '17 at 12:43
Adren
5,272319
answered Jan 12 '17 at 12:43
Adren
5,272319
answered Jan 12 '17 at 12:43
Adren
5,272319
5,272319
How did you see this magical decomposition: $fracx(1+xy)(1+x^2) = frac11+y^2 left( frac-y1+xy + fracx+y1+x^2 right)$ ?
– Sandeep Silwal
Mar 17 '17 at 20:37
add a comment |Â
How did you see this magical decomposition: $fracx(1+xy)(1+x^2) = frac11+y^2 left( frac-y1+xy + fracx+y1+x^2 right)$ ?
– Sandeep Silwal
Mar 17 '17 at 20:37
How did you see this magical decomposition: $fracx(1+xy)(1+x^2) = frac11+y^2 left( frac-y1+xy + fracx+y1+x^2 right)$ ?
– Sandeep Silwal
Mar 17 '17 at 20:37
How did you see this magical decomposition: $fracx(1+xy)(1+x^2) = frac11+y^2 left( frac-y1+xy + fracx+y1+x^2 right)$ ?
– Sandeep Silwal
Mar 17 '17 at 20:37
add a comment |Â
up vote
-2
down vote
you can use the residu theory to calculate the integral of ln|1+x|/(1+x^2) between - inf and + inf by putting f(z)=ln|1+z|/(1+z^2) amd we obtain (Pi/2)*ln(2)
then int(ln|1+x|/(1+x^2),x=-inf..inf)=
int(same function,x=-inf..-1)+int(same function,x=-1..0)+int(same function,x=0..1)+int(same function,x=1..inf) then by putting t=-x in the first and second integrals we obtain:
int(ln|1-x^2|/(1+x^2),x=0..1)+int(ln|1-x^2|/(1+x^2),x=1..inf)=(Pi/2)*ln(2)
then by putting t=1/x in the second integral we obtain:
int(ln|1-x^2|/(1+x^2),x=0..1)-int(ln(x)/(1+x^2),x=0..1)=(Pi/4)*ln(2)
and we put x=(1-t)/(1+t) in the first integral
we obtain after calculus :
int(ln(1+x)/(1+x^2),x=0..1)=(Pi/8)*ln(2)
answered Jul 9 '16 at 17:15
Michel
11
7
It's almost impossible to read this. You should find and read an introduction to MathJax, like this.
– Henrik
Jul 9 '16 at 17:39
In posting an Answer to a (by now four year old) Question having already an Accepted Answer, it is helpful to your Readers to highlight what new information you provide.
– hardmath
Jul 9 '16 at 19:48
why it's impossible ? 😳
– Michel
Jul 30 '16 at 9:15
@Michel I mean, compared to the accepted answer, which one is much easier to read? Also, you spelled "residue" wrong.
– Frank W.
May 20 at 4:18
add a comment |Â
up vote
-2
down vote
you can use the residu theory to calculate the integral of ln|1+x|/(1+x^2) between - inf and + inf by putting f(z)=ln|1+z|/(1+z^2) amd we obtain (Pi/2)*ln(2)
then int(ln|1+x|/(1+x^2),x=-inf..inf)=
int(same function,x=-inf..-1)+int(same function,x=-1..0)+int(same function,x=0..1)+int(same function,x=1..inf) then by putting t=-x in the first and second integrals we obtain:
int(ln|1-x^2|/(1+x^2),x=0..1)+int(ln|1-x^2|/(1+x^2),x=1..inf)=(Pi/2)*ln(2)
then by putting t=1/x in the second integral we obtain:
int(ln|1-x^2|/(1+x^2),x=0..1)-int(ln(x)/(1+x^2),x=0..1)=(Pi/4)*ln(2)
and we put x=(1-t)/(1+t) in the first integral
we obtain after calculus :
int(ln(1+x)/(1+x^2),x=0..1)=(Pi/8)*ln(2)
answered Jul 9 '16 at 17:15
Michel
11
7
It's almost impossible to read this. You should find and read an introduction to MathJax, like this.
– Henrik
Jul 9 '16 at 17:39
In posting an Answer to a (by now four year old) Question having already an Accepted Answer, it is helpful to your Readers to highlight what new information you provide.
– hardmath
Jul 9 '16 at 19:48
why it's impossible ? 😳
– Michel
Jul 30 '16 at 9:15
@Michel I mean, compared to the accepted answer, which one is much easier to read? Also, you spelled "residue" wrong.
– Frank W.
May 20 at 4:18
add a comment |Â
up vote
-2
down vote
up vote
-2
down vote
you can use the residu theory to calculate the integral of ln|1+x|/(1+x^2) between - inf and + inf by putting f(z)=ln|1+z|/(1+z^2) amd we obtain (Pi/2)*ln(2)
then int(ln|1+x|/(1+x^2),x=-inf..inf)=
int(same function,x=-inf..-1)+int(same function,x=-1..0)+int(same function,x=0..1)+int(same function,x=1..inf) then by putting t=-x in the first and second integrals we obtain:
int(ln|1-x^2|/(1+x^2),x=0..1)+int(ln|1-x^2|/(1+x^2),x=1..inf)=(Pi/2)*ln(2)
then by putting t=1/x in the second integral we obtain:
int(ln|1-x^2|/(1+x^2),x=0..1)-int(ln(x)/(1+x^2),x=0..1)=(Pi/4)*ln(2)
and we put x=(1-t)/(1+t) in the first integral
we obtain after calculus :
int(ln(1+x)/(1+x^2),x=0..1)=(Pi/8)*ln(2)
answered Jul 9 '16 at 17:15
Michel
11
you can use the residu theory to calculate the integral of ln|1+x|/(1+x^2) between - inf and + inf by putting f(z)=ln|1+z|/(1+z^2) amd we obtain (Pi/2)*ln(2)
then int(ln|1+x|/(1+x^2),x=-inf..inf)=
int(same function,x=-inf..-1)+int(same function,x=-1..0)+int(same function,x=0..1)+int(same function,x=1..inf) then by putting t=-x in the first and second integrals we obtain:
int(ln|1-x^2|/(1+x^2),x=0..1)+int(ln|1-x^2|/(1+x^2),x=1..inf)=(Pi/2)*ln(2)
then by putting t=1/x in the second integral we obtain:
int(ln|1-x^2|/(1+x^2),x=0..1)-int(ln(x)/(1+x^2),x=0..1)=(Pi/4)*ln(2)
and we put x=(1-t)/(1+t) in the first integral
we obtain after calculus :
int(ln(1+x)/(1+x^2),x=0..1)=(Pi/8)*ln(2)
answered Jul 9 '16 at 17:15
Michel
11
answered Jul 9 '16 at 17:15
Michel
11
answered Jul 9 '16 at 17:15
Michel
11
answered Jul 9 '16 at 17:15
Michel
11
11
7
It's almost impossible to read this. You should find and read an introduction to MathJax, like this.
– Henrik
Jul 9 '16 at 17:39
In posting an Answer to a (by now four year old) Question having already an Accepted Answer, it is helpful to your Readers to highlight what new information you provide.
– hardmath
Jul 9 '16 at 19:48
why it's impossible ? 😳
– Michel
Jul 30 '16 at 9:15
@Michel I mean, compared to the accepted answer, which one is much easier to read? Also, you spelled "residue" wrong.
– Frank W.
May 20 at 4:18
add a comment |Â
7
It's almost impossible to read this. You should find and read an introduction to MathJax, like this.
– Henrik
Jul 9 '16 at 17:39
In posting an Answer to a (by now four year old) Question having already an Accepted Answer, it is helpful to your Readers to highlight what new information you provide.
– hardmath
Jul 9 '16 at 19:48
why it's impossible ? 😳
– Michel
Jul 30 '16 at 9:15
@Michel I mean, compared to the accepted answer, which one is much easier to read? Also, you spelled "residue" wrong.
– Frank W.
May 20 at 4:18
7
7
It's almost impossible to read this. You should find and read an introduction to MathJax, like this.
– Henrik
Jul 9 '16 at 17:39
It's almost impossible to read this. You should find and read an introduction to MathJax, like this.
– Henrik
Jul 9 '16 at 17:39
In posting an Answer to a (by now four year old) Question having already an Accepted Answer, it is helpful to your Readers to highlight what new information you provide.
– hardmath
Jul 9 '16 at 19:48
In posting an Answer to a (by now four year old) Question having already an Accepted Answer, it is helpful to your Readers to highlight what new information you provide.
– hardmath
Jul 9 '16 at 19:48
why it's impossible ? 😳
– Michel
Jul 30 '16 at 9:15
why it's impossible ? 😳
– Michel
Jul 30 '16 at 9:15
@Michel I mean, compared to the accepted answer, which one is much easier to read? Also, you spelled "residue" wrong.
– Frank W.
May 20 at 4:18
@Michel I mean, compared to the accepted answer, which one is much easier to read? Also, you spelled "residue" wrong.
– Frank W.
May 20 at 4:18
add a comment |Â
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wolfram gives the answer as 0.272198
– user9413
Jun 9 '12 at 7:57
Mathematica gives the answer $fracpi8 log 2$. But I do not know how it computed this number...
– Siminore
Jun 9 '12 at 8:25
@Chris: Even, i thought of that only :)
– user9413
Jun 9 '12 at 8:26
7
Why dont you try and solve these Integrals yourself. Browsing through your most recent questions, you have had this type of question almost exclusively. Other users get downvoted for this. I see no reason not to hint the same to you and at least show some effort. -1
– CBenni
Mar 20 '13 at 17:01
3
Why don't you try to show some of your work....
– user210387
Jun 12 '15 at 10:29