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[Integral][Please identify problem] $displaystyleint cfrac11+x^4>mathrmd x$
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Here is my attempt. The result is not right. Please help identify the issue(s).
$displaystyle f(x)=intcfrac1x^4+1>mathrmdx$, let $x=tan t$, we have $ mathrmdx = sec^2 t>mathrmdt,> t=tan^-1 xinleft(-cfracpi2,cfracpi2right)$
beginalign
displaystyle f(tan t)&= intcfracsec^2 t> mathrmdt1+tan^4 t=intcfraccos^2 t> mathrmdtcos^4 t+sin^4 t=intcfraccfrac1+cos 2t2> mathrmdt(cos^2 t+sin^2 t)^2-2sin^2 tcos^2 t notag\
&=intcfrac1+cos 2t2-sin^2 2t >mathrmdt
=intcfracmathrmdt2-sin^2 2t + cfrac 12intcfracmathrmdsin 2t2-sin^2 2t notag\
&=intcfracsec^2 2t >mathrmdt2sec^2 2t-tan^2 2t + cfrac sqrt28intcfrac1sqrt2-sin 2t + cfrac1sqrt2+sin 2t>mathrmdsin 2t notag\
&=cfrac 12intcfracmathrmdtan 2t2+tan^2 2t +cfracsqrt28ln cfracsqrt2+sin 2tsqrt2-sin 2t notag\
&=cfrac sqrt24 tan^-1 cfractan 2tsqrt2 +cfracsqrt28ln cfracsqrt2+sin 2tsqrt2-sin 2t notag
endalign
As $tan 2t=cfrac2tan t1-tan^2 t=cfrac2x1-x^2, cfracsqrt2+sin 2tsqrt2-sin 2t=cfracsqrt2sec^2 t+tan tsqrt2sec^2 t-tan t=cfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x,$
$f(x)=cfrac sqrt24 tan^-1 cfracsqrt2x1+x^2 +cfracsqrt28ln cfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+c$
If the above holds, $displaystyle int_0^infty cfracmathrmd x1+x^4$ would be $0$, which is impossible(Should be $cfrac sqrt2pi4$).
indefinite-integrals
edited Aug 6 at 7:36
Bernard
110k635103
asked Aug 6 at 2:37
Lance
30818
add a comment |Â
up vote
5
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favorite
Here is my attempt. The result is not right. Please help identify the issue(s).
$displaystyle f(x)=intcfrac1x^4+1>mathrmdx$, let $x=tan t$, we have $ mathrmdx = sec^2 t>mathrmdt,> t=tan^-1 xinleft(-cfracpi2,cfracpi2right)$
beginalign
displaystyle f(tan t)&= intcfracsec^2 t> mathrmdt1+tan^4 t=intcfraccos^2 t> mathrmdtcos^4 t+sin^4 t=intcfraccfrac1+cos 2t2> mathrmdt(cos^2 t+sin^2 t)^2-2sin^2 tcos^2 t notag\
&=intcfrac1+cos 2t2-sin^2 2t >mathrmdt
=intcfracmathrmdt2-sin^2 2t + cfrac 12intcfracmathrmdsin 2t2-sin^2 2t notag\
&=intcfracsec^2 2t >mathrmdt2sec^2 2t-tan^2 2t + cfrac sqrt28intcfrac1sqrt2-sin 2t + cfrac1sqrt2+sin 2t>mathrmdsin 2t notag\
&=cfrac 12intcfracmathrmdtan 2t2+tan^2 2t +cfracsqrt28ln cfracsqrt2+sin 2tsqrt2-sin 2t notag\
&=cfrac sqrt24 tan^-1 cfractan 2tsqrt2 +cfracsqrt28ln cfracsqrt2+sin 2tsqrt2-sin 2t notag
endalign
As $tan 2t=cfrac2tan t1-tan^2 t=cfrac2x1-x^2, cfracsqrt2+sin 2tsqrt2-sin 2t=cfracsqrt2sec^2 t+tan tsqrt2sec^2 t-tan t=cfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x,$
$f(x)=cfrac sqrt24 tan^-1 cfracsqrt2x1+x^2 +cfracsqrt28ln cfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+c$
If the above holds, $displaystyle int_0^infty cfracmathrmd x1+x^4$ would be $0$, which is impossible(Should be $cfrac sqrt2pi4$).
indefinite-integrals
edited Aug 6 at 7:36
Bernard
110k635103
asked Aug 6 at 2:37
Lance
30818
If I recall correctly, this is one of those integrals best approaches with an integration by parts (with $dv=1$). From there a change of variables should be straightforward.
– Robert Wolfe
Aug 6 at 2:51
4
$x^4+1=x^4+2x^2+1-2x^2=(x^2+sqrt2x+1)(x^2-sqrt2+1)$ and you can use partial fractions
– saulspatz
Aug 6 at 2:55
2
Another approach is to use the fact that $$1/(xâ´+1)=frac1(x²+i)(x²-i)=frac1(x+sqrt-i)(x-sqrt-i)(x+sqrti)(x-sqrti)$$ and use partial fractions if you have the energy to work with ugly numbers
– Holo
Aug 6 at 2:58
@Holo but we should try to avoid complex no.s, right ?? I think it doesn't help in any way.
– Anik Bhowmick
Aug 6 at 3:27
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Here is my attempt. The result is not right. Please help identify the issue(s).
$displaystyle f(x)=intcfrac1x^4+1>mathrmdx$, let $x=tan t$, we have $ mathrmdx = sec^2 t>mathrmdt,> t=tan^-1 xinleft(-cfracpi2,cfracpi2right)$
beginalign
displaystyle f(tan t)&= intcfracsec^2 t> mathrmdt1+tan^4 t=intcfraccos^2 t> mathrmdtcos^4 t+sin^4 t=intcfraccfrac1+cos 2t2> mathrmdt(cos^2 t+sin^2 t)^2-2sin^2 tcos^2 t notag\
&=intcfrac1+cos 2t2-sin^2 2t >mathrmdt
=intcfracmathrmdt2-sin^2 2t + cfrac 12intcfracmathrmdsin 2t2-sin^2 2t notag\
&=intcfracsec^2 2t >mathrmdt2sec^2 2t-tan^2 2t + cfrac sqrt28intcfrac1sqrt2-sin 2t + cfrac1sqrt2+sin 2t>mathrmdsin 2t notag\
&=cfrac 12intcfracmathrmdtan 2t2+tan^2 2t +cfracsqrt28ln cfracsqrt2+sin 2tsqrt2-sin 2t notag\
&=cfrac sqrt24 tan^-1 cfractan 2tsqrt2 +cfracsqrt28ln cfracsqrt2+sin 2tsqrt2-sin 2t notag
endalign
As $tan 2t=cfrac2tan t1-tan^2 t=cfrac2x1-x^2, cfracsqrt2+sin 2tsqrt2-sin 2t=cfracsqrt2sec^2 t+tan tsqrt2sec^2 t-tan t=cfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x,$
$f(x)=cfrac sqrt24 tan^-1 cfracsqrt2x1+x^2 +cfracsqrt28ln cfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+c$
If the above holds, $displaystyle int_0^infty cfracmathrmd x1+x^4$ would be $0$, which is impossible(Should be $cfrac sqrt2pi4$).
indefinite-integrals
edited Aug 6 at 7:36
Bernard
110k635103
asked Aug 6 at 2:37
Lance
30818
Here is my attempt. The result is not right. Please help identify the issue(s).
$displaystyle f(x)=intcfrac1x^4+1>mathrmdx$, let $x=tan t$, we have $ mathrmdx = sec^2 t>mathrmdt,> t=tan^-1 xinleft(-cfracpi2,cfracpi2right)$
beginalign
displaystyle f(tan t)&= intcfracsec^2 t> mathrmdt1+tan^4 t=intcfraccos^2 t> mathrmdtcos^4 t+sin^4 t=intcfraccfrac1+cos 2t2> mathrmdt(cos^2 t+sin^2 t)^2-2sin^2 tcos^2 t notag\
&=intcfrac1+cos 2t2-sin^2 2t >mathrmdt
=intcfracmathrmdt2-sin^2 2t + cfrac 12intcfracmathrmdsin 2t2-sin^2 2t notag\
&=intcfracsec^2 2t >mathrmdt2sec^2 2t-tan^2 2t + cfrac sqrt28intcfrac1sqrt2-sin 2t + cfrac1sqrt2+sin 2t>mathrmdsin 2t notag\
&=cfrac 12intcfracmathrmdtan 2t2+tan^2 2t +cfracsqrt28ln cfracsqrt2+sin 2tsqrt2-sin 2t notag\
&=cfrac sqrt24 tan^-1 cfractan 2tsqrt2 +cfracsqrt28ln cfracsqrt2+sin 2tsqrt2-sin 2t notag
endalign
As $tan 2t=cfrac2tan t1-tan^2 t=cfrac2x1-x^2, cfracsqrt2+sin 2tsqrt2-sin 2t=cfracsqrt2sec^2 t+tan tsqrt2sec^2 t-tan t=cfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x,$
$f(x)=cfrac sqrt24 tan^-1 cfracsqrt2x1+x^2 +cfracsqrt28ln cfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+c$
If the above holds, $displaystyle int_0^infty cfracmathrmd x1+x^4$ would be $0$, which is impossible(Should be $cfrac sqrt2pi4$).
indefinite-integrals
edited Aug 6 at 7:36
Bernard
110k635103
asked Aug 6 at 2:37
Lance
30818
edited Aug 6 at 7:36
Bernard
110k635103
edited Aug 6 at 7:36
Bernard
110k635103
edited Aug 6 at 7:36
Bernard
110k635103
110k635103
asked Aug 6 at 2:37
Lance
30818
asked Aug 6 at 2:37
Lance
30818
asked Aug 6 at 2:37
Lance
30818
30818
If I recall correctly, this is one of those integrals best approaches with an integration by parts (with $dv=1$). From there a change of variables should be straightforward.
– Robert Wolfe
Aug 6 at 2:51
4
$x^4+1=x^4+2x^2+1-2x^2=(x^2+sqrt2x+1)(x^2-sqrt2+1)$ and you can use partial fractions
– saulspatz
Aug 6 at 2:55
2
Another approach is to use the fact that $$1/(xâ´+1)=frac1(x²+i)(x²-i)=frac1(x+sqrt-i)(x-sqrt-i)(x+sqrti)(x-sqrti)$$ and use partial fractions if you have the energy to work with ugly numbers
– Holo
Aug 6 at 2:58
@Holo but we should try to avoid complex no.s, right ?? I think it doesn't help in any way.
– Anik Bhowmick
Aug 6 at 3:27
add a comment |Â
If I recall correctly, this is one of those integrals best approaches with an integration by parts (with $dv=1$). From there a change of variables should be straightforward.
– Robert Wolfe
Aug 6 at 2:51
4
$x^4+1=x^4+2x^2+1-2x^2=(x^2+sqrt2x+1)(x^2-sqrt2+1)$ and you can use partial fractions
– saulspatz
Aug 6 at 2:55
2
Another approach is to use the fact that $$1/(xâ´+1)=frac1(x²+i)(x²-i)=frac1(x+sqrt-i)(x-sqrt-i)(x+sqrti)(x-sqrti)$$ and use partial fractions if you have the energy to work with ugly numbers
– Holo
Aug 6 at 2:58
@Holo but we should try to avoid complex no.s, right ?? I think it doesn't help in any way.
– Anik Bhowmick
Aug 6 at 3:27
If I recall correctly, this is one of those integrals best approaches with an integration by parts (with $dv=1$). From there a change of variables should be straightforward.
– Robert Wolfe
Aug 6 at 2:51
If I recall correctly, this is one of those integrals best approaches with an integration by parts (with $dv=1$). From there a change of variables should be straightforward.
– Robert Wolfe
Aug 6 at 2:51
4
4
$x^4+1=x^4+2x^2+1-2x^2=(x^2+sqrt2x+1)(x^2-sqrt2+1)$ and you can use partial fractions
– saulspatz
Aug 6 at 2:55
$x^4+1=x^4+2x^2+1-2x^2=(x^2+sqrt2x+1)(x^2-sqrt2+1)$ and you can use partial fractions
– saulspatz
Aug 6 at 2:55
2
2
Another approach is to use the fact that $$1/(xâ´+1)=frac1(x²+i)(x²-i)=frac1(x+sqrt-i)(x-sqrt-i)(x+sqrti)(x-sqrti)$$ and use partial fractions if you have the energy to work with ugly numbers
– Holo
Aug 6 at 2:58
Another approach is to use the fact that $$1/(xâ´+1)=frac1(x²+i)(x²-i)=frac1(x+sqrt-i)(x-sqrt-i)(x+sqrti)(x-sqrti)$$ and use partial fractions if you have the energy to work with ugly numbers
– Holo
Aug 6 at 2:58
@Holo but we should try to avoid complex no.s, right ?? I think it doesn't help in any way.
– Anik Bhowmick
Aug 6 at 3:27
@Holo but we should try to avoid complex no.s, right ?? I think it doesn't help in any way.
– Anik Bhowmick
Aug 6 at 3:27
add a comment |Â
2 Answers
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up vote
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First of all, you made a typo in the final answer — the correct answer must be
$$f(x)=fracsqrt24tan^-1fracsqrt2x1colorred-x^2+fracsqrt28lnfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C.$$
The next issue is the introduction of $sec(2t)$ and $tan(2t)$ when you switched to
$$intfracsec^2 2t,mathrmdt2sec^2 2t-tan^2 2t$$
(as part of an expression). Both $sec(2t)$ and $tan(2t)$ are undefined at some points within the domain $displaystyle tinleft(-fracpi2,fracpi2right)$, viz. at $displaystyle t=pmfracpi4$. Therefore, the antiderivative you find in terms of $t$ is in fact a piecewise-defined function:
$$f(x(t))=begincases
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_1, text if tinleft(-cfracpi2,-cfracpi4right); \
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_2, text if tinleft(-cfracpi4,cfracpi4right); \
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_3, text if tinleft(cfracpi4,cfracpi2right).
endcases$$
Switching back to $x$ still creates a piecewise-defined function:
$$f(x)=begincases
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_1, text if xin(-infty,-1); \
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_2, text if xin(-1,1); \
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_3, text if xin(1,+infty).
endcases$$
At the points $x=pm1$, these expressions are undefined, and so the corresponding integrals have to be treated as improper. In your case, the integral $displaystyle int_0^+infty$ has to be split at the discontinuity at $x=1$:
$$int_0^+inftycdots,mathrmdx=int_0^1cdots,mathrmdx+int_1^+inftycdots,mathrmdx,$$
and then, when evaluating the antiderivative that you found, you'll have to take the one-sided limits from the left and from the right at $x=1$, which are NOT equal to each other! And that's probably the source of your wrong answer.
More specifically:
$$lim_xto1^-fracsqrt2x1-x^2=+infty implies lim_xto1^-arctanfracsqrt2x1-x^2=fracpi2,$$
while
$$lim_xto1^+fracsqrt2x1-x^2=-infty implies lim_xto1^-arctanfracsqrt2x1-x^2=-fracpi2.$$
answered Aug 6 at 4:26
zipirovich
10.1k11630
1
This problem of branch cuts can be fixed by writing $arctan(sqrt2x +1) + arctan(sqrt2x - 1)$ instead of $arctan[sqrt2x/(1-x^2)]$, which gives the correct function.
– eyeballfrog
Aug 6 at 4:32
@eyeballfrog: I agree. However, the OP's question wasn't asking for a different solution, but to help find out where the proposed solution went wrong. So I specifically limited my answer to an analysis of the proposed solution only.
– zipirovich
Aug 6 at 4:43
@zipirovich That's very considerate. I was having hard time to pick the "correct answer". Both of you pointed out my error. I chose yours over xbh's because you pointed out how the error is introduced. Thank you!
– Lance
Aug 6 at 13:47
add a comment |Â
up vote
4
down vote
Comparing to the method I used in the following, maybe the issue occurs when computing
$$
int frac mathrm dt 2-sin^2(2t).
$$
Then from now on $t$ cannot take the value $pm pi /4$ if we want to devide the numerator and the denominator by $cos^2(2t)$. Now to compute the improper integral, we should take the limit $x to 1^-$ and $xto 1^+$ separately, since the result is discontinuous at $1$. Fundamental Theorem of Calculus may deduce the wrong result if we apply it to a discontinued antiderivative. So if we use the OP as the antiderivative, we should compute
$$
f(+infty) - f(1^+) + f(1^-) - f(0),
$$
which would give the right result $sqrt 2 pi/4$.
Conclusion: the computation in the OP is right, but when apply it to compute definite integral, we should split the interval at the point $1$.
Appendix
I'm here to give another approach. We would introduce a conjugate pair. Assume $x neq 0$.
beginalign*
int frac mathrm d x 1+x^4
&= frac 12 int frac 1-x^21+x^4 mathrm dx + int frac 1+x^21+x^4mathrm dx \
&= frac 12 int frac x^-2 - 1x^2+x^-2 mathrm dx + frac 12 int frac x^-2 + 1x^2+x^-2 mathrm dx\
&= -frac 12 int frac mathrm d (x + x^-1) (x+x^-1 )^2 -2 mathrm dx+frac 12 int frac mathrm d(x-x^-1) (x - x^-1)^2 +2\
&= -frac sqrt 28 int left( frac 1 x +x^-1-sqrt 2 - frac 1x+x^-1+ sqrt 2right)mathrm d(x+x^-1) \
&phantom==+frac sqrt24 int frac mathrm d(x - x^-1)/sqrt 2 ((x-x^-1)/sqrt 2)^2 +1 \
&= frac sqrt 28 log left( frac x + x^-1+sqrt 2 x +x^-1-sqrt 2right) + frac sqrt 24 mathrm arctanleft( frac x-x^-1 sqrt 2right) + C \
&= frac sqrt 28 log left(frac x^2 +sqrt 2 x + 1 x^2 - sqrt2 x +1right) +frac sqrt 24 mathrmarctan left(frac x^2 -1sqrt 2 xright) + C.
endalign*
If we use this as the result, then
$$
f(+infty) - f(0) = frac sqrt 2 4 left( frac pi 2 + frac pi 2right) = frac sqrt 24 pi.
$$
Also note that when $x neq 0$,
$$
arctan(x) + mathrmarccot (x) = mathrm sgn (x)frac pi 2 implies arctan (x) = mathrm sgn (x)frac pi 2 + arctan left(-frac 1xright),
$$
so the OP is correct.
answered Aug 6 at 3:27
xbh
1,6079
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
First of all, you made a typo in the final answer — the correct answer must be
$$f(x)=fracsqrt24tan^-1fracsqrt2x1colorred-x^2+fracsqrt28lnfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C.$$
The next issue is the introduction of $sec(2t)$ and $tan(2t)$ when you switched to
$$intfracsec^2 2t,mathrmdt2sec^2 2t-tan^2 2t$$
(as part of an expression). Both $sec(2t)$ and $tan(2t)$ are undefined at some points within the domain $displaystyle tinleft(-fracpi2,fracpi2right)$, viz. at $displaystyle t=pmfracpi4$. Therefore, the antiderivative you find in terms of $t$ is in fact a piecewise-defined function:
$$f(x(t))=begincases
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_1, text if tinleft(-cfracpi2,-cfracpi4right); \
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_2, text if tinleft(-cfracpi4,cfracpi4right); \
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_3, text if tinleft(cfracpi4,cfracpi2right).
endcases$$
Switching back to $x$ still creates a piecewise-defined function:
$$f(x)=begincases
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_1, text if xin(-infty,-1); \
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_2, text if xin(-1,1); \
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_3, text if xin(1,+infty).
endcases$$
At the points $x=pm1$, these expressions are undefined, and so the corresponding integrals have to be treated as improper. In your case, the integral $displaystyle int_0^+infty$ has to be split at the discontinuity at $x=1$:
$$int_0^+inftycdots,mathrmdx=int_0^1cdots,mathrmdx+int_1^+inftycdots,mathrmdx,$$
and then, when evaluating the antiderivative that you found, you'll have to take the one-sided limits from the left and from the right at $x=1$, which are NOT equal to each other! And that's probably the source of your wrong answer.
More specifically:
$$lim_xto1^-fracsqrt2x1-x^2=+infty implies lim_xto1^-arctanfracsqrt2x1-x^2=fracpi2,$$
while
$$lim_xto1^+fracsqrt2x1-x^2=-infty implies lim_xto1^-arctanfracsqrt2x1-x^2=-fracpi2.$$
answered Aug 6 at 4:26
zipirovich
10.1k11630
1
This problem of branch cuts can be fixed by writing $arctan(sqrt2x +1) + arctan(sqrt2x - 1)$ instead of $arctan[sqrt2x/(1-x^2)]$, which gives the correct function.
– eyeballfrog
Aug 6 at 4:32
@eyeballfrog: I agree. However, the OP's question wasn't asking for a different solution, but to help find out where the proposed solution went wrong. So I specifically limited my answer to an analysis of the proposed solution only.
– zipirovich
Aug 6 at 4:43
@zipirovich That's very considerate. I was having hard time to pick the "correct answer". Both of you pointed out my error. I chose yours over xbh's because you pointed out how the error is introduced. Thank you!
– Lance
Aug 6 at 13:47
add a comment |Â
up vote
1
down vote
accepted
First of all, you made a typo in the final answer — the correct answer must be
$$f(x)=fracsqrt24tan^-1fracsqrt2x1colorred-x^2+fracsqrt28lnfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C.$$
The next issue is the introduction of $sec(2t)$ and $tan(2t)$ when you switched to
$$intfracsec^2 2t,mathrmdt2sec^2 2t-tan^2 2t$$
(as part of an expression). Both $sec(2t)$ and $tan(2t)$ are undefined at some points within the domain $displaystyle tinleft(-fracpi2,fracpi2right)$, viz. at $displaystyle t=pmfracpi4$. Therefore, the antiderivative you find in terms of $t$ is in fact a piecewise-defined function:
$$f(x(t))=begincases
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_1, text if tinleft(-cfracpi2,-cfracpi4right); \
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_2, text if tinleft(-cfracpi4,cfracpi4right); \
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_3, text if tinleft(cfracpi4,cfracpi2right).
endcases$$
Switching back to $x$ still creates a piecewise-defined function:
$$f(x)=begincases
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_1, text if xin(-infty,-1); \
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_2, text if xin(-1,1); \
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_3, text if xin(1,+infty).
endcases$$
At the points $x=pm1$, these expressions are undefined, and so the corresponding integrals have to be treated as improper. In your case, the integral $displaystyle int_0^+infty$ has to be split at the discontinuity at $x=1$:
$$int_0^+inftycdots,mathrmdx=int_0^1cdots,mathrmdx+int_1^+inftycdots,mathrmdx,$$
and then, when evaluating the antiderivative that you found, you'll have to take the one-sided limits from the left and from the right at $x=1$, which are NOT equal to each other! And that's probably the source of your wrong answer.
More specifically:
$$lim_xto1^-fracsqrt2x1-x^2=+infty implies lim_xto1^-arctanfracsqrt2x1-x^2=fracpi2,$$
while
$$lim_xto1^+fracsqrt2x1-x^2=-infty implies lim_xto1^-arctanfracsqrt2x1-x^2=-fracpi2.$$
answered Aug 6 at 4:26
zipirovich
10.1k11630
1
This problem of branch cuts can be fixed by writing $arctan(sqrt2x +1) + arctan(sqrt2x - 1)$ instead of $arctan[sqrt2x/(1-x^2)]$, which gives the correct function.
– eyeballfrog
Aug 6 at 4:32
@eyeballfrog: I agree. However, the OP's question wasn't asking for a different solution, but to help find out where the proposed solution went wrong. So I specifically limited my answer to an analysis of the proposed solution only.
– zipirovich
Aug 6 at 4:43
@zipirovich That's very considerate. I was having hard time to pick the "correct answer". Both of you pointed out my error. I chose yours over xbh's because you pointed out how the error is introduced. Thank you!
– Lance
Aug 6 at 13:47
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First of all, you made a typo in the final answer — the correct answer must be
$$f(x)=fracsqrt24tan^-1fracsqrt2x1colorred-x^2+fracsqrt28lnfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C.$$
The next issue is the introduction of $sec(2t)$ and $tan(2t)$ when you switched to
$$intfracsec^2 2t,mathrmdt2sec^2 2t-tan^2 2t$$
(as part of an expression). Both $sec(2t)$ and $tan(2t)$ are undefined at some points within the domain $displaystyle tinleft(-fracpi2,fracpi2right)$, viz. at $displaystyle t=pmfracpi4$. Therefore, the antiderivative you find in terms of $t$ is in fact a piecewise-defined function:
$$f(x(t))=begincases
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_1, text if tinleft(-cfracpi2,-cfracpi4right); \
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_2, text if tinleft(-cfracpi4,cfracpi4right); \
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_3, text if tinleft(cfracpi4,cfracpi2right).
endcases$$
Switching back to $x$ still creates a piecewise-defined function:
$$f(x)=begincases
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_1, text if xin(-infty,-1); \
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_2, text if xin(-1,1); \
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_3, text if xin(1,+infty).
endcases$$
At the points $x=pm1$, these expressions are undefined, and so the corresponding integrals have to be treated as improper. In your case, the integral $displaystyle int_0^+infty$ has to be split at the discontinuity at $x=1$:
$$int_0^+inftycdots,mathrmdx=int_0^1cdots,mathrmdx+int_1^+inftycdots,mathrmdx,$$
and then, when evaluating the antiderivative that you found, you'll have to take the one-sided limits from the left and from the right at $x=1$, which are NOT equal to each other! And that's probably the source of your wrong answer.
More specifically:
$$lim_xto1^-fracsqrt2x1-x^2=+infty implies lim_xto1^-arctanfracsqrt2x1-x^2=fracpi2,$$
while
$$lim_xto1^+fracsqrt2x1-x^2=-infty implies lim_xto1^-arctanfracsqrt2x1-x^2=-fracpi2.$$
answered Aug 6 at 4:26
zipirovich
10.1k11630
First of all, you made a typo in the final answer — the correct answer must be
$$f(x)=fracsqrt24tan^-1fracsqrt2x1colorred-x^2+fracsqrt28lnfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C.$$
The next issue is the introduction of $sec(2t)$ and $tan(2t)$ when you switched to
$$intfracsec^2 2t,mathrmdt2sec^2 2t-tan^2 2t$$
(as part of an expression). Both $sec(2t)$ and $tan(2t)$ are undefined at some points within the domain $displaystyle tinleft(-fracpi2,fracpi2right)$, viz. at $displaystyle t=pmfracpi4$. Therefore, the antiderivative you find in terms of $t$ is in fact a piecewise-defined function:
$$f(x(t))=begincases
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_1, text if tinleft(-cfracpi2,-cfracpi4right); \
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_2, text if tinleft(-cfracpi4,cfracpi4right); \
cfracsqrt24tan^-1cfractan2tsqrt2+cfracsqrt28lncfracsqrt2+sin 2tsqrt2-sin 2t+C_3, text if tinleft(cfracpi4,cfracpi2right).
endcases$$
Switching back to $x$ still creates a piecewise-defined function:
$$f(x)=begincases
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_1, text if xin(-infty,-1); \
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_2, text if xin(-1,1); \
cfracsqrt24tan^-1cfracsqrt2x1colorred-x^2+cfracsqrt28lncfracsqrt2(x^2+1)+xsqrt2(x^2+1)-x+C_3, text if xin(1,+infty).
endcases$$
At the points $x=pm1$, these expressions are undefined, and so the corresponding integrals have to be treated as improper. In your case, the integral $displaystyle int_0^+infty$ has to be split at the discontinuity at $x=1$:
$$int_0^+inftycdots,mathrmdx=int_0^1cdots,mathrmdx+int_1^+inftycdots,mathrmdx,$$
and then, when evaluating the antiderivative that you found, you'll have to take the one-sided limits from the left and from the right at $x=1$, which are NOT equal to each other! And that's probably the source of your wrong answer.
More specifically:
$$lim_xto1^-fracsqrt2x1-x^2=+infty implies lim_xto1^-arctanfracsqrt2x1-x^2=fracpi2,$$
while
$$lim_xto1^+fracsqrt2x1-x^2=-infty implies lim_xto1^-arctanfracsqrt2x1-x^2=-fracpi2.$$
answered Aug 6 at 4:26
zipirovich
10.1k11630
answered Aug 6 at 4:26
zipirovich
10.1k11630
answered Aug 6 at 4:26
zipirovich
10.1k11630
answered Aug 6 at 4:26
zipirovich
10.1k11630
10.1k11630
1
This problem of branch cuts can be fixed by writing $arctan(sqrt2x +1) + arctan(sqrt2x - 1)$ instead of $arctan[sqrt2x/(1-x^2)]$, which gives the correct function.
– eyeballfrog
Aug 6 at 4:32
@eyeballfrog: I agree. However, the OP's question wasn't asking for a different solution, but to help find out where the proposed solution went wrong. So I specifically limited my answer to an analysis of the proposed solution only.
– zipirovich
Aug 6 at 4:43
@zipirovich That's very considerate. I was having hard time to pick the "correct answer". Both of you pointed out my error. I chose yours over xbh's because you pointed out how the error is introduced. Thank you!
– Lance
Aug 6 at 13:47
add a comment |Â
1
This problem of branch cuts can be fixed by writing $arctan(sqrt2x +1) + arctan(sqrt2x - 1)$ instead of $arctan[sqrt2x/(1-x^2)]$, which gives the correct function.
– eyeballfrog
Aug 6 at 4:32
@eyeballfrog: I agree. However, the OP's question wasn't asking for a different solution, but to help find out where the proposed solution went wrong. So I specifically limited my answer to an analysis of the proposed solution only.
– zipirovich
Aug 6 at 4:43
@zipirovich That's very considerate. I was having hard time to pick the "correct answer". Both of you pointed out my error. I chose yours over xbh's because you pointed out how the error is introduced. Thank you!
– Lance
Aug 6 at 13:47
1
1
This problem of branch cuts can be fixed by writing $arctan(sqrt2x +1) + arctan(sqrt2x - 1)$ instead of $arctan[sqrt2x/(1-x^2)]$, which gives the correct function.
– eyeballfrog
Aug 6 at 4:32
This problem of branch cuts can be fixed by writing $arctan(sqrt2x +1) + arctan(sqrt2x - 1)$ instead of $arctan[sqrt2x/(1-x^2)]$, which gives the correct function.
– eyeballfrog
Aug 6 at 4:32
@eyeballfrog: I agree. However, the OP's question wasn't asking for a different solution, but to help find out where the proposed solution went wrong. So I specifically limited my answer to an analysis of the proposed solution only.
– zipirovich
Aug 6 at 4:43
@eyeballfrog: I agree. However, the OP's question wasn't asking for a different solution, but to help find out where the proposed solution went wrong. So I specifically limited my answer to an analysis of the proposed solution only.
– zipirovich
Aug 6 at 4:43
@zipirovich That's very considerate. I was having hard time to pick the "correct answer". Both of you pointed out my error. I chose yours over xbh's because you pointed out how the error is introduced. Thank you!
– Lance
Aug 6 at 13:47
@zipirovich That's very considerate. I was having hard time to pick the "correct answer". Both of you pointed out my error. I chose yours over xbh's because you pointed out how the error is introduced. Thank you!
– Lance
Aug 6 at 13:47
add a comment |Â
up vote
4
down vote
Comparing to the method I used in the following, maybe the issue occurs when computing
$$
int frac mathrm dt 2-sin^2(2t).
$$
Then from now on $t$ cannot take the value $pm pi /4$ if we want to devide the numerator and the denominator by $cos^2(2t)$. Now to compute the improper integral, we should take the limit $x to 1^-$ and $xto 1^+$ separately, since the result is discontinuous at $1$. Fundamental Theorem of Calculus may deduce the wrong result if we apply it to a discontinued antiderivative. So if we use the OP as the antiderivative, we should compute
$$
f(+infty) - f(1^+) + f(1^-) - f(0),
$$
which would give the right result $sqrt 2 pi/4$.
Conclusion: the computation in the OP is right, but when apply it to compute definite integral, we should split the interval at the point $1$.
Appendix
I'm here to give another approach. We would introduce a conjugate pair. Assume $x neq 0$.
beginalign*
int frac mathrm d x 1+x^4
&= frac 12 int frac 1-x^21+x^4 mathrm dx + int frac 1+x^21+x^4mathrm dx \
&= frac 12 int frac x^-2 - 1x^2+x^-2 mathrm dx + frac 12 int frac x^-2 + 1x^2+x^-2 mathrm dx\
&= -frac 12 int frac mathrm d (x + x^-1) (x+x^-1 )^2 -2 mathrm dx+frac 12 int frac mathrm d(x-x^-1) (x - x^-1)^2 +2\
&= -frac sqrt 28 int left( frac 1 x +x^-1-sqrt 2 - frac 1x+x^-1+ sqrt 2right)mathrm d(x+x^-1) \
&phantom==+frac sqrt24 int frac mathrm d(x - x^-1)/sqrt 2 ((x-x^-1)/sqrt 2)^2 +1 \
&= frac sqrt 28 log left( frac x + x^-1+sqrt 2 x +x^-1-sqrt 2right) + frac sqrt 24 mathrm arctanleft( frac x-x^-1 sqrt 2right) + C \
&= frac sqrt 28 log left(frac x^2 +sqrt 2 x + 1 x^2 - sqrt2 x +1right) +frac sqrt 24 mathrmarctan left(frac x^2 -1sqrt 2 xright) + C.
endalign*
If we use this as the result, then
$$
f(+infty) - f(0) = frac sqrt 2 4 left( frac pi 2 + frac pi 2right) = frac sqrt 24 pi.
$$
Also note that when $x neq 0$,
$$
arctan(x) + mathrmarccot (x) = mathrm sgn (x)frac pi 2 implies arctan (x) = mathrm sgn (x)frac pi 2 + arctan left(-frac 1xright),
$$
so the OP is correct.
answered Aug 6 at 3:27
xbh
1,6079
add a comment |Â
up vote
4
down vote
Comparing to the method I used in the following, maybe the issue occurs when computing
$$
int frac mathrm dt 2-sin^2(2t).
$$
Then from now on $t$ cannot take the value $pm pi /4$ if we want to devide the numerator and the denominator by $cos^2(2t)$. Now to compute the improper integral, we should take the limit $x to 1^-$ and $xto 1^+$ separately, since the result is discontinuous at $1$. Fundamental Theorem of Calculus may deduce the wrong result if we apply it to a discontinued antiderivative. So if we use the OP as the antiderivative, we should compute
$$
f(+infty) - f(1^+) + f(1^-) - f(0),
$$
which would give the right result $sqrt 2 pi/4$.
Conclusion: the computation in the OP is right, but when apply it to compute definite integral, we should split the interval at the point $1$.
Appendix
I'm here to give another approach. We would introduce a conjugate pair. Assume $x neq 0$.
beginalign*
int frac mathrm d x 1+x^4
&= frac 12 int frac 1-x^21+x^4 mathrm dx + int frac 1+x^21+x^4mathrm dx \
&= frac 12 int frac x^-2 - 1x^2+x^-2 mathrm dx + frac 12 int frac x^-2 + 1x^2+x^-2 mathrm dx\
&= -frac 12 int frac mathrm d (x + x^-1) (x+x^-1 )^2 -2 mathrm dx+frac 12 int frac mathrm d(x-x^-1) (x - x^-1)^2 +2\
&= -frac sqrt 28 int left( frac 1 x +x^-1-sqrt 2 - frac 1x+x^-1+ sqrt 2right)mathrm d(x+x^-1) \
&phantom==+frac sqrt24 int frac mathrm d(x - x^-1)/sqrt 2 ((x-x^-1)/sqrt 2)^2 +1 \
&= frac sqrt 28 log left( frac x + x^-1+sqrt 2 x +x^-1-sqrt 2right) + frac sqrt 24 mathrm arctanleft( frac x-x^-1 sqrt 2right) + C \
&= frac sqrt 28 log left(frac x^2 +sqrt 2 x + 1 x^2 - sqrt2 x +1right) +frac sqrt 24 mathrmarctan left(frac x^2 -1sqrt 2 xright) + C.
endalign*
If we use this as the result, then
$$
f(+infty) - f(0) = frac sqrt 2 4 left( frac pi 2 + frac pi 2right) = frac sqrt 24 pi.
$$
Also note that when $x neq 0$,
$$
arctan(x) + mathrmarccot (x) = mathrm sgn (x)frac pi 2 implies arctan (x) = mathrm sgn (x)frac pi 2 + arctan left(-frac 1xright),
$$
so the OP is correct.
answered Aug 6 at 3:27
xbh
1,6079
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Comparing to the method I used in the following, maybe the issue occurs when computing
$$
int frac mathrm dt 2-sin^2(2t).
$$
Then from now on $t$ cannot take the value $pm pi /4$ if we want to devide the numerator and the denominator by $cos^2(2t)$. Now to compute the improper integral, we should take the limit $x to 1^-$ and $xto 1^+$ separately, since the result is discontinuous at $1$. Fundamental Theorem of Calculus may deduce the wrong result if we apply it to a discontinued antiderivative. So if we use the OP as the antiderivative, we should compute
$$
f(+infty) - f(1^+) + f(1^-) - f(0),
$$
which would give the right result $sqrt 2 pi/4$.
Conclusion: the computation in the OP is right, but when apply it to compute definite integral, we should split the interval at the point $1$.
Appendix
I'm here to give another approach. We would introduce a conjugate pair. Assume $x neq 0$.
beginalign*
int frac mathrm d x 1+x^4
&= frac 12 int frac 1-x^21+x^4 mathrm dx + int frac 1+x^21+x^4mathrm dx \
&= frac 12 int frac x^-2 - 1x^2+x^-2 mathrm dx + frac 12 int frac x^-2 + 1x^2+x^-2 mathrm dx\
&= -frac 12 int frac mathrm d (x + x^-1) (x+x^-1 )^2 -2 mathrm dx+frac 12 int frac mathrm d(x-x^-1) (x - x^-1)^2 +2\
&= -frac sqrt 28 int left( frac 1 x +x^-1-sqrt 2 - frac 1x+x^-1+ sqrt 2right)mathrm d(x+x^-1) \
&phantom==+frac sqrt24 int frac mathrm d(x - x^-1)/sqrt 2 ((x-x^-1)/sqrt 2)^2 +1 \
&= frac sqrt 28 log left( frac x + x^-1+sqrt 2 x +x^-1-sqrt 2right) + frac sqrt 24 mathrm arctanleft( frac x-x^-1 sqrt 2right) + C \
&= frac sqrt 28 log left(frac x^2 +sqrt 2 x + 1 x^2 - sqrt2 x +1right) +frac sqrt 24 mathrmarctan left(frac x^2 -1sqrt 2 xright) + C.
endalign*
If we use this as the result, then
$$
f(+infty) - f(0) = frac sqrt 2 4 left( frac pi 2 + frac pi 2right) = frac sqrt 24 pi.
$$
Also note that when $x neq 0$,
$$
arctan(x) + mathrmarccot (x) = mathrm sgn (x)frac pi 2 implies arctan (x) = mathrm sgn (x)frac pi 2 + arctan left(-frac 1xright),
$$
so the OP is correct.
answered Aug 6 at 3:27
xbh
1,6079
Comparing to the method I used in the following, maybe the issue occurs when computing
$$
int frac mathrm dt 2-sin^2(2t).
$$
Then from now on $t$ cannot take the value $pm pi /4$ if we want to devide the numerator and the denominator by $cos^2(2t)$. Now to compute the improper integral, we should take the limit $x to 1^-$ and $xto 1^+$ separately, since the result is discontinuous at $1$. Fundamental Theorem of Calculus may deduce the wrong result if we apply it to a discontinued antiderivative. So if we use the OP as the antiderivative, we should compute
$$
f(+infty) - f(1^+) + f(1^-) - f(0),
$$
which would give the right result $sqrt 2 pi/4$.
Conclusion: the computation in the OP is right, but when apply it to compute definite integral, we should split the interval at the point $1$.
Appendix
I'm here to give another approach. We would introduce a conjugate pair. Assume $x neq 0$.
beginalign*
int frac mathrm d x 1+x^4
&= frac 12 int frac 1-x^21+x^4 mathrm dx + int frac 1+x^21+x^4mathrm dx \
&= frac 12 int frac x^-2 - 1x^2+x^-2 mathrm dx + frac 12 int frac x^-2 + 1x^2+x^-2 mathrm dx\
&= -frac 12 int frac mathrm d (x + x^-1) (x+x^-1 )^2 -2 mathrm dx+frac 12 int frac mathrm d(x-x^-1) (x - x^-1)^2 +2\
&= -frac sqrt 28 int left( frac 1 x +x^-1-sqrt 2 - frac 1x+x^-1+ sqrt 2right)mathrm d(x+x^-1) \
&phantom==+frac sqrt24 int frac mathrm d(x - x^-1)/sqrt 2 ((x-x^-1)/sqrt 2)^2 +1 \
&= frac sqrt 28 log left( frac x + x^-1+sqrt 2 x +x^-1-sqrt 2right) + frac sqrt 24 mathrm arctanleft( frac x-x^-1 sqrt 2right) + C \
&= frac sqrt 28 log left(frac x^2 +sqrt 2 x + 1 x^2 - sqrt2 x +1right) +frac sqrt 24 mathrmarctan left(frac x^2 -1sqrt 2 xright) + C.
endalign*
If we use this as the result, then
$$
f(+infty) - f(0) = frac sqrt 2 4 left( frac pi 2 + frac pi 2right) = frac sqrt 24 pi.
$$
Also note that when $x neq 0$,
$$
arctan(x) + mathrmarccot (x) = mathrm sgn (x)frac pi 2 implies arctan (x) = mathrm sgn (x)frac pi 2 + arctan left(-frac 1xright),
$$
so the OP is correct.
answered Aug 6 at 3:27
xbh
1,6079
edited Aug 6 at 4:10
answered Aug 6 at 3:27
xbh
1,6079
answered Aug 6 at 3:27
xbh
1,6079
answered Aug 6 at 3:27
xbh
1,6079
1,6079
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If I recall correctly, this is one of those integrals best approaches with an integration by parts (with $dv=1$). From there a change of variables should be straightforward.
– Robert Wolfe
Aug 6 at 2:51
4
$x^4+1=x^4+2x^2+1-2x^2=(x^2+sqrt2x+1)(x^2-sqrt2+1)$ and you can use partial fractions
– saulspatz
Aug 6 at 2:55
2
Another approach is to use the fact that $$1/(xâ´+1)=frac1(x²+i)(x²-i)=frac1(x+sqrt-i)(x-sqrt-i)(x+sqrti)(x-sqrti)$$ and use partial fractions if you have the energy to work with ugly numbers
– Holo
Aug 6 at 2:58
@Holo but we should try to avoid complex no.s, right ?? I think it doesn't help in any way.
– Anik Bhowmick
Aug 6 at 3:27