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Evaluating $int_0^inftyfraccos(x/a)-1xcdotcos(x)ln(x) dx$
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$$I=large int_0^inftyfraccos(x/a)-1xcdotcos(x)ln(x)mathrm dx$$
Attempt:
Split them into
$$large int_0^inftyfraccos(x/a)xcdotcos(x)ln(x)mathrm dx-int_0^inftyfrac1xcdotcos(x)ln(x)mathrm dx=J-K$$
Applying integration by parts, $large u=cos(x)$
$large mathrm du=-sin(x)$
$large mathrm dv=fracln(x)xmathrm dx$
$large v=fracln^2(x)2$
$$K=fraccos(x)ln^2(x)2+frac12intsin(x)ln^2(x)mathrm dx$$
Try another integration by parts,
$large u=sin(x)ln(x)$
$large mathrm du=fracsin(x)x+cos(x)ln(x)$
$large mathrm dv=sin(x)mathrm dx$
$large v=sin(x)$
$$intsin(x)ln^2(x)mathrm dx=-cos(x)sin(x)ln(x)+int left(fracsin(2x)2x+ln(x)-sin^2(x)ln(x)right)mathrm dx$$
$$intsin(x)ln^2(x)mathrm dx=-cos(x)sin(x)ln(x)+xln(x)-x+int left(fracsin(2x)2x-sin^2(x)ln(x)right)mathrm dx$$
I have try doing integration by parts to reduce it but, it is not working.
I have try and look up standard table of integrals but can't find much to help me.
I am unable to proceed. Can anyone please lead the way?
calculus integration definite-integrals trigonometric-integrals
edited Jul 20 at 10:30
Abcd
2,3811624
add a comment |Â
up vote
4
down vote
favorite
$$I=large int_0^inftyfraccos(x/a)-1xcdotcos(x)ln(x)mathrm dx$$
Attempt:
Split them into
$$large int_0^inftyfraccos(x/a)xcdotcos(x)ln(x)mathrm dx-int_0^inftyfrac1xcdotcos(x)ln(x)mathrm dx=J-K$$
Applying integration by parts, $large u=cos(x)$
$large mathrm du=-sin(x)$
$large mathrm dv=fracln(x)xmathrm dx$
$large v=fracln^2(x)2$
$$K=fraccos(x)ln^2(x)2+frac12intsin(x)ln^2(x)mathrm dx$$
Try another integration by parts,
$large u=sin(x)ln(x)$
$large mathrm du=fracsin(x)x+cos(x)ln(x)$
$large mathrm dv=sin(x)mathrm dx$
$large v=sin(x)$
$$intsin(x)ln^2(x)mathrm dx=-cos(x)sin(x)ln(x)+int left(fracsin(2x)2x+ln(x)-sin^2(x)ln(x)right)mathrm dx$$
$$intsin(x)ln^2(x)mathrm dx=-cos(x)sin(x)ln(x)+xln(x)-x+int left(fracsin(2x)2x-sin^2(x)ln(x)right)mathrm dx$$
I have try doing integration by parts to reduce it but, it is not working.
I have try and look up standard table of integrals but can't find much to help me.
I am unable to proceed. Can anyone please lead the way?
calculus integration definite-integrals trigonometric-integrals
edited Jul 20 at 10:30
Abcd
2,3811624
I have a problem with $K$ which can not integrated if the lower bound is $0$
– Claude Leibovici
Jul 20 at 8:47
sorry for the mistake, the limit is $int_0^infty$
– user565198
Jul 20 at 8:53
The problem remains with $K$ around $x=0$. I do not see the trick but I suppose that we must not split as $I=J-K$.
– Claude Leibovici
Jul 20 at 9:00
When I read answers such as the ones you received, I realize how bad I am !
– Claude Leibovici
Jul 21 at 2:07
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
$$I=large int_0^inftyfraccos(x/a)-1xcdotcos(x)ln(x)mathrm dx$$
Attempt:
Split them into
$$large int_0^inftyfraccos(x/a)xcdotcos(x)ln(x)mathrm dx-int_0^inftyfrac1xcdotcos(x)ln(x)mathrm dx=J-K$$
Applying integration by parts, $large u=cos(x)$
$large mathrm du=-sin(x)$
$large mathrm dv=fracln(x)xmathrm dx$
$large v=fracln^2(x)2$
$$K=fraccos(x)ln^2(x)2+frac12intsin(x)ln^2(x)mathrm dx$$
Try another integration by parts,
$large u=sin(x)ln(x)$
$large mathrm du=fracsin(x)x+cos(x)ln(x)$
$large mathrm dv=sin(x)mathrm dx$
$large v=sin(x)$
$$intsin(x)ln^2(x)mathrm dx=-cos(x)sin(x)ln(x)+int left(fracsin(2x)2x+ln(x)-sin^2(x)ln(x)right)mathrm dx$$
$$intsin(x)ln^2(x)mathrm dx=-cos(x)sin(x)ln(x)+xln(x)-x+int left(fracsin(2x)2x-sin^2(x)ln(x)right)mathrm dx$$
I have try doing integration by parts to reduce it but, it is not working.
I have try and look up standard table of integrals but can't find much to help me.
I am unable to proceed. Can anyone please lead the way?
calculus integration definite-integrals trigonometric-integrals
edited Jul 20 at 10:30
Abcd
2,3811624
$$I=large int_0^inftyfraccos(x/a)-1xcdotcos(x)ln(x)mathrm dx$$
Attempt:
Split them into
$$large int_0^inftyfraccos(x/a)xcdotcos(x)ln(x)mathrm dx-int_0^inftyfrac1xcdotcos(x)ln(x)mathrm dx=J-K$$
Applying integration by parts, $large u=cos(x)$
$large mathrm du=-sin(x)$
$large mathrm dv=fracln(x)xmathrm dx$
$large v=fracln^2(x)2$
$$K=fraccos(x)ln^2(x)2+frac12intsin(x)ln^2(x)mathrm dx$$
Try another integration by parts,
$large u=sin(x)ln(x)$
$large mathrm du=fracsin(x)x+cos(x)ln(x)$
$large mathrm dv=sin(x)mathrm dx$
$large v=sin(x)$
$$intsin(x)ln^2(x)mathrm dx=-cos(x)sin(x)ln(x)+int left(fracsin(2x)2x+ln(x)-sin^2(x)ln(x)right)mathrm dx$$
$$intsin(x)ln^2(x)mathrm dx=-cos(x)sin(x)ln(x)+xln(x)-x+int left(fracsin(2x)2x-sin^2(x)ln(x)right)mathrm dx$$
I have try doing integration by parts to reduce it but, it is not working.
I have try and look up standard table of integrals but can't find much to help me.
I am unable to proceed. Can anyone please lead the way?
calculus integration definite-integrals trigonometric-integrals
edited Jul 20 at 10:30
Abcd
2,3811624
edited Jul 20 at 10:30
Abcd
2,3811624
edited Jul 20 at 10:30
Abcd
2,3811624
edited Jul 20 at 10:30
Abcd
2,3811624
2,3811624
asked Jul 20 at 8:10
user565198
I have a problem with $K$ which can not integrated if the lower bound is $0$
– Claude Leibovici
Jul 20 at 8:47
sorry for the mistake, the limit is $int_0^infty$
– user565198
Jul 20 at 8:53
The problem remains with $K$ around $x=0$. I do not see the trick but I suppose that we must not split as $I=J-K$.
– Claude Leibovici
Jul 20 at 9:00
When I read answers such as the ones you received, I realize how bad I am !
– Claude Leibovici
Jul 21 at 2:07
add a comment |Â
I have a problem with $K$ which can not integrated if the lower bound is $0$
– Claude Leibovici
Jul 20 at 8:47
sorry for the mistake, the limit is $int_0^infty$
– user565198
Jul 20 at 8:53
The problem remains with $K$ around $x=0$. I do not see the trick but I suppose that we must not split as $I=J-K$.
– Claude Leibovici
Jul 20 at 9:00
When I read answers such as the ones you received, I realize how bad I am !
– Claude Leibovici
Jul 21 at 2:07
I have a problem with $K$ which can not integrated if the lower bound is $0$
– Claude Leibovici
Jul 20 at 8:47
I have a problem with $K$ which can not integrated if the lower bound is $0$
– Claude Leibovici
Jul 20 at 8:47
sorry for the mistake, the limit is $int_0^infty$
– user565198
Jul 20 at 8:53
sorry for the mistake, the limit is $int_0^infty$
– user565198
Jul 20 at 8:53
The problem remains with $K$ around $x=0$. I do not see the trick but I suppose that we must not split as $I=J-K$.
– Claude Leibovici
Jul 20 at 9:00
The problem remains with $K$ around $x=0$. I do not see the trick but I suppose that we must not split as $I=J-K$.
– Claude Leibovici
Jul 20 at 9:00
When I read answers such as the ones you received, I realize how bad I am !
– Claude Leibovici
Jul 21 at 2:07
When I read answers such as the ones you received, I realize how bad I am !
– Claude Leibovici
Jul 21 at 2:07
add a comment |Â
4 Answers
4
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oldest
votes
up vote
3
down vote
Lemma. For $a > 0$, we have
$$ f(a) := int_0^infty fraccos (ax) - cos xxlog x , dx
= frac(2gamma + log a)log a2. $$
Proof. Consider the function $I(s) = int_0^infty fraccos(ax) - cos xx^1+s , dx$. Then $I(s)$ defines an analytic function on the strip $operatornameRe(s) in (-1, 1)$. Also, for $s in (-1, 0)$ we have
beginalign*
I(s)
&= int_0^infty left( frac1Gamma(1+s) int_0^infty u^s e^-xu , du right) (cos(ax) - cos x) , dx \
&= frac1Gamma(1+s) int_0^infty u^s left( int_0^infty e^-xu (cos(ax) - cos x) , dx right) , du \
&= frac1Gamma(1+s) int_0^infty left( fracu^s+1u^2+a^2 - fracu^s+1u^2 + 1 right) , du \
&= frac(a^s - 1)mathrmB(fracs2+1,-fracs2)2Gamma(1+s) \
&= -fracpi (a^s - 1)2sin(fracpi s2)Gamma(1+s).
endalign*
This formula continues to hold on all of the strip by the principle of analytic continuation. So it follows that
$$ f(a)
= -I'(0) = -lim_s to 0 I'(s) = frac(2gamma + log a)log a2. $$
Using the lemma and $cos(x/a)cos x = frac12left( cos (fraca+1ax) + cos(|fraca-1a|x) right)$, we obtain
$$ int_0^infty fraccos (ax) - 1xcos xlog x , dx
= fracfraca-1aright2. $$
answered Jul 20 at 10:45
Sangchul Lee
85.5k12155253
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up vote
2
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Im in a hurry so Il only give a sketch of my ideea and maybe someone can finish it, or Il try later. Let us consider: $$I(a)=int_0^infty fraccos left(fracxaright)-1xcos x ln x,dx$$ We differentiate w.r.t a (see here: https://en.wikipedia.org/wiki/Leibniz_integral_rule)
$$I'(a)=int_0^infty frac1a^2sinleft(fracxaright)cos x ln x ,fracxx,dx $$ In case you wonder why I added $fracxx=1$ is because the next step is to use https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_integrals_over_the_positive_real_axis but first lets adjust the integral usign the product to sum formula then apply the laplace transform. $$I'(a)=frac12a^2int_0^infty left(xsinleft(x+fracxaright)-xsinleft(x-fracxaright)right)left(fracln x xright),dx$$ The inverse laplace of $fracln ss=-(ln t-gamma),$ which can be tackled down working backwards by computing the laplace transform of $ln x$ using the derivate of the gamma function and for the first two its just the laplace of the sine part differentiated once.$$I'(a)=-frac12a^2int_0^infty left(frac2xleft(1+frac1aright)left(x^2+left(1+frac1aright)^2right)^2 -frac2xleft(1-frac1aright)left(x^2+left(1-frac1aright)^2right)^2right)(ln x +gamma),dx $$ Now what's next is to split into $4$ integrals. Since wolfram evaluates $$int_0^infty fracbxln x(x^2+b^2)^2dx=frac14bln left(frac1b^2right)$$ and closed form for other parts are not hard to find, also its pretty easy to integrate back but this time dont forget its $ da,$ instead of $dx,$ then for $a=1$ we should find the constant and the final result gave by @Claude Leibovici. Let me know if you can finish it.
answered Jul 20 at 10:50
Zacky
2,2251326
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up vote
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Let $b=frac1a > 0$ to avoid fractions. Using trigonometric identities we have
$$ [cos(b x) - 1]cos(x) = - sinleft(fracb2xright) leftsinleft[left(fracb2-1right)xright] + sinleft[left(fracb2+1right)xright]right , . $$
In my answer to this question I have shown that
$$ int limits_0^infty fracsin(alpha x) sin(beta x)x ln(x) , mathrmd x = frac14 ln left(fraclvert alpha - betarvertalpha + betaright)[ln(lvert alpha^2-beta^2rvert) + 2 gamma]$$
holds for $alpha, beta > 0$ if $alpha neq beta$ . Therefore we obtain for $a neq 1 Leftrightarrow b neq 1$
beginalign
I &= int limits_0^infty frac[cos(b x)-1]cos(x)x ln(x) , mathrmd x \
&= - int limits_0^infty fracsinleft(fracb2xright) leftsinleft[left(fracb2-1right)xright] + sinleft[left(fracb2+1right)xright]rightx ln(x) , mathrmd x \
&= frac14 ln(lvert 1-brvert) [ln(lvert 1-brvert) + 2 gamma] + frac14 ln( 1+b) [ln( 1+b) + 2 gamma] \
&= frac14 [ln^2 (lvert 1-brvert) + ln^2 (1+b) + 2 gamma ln (lvert 1-b^2 rvert)] , .
endalign
answered Jul 20 at 10:54
ComplexYetTrivial
2,607624
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up vote
1
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This is not an answer but it is too long for a comment.
As I wrote in a comment, I suppose that there is a trick but I suppose that we must not split as $I=J−K$ since there is a serious problem with $K$ around $x=0$.
Close to zero
$$fraccos left(fracxaright)-1x,cos (x),log (x)=-fracx log (x)2 a^2+Oleft(x^3right) $$
$$int fraccos left(fracxaright)-1x,cos (x),log (x),dx=fracx^2 (1-2 log (x))8 a^2+Oleft(x^4right)$$ does not make any problem.
Being short of ideas and lazy, just by curiosity, I gave it to a CAS and obtained the result
$$I=frac116 left(log ^2left(frac(a-1)^2a^2right)+4 gamma log
left(frac(a-1)^2a^2right)+4 log left(frac1+aaright) left(log
left(frac1+aaright)+2 gamma right)right)$$ How to get it, this is the question !
answered Jul 20 at 9:57
Claude Leibovici
111k1055126
I have simplified to $fracln^2(fracaa-1)-ln^2(fracaa+1)+2gammaln(fraca^2-1a^2)4$
– user565198
Jul 20 at 10:06
@Bonjour. I did not try since I have to go now. Please, could you let me know if you get an answer ? By the way, do you agree with me that splitting the integral does not seem to be a good idea . Cheers.
– Claude Leibovici
Jul 20 at 10:16
ok, thank you, I agree. See you.
– user565198
Jul 20 at 10:40
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Lemma. For $a > 0$, we have
$$ f(a) := int_0^infty fraccos (ax) - cos xxlog x , dx
= frac(2gamma + log a)log a2. $$
Proof. Consider the function $I(s) = int_0^infty fraccos(ax) - cos xx^1+s , dx$. Then $I(s)$ defines an analytic function on the strip $operatornameRe(s) in (-1, 1)$. Also, for $s in (-1, 0)$ we have
beginalign*
I(s)
&= int_0^infty left( frac1Gamma(1+s) int_0^infty u^s e^-xu , du right) (cos(ax) - cos x) , dx \
&= frac1Gamma(1+s) int_0^infty u^s left( int_0^infty e^-xu (cos(ax) - cos x) , dx right) , du \
&= frac1Gamma(1+s) int_0^infty left( fracu^s+1u^2+a^2 - fracu^s+1u^2 + 1 right) , du \
&= frac(a^s - 1)mathrmB(fracs2+1,-fracs2)2Gamma(1+s) \
&= -fracpi (a^s - 1)2sin(fracpi s2)Gamma(1+s).
endalign*
This formula continues to hold on all of the strip by the principle of analytic continuation. So it follows that
$$ f(a)
= -I'(0) = -lim_s to 0 I'(s) = frac(2gamma + log a)log a2. $$
Using the lemma and $cos(x/a)cos x = frac12left( cos (fraca+1ax) + cos(|fraca-1a|x) right)$, we obtain
$$ int_0^infty fraccos (ax) - 1xcos xlog x , dx
= fracfraca-1aright2. $$
answered Jul 20 at 10:45
Sangchul Lee
85.5k12155253
add a comment |Â
up vote
3
down vote
Lemma. For $a > 0$, we have
$$ f(a) := int_0^infty fraccos (ax) - cos xxlog x , dx
= frac(2gamma + log a)log a2. $$
Proof. Consider the function $I(s) = int_0^infty fraccos(ax) - cos xx^1+s , dx$. Then $I(s)$ defines an analytic function on the strip $operatornameRe(s) in (-1, 1)$. Also, for $s in (-1, 0)$ we have
beginalign*
I(s)
&= int_0^infty left( frac1Gamma(1+s) int_0^infty u^s e^-xu , du right) (cos(ax) - cos x) , dx \
&= frac1Gamma(1+s) int_0^infty u^s left( int_0^infty e^-xu (cos(ax) - cos x) , dx right) , du \
&= frac1Gamma(1+s) int_0^infty left( fracu^s+1u^2+a^2 - fracu^s+1u^2 + 1 right) , du \
&= frac(a^s - 1)mathrmB(fracs2+1,-fracs2)2Gamma(1+s) \
&= -fracpi (a^s - 1)2sin(fracpi s2)Gamma(1+s).
endalign*
This formula continues to hold on all of the strip by the principle of analytic continuation. So it follows that
$$ f(a)
= -I'(0) = -lim_s to 0 I'(s) = frac(2gamma + log a)log a2. $$
Using the lemma and $cos(x/a)cos x = frac12left( cos (fraca+1ax) + cos(|fraca-1a|x) right)$, we obtain
$$ int_0^infty fraccos (ax) - 1xcos xlog x , dx
= fracfraca-1aright2. $$
answered Jul 20 at 10:45
Sangchul Lee
85.5k12155253
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Lemma. For $a > 0$, we have
$$ f(a) := int_0^infty fraccos (ax) - cos xxlog x , dx
= frac(2gamma + log a)log a2. $$
Proof. Consider the function $I(s) = int_0^infty fraccos(ax) - cos xx^1+s , dx$. Then $I(s)$ defines an analytic function on the strip $operatornameRe(s) in (-1, 1)$. Also, for $s in (-1, 0)$ we have
beginalign*
I(s)
&= int_0^infty left( frac1Gamma(1+s) int_0^infty u^s e^-xu , du right) (cos(ax) - cos x) , dx \
&= frac1Gamma(1+s) int_0^infty u^s left( int_0^infty e^-xu (cos(ax) - cos x) , dx right) , du \
&= frac1Gamma(1+s) int_0^infty left( fracu^s+1u^2+a^2 - fracu^s+1u^2 + 1 right) , du \
&= frac(a^s - 1)mathrmB(fracs2+1,-fracs2)2Gamma(1+s) \
&= -fracpi (a^s - 1)2sin(fracpi s2)Gamma(1+s).
endalign*
This formula continues to hold on all of the strip by the principle of analytic continuation. So it follows that
$$ f(a)
= -I'(0) = -lim_s to 0 I'(s) = frac(2gamma + log a)log a2. $$
Using the lemma and $cos(x/a)cos x = frac12left( cos (fraca+1ax) + cos(|fraca-1a|x) right)$, we obtain
$$ int_0^infty fraccos (ax) - 1xcos xlog x , dx
= fracfraca-1aright2. $$
answered Jul 20 at 10:45
Sangchul Lee
85.5k12155253
Lemma. For $a > 0$, we have
$$ f(a) := int_0^infty fraccos (ax) - cos xxlog x , dx
= frac(2gamma + log a)log a2. $$
Proof. Consider the function $I(s) = int_0^infty fraccos(ax) - cos xx^1+s , dx$. Then $I(s)$ defines an analytic function on the strip $operatornameRe(s) in (-1, 1)$. Also, for $s in (-1, 0)$ we have
beginalign*
I(s)
&= int_0^infty left( frac1Gamma(1+s) int_0^infty u^s e^-xu , du right) (cos(ax) - cos x) , dx \
&= frac1Gamma(1+s) int_0^infty u^s left( int_0^infty e^-xu (cos(ax) - cos x) , dx right) , du \
&= frac1Gamma(1+s) int_0^infty left( fracu^s+1u^2+a^2 - fracu^s+1u^2 + 1 right) , du \
&= frac(a^s - 1)mathrmB(fracs2+1,-fracs2)2Gamma(1+s) \
&= -fracpi (a^s - 1)2sin(fracpi s2)Gamma(1+s).
endalign*
This formula continues to hold on all of the strip by the principle of analytic continuation. So it follows that
$$ f(a)
= -I'(0) = -lim_s to 0 I'(s) = frac(2gamma + log a)log a2. $$
Using the lemma and $cos(x/a)cos x = frac12left( cos (fraca+1ax) + cos(|fraca-1a|x) right)$, we obtain
$$ int_0^infty fraccos (ax) - 1xcos xlog x , dx
= fracfraca-1aright2. $$
answered Jul 20 at 10:45
Sangchul Lee
85.5k12155253
answered Jul 20 at 10:45
Sangchul Lee
85.5k12155253
answered Jul 20 at 10:45
Sangchul Lee
85.5k12155253
answered Jul 20 at 10:45
Sangchul Lee
85.5k12155253
85.5k12155253
add a comment |Â
add a comment |Â
up vote
2
down vote
Im in a hurry so Il only give a sketch of my ideea and maybe someone can finish it, or Il try later. Let us consider: $$I(a)=int_0^infty fraccos left(fracxaright)-1xcos x ln x,dx$$ We differentiate w.r.t a (see here: https://en.wikipedia.org/wiki/Leibniz_integral_rule)
$$I'(a)=int_0^infty frac1a^2sinleft(fracxaright)cos x ln x ,fracxx,dx $$ In case you wonder why I added $fracxx=1$ is because the next step is to use https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_integrals_over_the_positive_real_axis but first lets adjust the integral usign the product to sum formula then apply the laplace transform. $$I'(a)=frac12a^2int_0^infty left(xsinleft(x+fracxaright)-xsinleft(x-fracxaright)right)left(fracln x xright),dx$$ The inverse laplace of $fracln ss=-(ln t-gamma),$ which can be tackled down working backwards by computing the laplace transform of $ln x$ using the derivate of the gamma function and for the first two its just the laplace of the sine part differentiated once.$$I'(a)=-frac12a^2int_0^infty left(frac2xleft(1+frac1aright)left(x^2+left(1+frac1aright)^2right)^2 -frac2xleft(1-frac1aright)left(x^2+left(1-frac1aright)^2right)^2right)(ln x +gamma),dx $$ Now what's next is to split into $4$ integrals. Since wolfram evaluates $$int_0^infty fracbxln x(x^2+b^2)^2dx=frac14bln left(frac1b^2right)$$ and closed form for other parts are not hard to find, also its pretty easy to integrate back but this time dont forget its $ da,$ instead of $dx,$ then for $a=1$ we should find the constant and the final result gave by @Claude Leibovici. Let me know if you can finish it.
answered Jul 20 at 10:50
Zacky
2,2251326
add a comment |Â
up vote
2
down vote
Im in a hurry so Il only give a sketch of my ideea and maybe someone can finish it, or Il try later. Let us consider: $$I(a)=int_0^infty fraccos left(fracxaright)-1xcos x ln x,dx$$ We differentiate w.r.t a (see here: https://en.wikipedia.org/wiki/Leibniz_integral_rule)
$$I'(a)=int_0^infty frac1a^2sinleft(fracxaright)cos x ln x ,fracxx,dx $$ In case you wonder why I added $fracxx=1$ is because the next step is to use https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_integrals_over_the_positive_real_axis but first lets adjust the integral usign the product to sum formula then apply the laplace transform. $$I'(a)=frac12a^2int_0^infty left(xsinleft(x+fracxaright)-xsinleft(x-fracxaright)right)left(fracln x xright),dx$$ The inverse laplace of $fracln ss=-(ln t-gamma),$ which can be tackled down working backwards by computing the laplace transform of $ln x$ using the derivate of the gamma function and for the first two its just the laplace of the sine part differentiated once.$$I'(a)=-frac12a^2int_0^infty left(frac2xleft(1+frac1aright)left(x^2+left(1+frac1aright)^2right)^2 -frac2xleft(1-frac1aright)left(x^2+left(1-frac1aright)^2right)^2right)(ln x +gamma),dx $$ Now what's next is to split into $4$ integrals. Since wolfram evaluates $$int_0^infty fracbxln x(x^2+b^2)^2dx=frac14bln left(frac1b^2right)$$ and closed form for other parts are not hard to find, also its pretty easy to integrate back but this time dont forget its $ da,$ instead of $dx,$ then for $a=1$ we should find the constant and the final result gave by @Claude Leibovici. Let me know if you can finish it.
answered Jul 20 at 10:50
Zacky
2,2251326
add a comment |Â
up vote
2
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up vote
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down vote
Im in a hurry so Il only give a sketch of my ideea and maybe someone can finish it, or Il try later. Let us consider: $$I(a)=int_0^infty fraccos left(fracxaright)-1xcos x ln x,dx$$ We differentiate w.r.t a (see here: https://en.wikipedia.org/wiki/Leibniz_integral_rule)
$$I'(a)=int_0^infty frac1a^2sinleft(fracxaright)cos x ln x ,fracxx,dx $$ In case you wonder why I added $fracxx=1$ is because the next step is to use https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_integrals_over_the_positive_real_axis but first lets adjust the integral usign the product to sum formula then apply the laplace transform. $$I'(a)=frac12a^2int_0^infty left(xsinleft(x+fracxaright)-xsinleft(x-fracxaright)right)left(fracln x xright),dx$$ The inverse laplace of $fracln ss=-(ln t-gamma),$ which can be tackled down working backwards by computing the laplace transform of $ln x$ using the derivate of the gamma function and for the first two its just the laplace of the sine part differentiated once.$$I'(a)=-frac12a^2int_0^infty left(frac2xleft(1+frac1aright)left(x^2+left(1+frac1aright)^2right)^2 -frac2xleft(1-frac1aright)left(x^2+left(1-frac1aright)^2right)^2right)(ln x +gamma),dx $$ Now what's next is to split into $4$ integrals. Since wolfram evaluates $$int_0^infty fracbxln x(x^2+b^2)^2dx=frac14bln left(frac1b^2right)$$ and closed form for other parts are not hard to find, also its pretty easy to integrate back but this time dont forget its $ da,$ instead of $dx,$ then for $a=1$ we should find the constant and the final result gave by @Claude Leibovici. Let me know if you can finish it.
answered Jul 20 at 10:50
Zacky
2,2251326
Im in a hurry so Il only give a sketch of my ideea and maybe someone can finish it, or Il try later. Let us consider: $$I(a)=int_0^infty fraccos left(fracxaright)-1xcos x ln x,dx$$ We differentiate w.r.t a (see here: https://en.wikipedia.org/wiki/Leibniz_integral_rule)
$$I'(a)=int_0^infty frac1a^2sinleft(fracxaright)cos x ln x ,fracxx,dx $$ In case you wonder why I added $fracxx=1$ is because the next step is to use https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_integrals_over_the_positive_real_axis but first lets adjust the integral usign the product to sum formula then apply the laplace transform. $$I'(a)=frac12a^2int_0^infty left(xsinleft(x+fracxaright)-xsinleft(x-fracxaright)right)left(fracln x xright),dx$$ The inverse laplace of $fracln ss=-(ln t-gamma),$ which can be tackled down working backwards by computing the laplace transform of $ln x$ using the derivate of the gamma function and for the first two its just the laplace of the sine part differentiated once.$$I'(a)=-frac12a^2int_0^infty left(frac2xleft(1+frac1aright)left(x^2+left(1+frac1aright)^2right)^2 -frac2xleft(1-frac1aright)left(x^2+left(1-frac1aright)^2right)^2right)(ln x +gamma),dx $$ Now what's next is to split into $4$ integrals. Since wolfram evaluates $$int_0^infty fracbxln x(x^2+b^2)^2dx=frac14bln left(frac1b^2right)$$ and closed form for other parts are not hard to find, also its pretty easy to integrate back but this time dont forget its $ da,$ instead of $dx,$ then for $a=1$ we should find the constant and the final result gave by @Claude Leibovici. Let me know if you can finish it.
answered Jul 20 at 10:50
Zacky
2,2251326
answered Jul 20 at 10:50
Zacky
2,2251326
answered Jul 20 at 10:50
Zacky
2,2251326
answered Jul 20 at 10:50
Zacky
2,2251326
2,2251326
add a comment |Â
add a comment |Â
up vote
2
down vote
Let $b=frac1a > 0$ to avoid fractions. Using trigonometric identities we have
$$ [cos(b x) - 1]cos(x) = - sinleft(fracb2xright) leftsinleft[left(fracb2-1right)xright] + sinleft[left(fracb2+1right)xright]right , . $$
In my answer to this question I have shown that
$$ int limits_0^infty fracsin(alpha x) sin(beta x)x ln(x) , mathrmd x = frac14 ln left(fraclvert alpha - betarvertalpha + betaright)[ln(lvert alpha^2-beta^2rvert) + 2 gamma]$$
holds for $alpha, beta > 0$ if $alpha neq beta$ . Therefore we obtain for $a neq 1 Leftrightarrow b neq 1$
beginalign
I &= int limits_0^infty frac[cos(b x)-1]cos(x)x ln(x) , mathrmd x \
&= - int limits_0^infty fracsinleft(fracb2xright) leftsinleft[left(fracb2-1right)xright] + sinleft[left(fracb2+1right)xright]rightx ln(x) , mathrmd x \
&= frac14 ln(lvert 1-brvert) [ln(lvert 1-brvert) + 2 gamma] + frac14 ln( 1+b) [ln( 1+b) + 2 gamma] \
&= frac14 [ln^2 (lvert 1-brvert) + ln^2 (1+b) + 2 gamma ln (lvert 1-b^2 rvert)] , .
endalign
answered Jul 20 at 10:54
ComplexYetTrivial
2,607624
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up vote
2
down vote
Let $b=frac1a > 0$ to avoid fractions. Using trigonometric identities we have
$$ [cos(b x) - 1]cos(x) = - sinleft(fracb2xright) leftsinleft[left(fracb2-1right)xright] + sinleft[left(fracb2+1right)xright]right , . $$
In my answer to this question I have shown that
$$ int limits_0^infty fracsin(alpha x) sin(beta x)x ln(x) , mathrmd x = frac14 ln left(fraclvert alpha - betarvertalpha + betaright)[ln(lvert alpha^2-beta^2rvert) + 2 gamma]$$
holds for $alpha, beta > 0$ if $alpha neq beta$ . Therefore we obtain for $a neq 1 Leftrightarrow b neq 1$
beginalign
I &= int limits_0^infty frac[cos(b x)-1]cos(x)x ln(x) , mathrmd x \
&= - int limits_0^infty fracsinleft(fracb2xright) leftsinleft[left(fracb2-1right)xright] + sinleft[left(fracb2+1right)xright]rightx ln(x) , mathrmd x \
&= frac14 ln(lvert 1-brvert) [ln(lvert 1-brvert) + 2 gamma] + frac14 ln( 1+b) [ln( 1+b) + 2 gamma] \
&= frac14 [ln^2 (lvert 1-brvert) + ln^2 (1+b) + 2 gamma ln (lvert 1-b^2 rvert)] , .
endalign
answered Jul 20 at 10:54
ComplexYetTrivial
2,607624
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $b=frac1a > 0$ to avoid fractions. Using trigonometric identities we have
$$ [cos(b x) - 1]cos(x) = - sinleft(fracb2xright) leftsinleft[left(fracb2-1right)xright] + sinleft[left(fracb2+1right)xright]right , . $$
In my answer to this question I have shown that
$$ int limits_0^infty fracsin(alpha x) sin(beta x)x ln(x) , mathrmd x = frac14 ln left(fraclvert alpha - betarvertalpha + betaright)[ln(lvert alpha^2-beta^2rvert) + 2 gamma]$$
holds for $alpha, beta > 0$ if $alpha neq beta$ . Therefore we obtain for $a neq 1 Leftrightarrow b neq 1$
beginalign
I &= int limits_0^infty frac[cos(b x)-1]cos(x)x ln(x) , mathrmd x \
&= - int limits_0^infty fracsinleft(fracb2xright) leftsinleft[left(fracb2-1right)xright] + sinleft[left(fracb2+1right)xright]rightx ln(x) , mathrmd x \
&= frac14 ln(lvert 1-brvert) [ln(lvert 1-brvert) + 2 gamma] + frac14 ln( 1+b) [ln( 1+b) + 2 gamma] \
&= frac14 [ln^2 (lvert 1-brvert) + ln^2 (1+b) + 2 gamma ln (lvert 1-b^2 rvert)] , .
endalign
answered Jul 20 at 10:54
ComplexYetTrivial
2,607624
Let $b=frac1a > 0$ to avoid fractions. Using trigonometric identities we have
$$ [cos(b x) - 1]cos(x) = - sinleft(fracb2xright) leftsinleft[left(fracb2-1right)xright] + sinleft[left(fracb2+1right)xright]right , . $$
In my answer to this question I have shown that
$$ int limits_0^infty fracsin(alpha x) sin(beta x)x ln(x) , mathrmd x = frac14 ln left(fraclvert alpha - betarvertalpha + betaright)[ln(lvert alpha^2-beta^2rvert) + 2 gamma]$$
holds for $alpha, beta > 0$ if $alpha neq beta$ . Therefore we obtain for $a neq 1 Leftrightarrow b neq 1$
beginalign
I &= int limits_0^infty frac[cos(b x)-1]cos(x)x ln(x) , mathrmd x \
&= - int limits_0^infty fracsinleft(fracb2xright) leftsinleft[left(fracb2-1right)xright] + sinleft[left(fracb2+1right)xright]rightx ln(x) , mathrmd x \
&= frac14 ln(lvert 1-brvert) [ln(lvert 1-brvert) + 2 gamma] + frac14 ln( 1+b) [ln( 1+b) + 2 gamma] \
&= frac14 [ln^2 (lvert 1-brvert) + ln^2 (1+b) + 2 gamma ln (lvert 1-b^2 rvert)] , .
endalign
answered Jul 20 at 10:54
ComplexYetTrivial
2,607624
answered Jul 20 at 10:54
ComplexYetTrivial
2,607624
answered Jul 20 at 10:54
ComplexYetTrivial
2,607624
answered Jul 20 at 10:54
ComplexYetTrivial
2,607624
2,607624
add a comment |Â
add a comment |Â
up vote
1
down vote
This is not an answer but it is too long for a comment.
As I wrote in a comment, I suppose that there is a trick but I suppose that we must not split as $I=J−K$ since there is a serious problem with $K$ around $x=0$.
Close to zero
$$fraccos left(fracxaright)-1x,cos (x),log (x)=-fracx log (x)2 a^2+Oleft(x^3right) $$
$$int fraccos left(fracxaright)-1x,cos (x),log (x),dx=fracx^2 (1-2 log (x))8 a^2+Oleft(x^4right)$$ does not make any problem.
Being short of ideas and lazy, just by curiosity, I gave it to a CAS and obtained the result
$$I=frac116 left(log ^2left(frac(a-1)^2a^2right)+4 gamma log
left(frac(a-1)^2a^2right)+4 log left(frac1+aaright) left(log
left(frac1+aaright)+2 gamma right)right)$$ How to get it, this is the question !
answered Jul 20 at 9:57
Claude Leibovici
111k1055126
I have simplified to $fracln^2(fracaa-1)-ln^2(fracaa+1)+2gammaln(fraca^2-1a^2)4$
– user565198
Jul 20 at 10:06
@Bonjour. I did not try since I have to go now. Please, could you let me know if you get an answer ? By the way, do you agree with me that splitting the integral does not seem to be a good idea . Cheers.
– Claude Leibovici
Jul 20 at 10:16
ok, thank you, I agree. See you.
– user565198
Jul 20 at 10:40
add a comment |Â
up vote
1
down vote
This is not an answer but it is too long for a comment.
As I wrote in a comment, I suppose that there is a trick but I suppose that we must not split as $I=J−K$ since there is a serious problem with $K$ around $x=0$.
Close to zero
$$fraccos left(fracxaright)-1x,cos (x),log (x)=-fracx log (x)2 a^2+Oleft(x^3right) $$
$$int fraccos left(fracxaright)-1x,cos (x),log (x),dx=fracx^2 (1-2 log (x))8 a^2+Oleft(x^4right)$$ does not make any problem.
Being short of ideas and lazy, just by curiosity, I gave it to a CAS and obtained the result
$$I=frac116 left(log ^2left(frac(a-1)^2a^2right)+4 gamma log
left(frac(a-1)^2a^2right)+4 log left(frac1+aaright) left(log
left(frac1+aaright)+2 gamma right)right)$$ How to get it, this is the question !
answered Jul 20 at 9:57
Claude Leibovici
111k1055126
I have simplified to $fracln^2(fracaa-1)-ln^2(fracaa+1)+2gammaln(fraca^2-1a^2)4$
– user565198
Jul 20 at 10:06
@Bonjour. I did not try since I have to go now. Please, could you let me know if you get an answer ? By the way, do you agree with me that splitting the integral does not seem to be a good idea . Cheers.
– Claude Leibovici
Jul 20 at 10:16
ok, thank you, I agree. See you.
– user565198
Jul 20 at 10:40
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is not an answer but it is too long for a comment.
As I wrote in a comment, I suppose that there is a trick but I suppose that we must not split as $I=J−K$ since there is a serious problem with $K$ around $x=0$.
Close to zero
$$fraccos left(fracxaright)-1x,cos (x),log (x)=-fracx log (x)2 a^2+Oleft(x^3right) $$
$$int fraccos left(fracxaright)-1x,cos (x),log (x),dx=fracx^2 (1-2 log (x))8 a^2+Oleft(x^4right)$$ does not make any problem.
Being short of ideas and lazy, just by curiosity, I gave it to a CAS and obtained the result
$$I=frac116 left(log ^2left(frac(a-1)^2a^2right)+4 gamma log
left(frac(a-1)^2a^2right)+4 log left(frac1+aaright) left(log
left(frac1+aaright)+2 gamma right)right)$$ How to get it, this is the question !
answered Jul 20 at 9:57
Claude Leibovici
111k1055126
This is not an answer but it is too long for a comment.
As I wrote in a comment, I suppose that there is a trick but I suppose that we must not split as $I=J−K$ since there is a serious problem with $K$ around $x=0$.
Close to zero
$$fraccos left(fracxaright)-1x,cos (x),log (x)=-fracx log (x)2 a^2+Oleft(x^3right) $$
$$int fraccos left(fracxaright)-1x,cos (x),log (x),dx=fracx^2 (1-2 log (x))8 a^2+Oleft(x^4right)$$ does not make any problem.
Being short of ideas and lazy, just by curiosity, I gave it to a CAS and obtained the result
$$I=frac116 left(log ^2left(frac(a-1)^2a^2right)+4 gamma log
left(frac(a-1)^2a^2right)+4 log left(frac1+aaright) left(log
left(frac1+aaright)+2 gamma right)right)$$ How to get it, this is the question !
answered Jul 20 at 9:57
Claude Leibovici
111k1055126
answered Jul 20 at 9:57
Claude Leibovici
111k1055126
answered Jul 20 at 9:57
Claude Leibovici
111k1055126
answered Jul 20 at 9:57
Claude Leibovici
111k1055126
111k1055126
I have simplified to $fracln^2(fracaa-1)-ln^2(fracaa+1)+2gammaln(fraca^2-1a^2)4$
– user565198
Jul 20 at 10:06
@Bonjour. I did not try since I have to go now. Please, could you let me know if you get an answer ? By the way, do you agree with me that splitting the integral does not seem to be a good idea . Cheers.
– Claude Leibovici
Jul 20 at 10:16
ok, thank you, I agree. See you.
– user565198
Jul 20 at 10:40
add a comment |Â
I have simplified to $fracln^2(fracaa-1)-ln^2(fracaa+1)+2gammaln(fraca^2-1a^2)4$
– user565198
Jul 20 at 10:06
@Bonjour. I did not try since I have to go now. Please, could you let me know if you get an answer ? By the way, do you agree with me that splitting the integral does not seem to be a good idea . Cheers.
– Claude Leibovici
Jul 20 at 10:16
ok, thank you, I agree. See you.
– user565198
Jul 20 at 10:40
I have simplified to $fracln^2(fracaa-1)-ln^2(fracaa+1)+2gammaln(fraca^2-1a^2)4$
– user565198
Jul 20 at 10:06
I have simplified to $fracln^2(fracaa-1)-ln^2(fracaa+1)+2gammaln(fraca^2-1a^2)4$
– user565198
Jul 20 at 10:06
@Bonjour. I did not try since I have to go now. Please, could you let me know if you get an answer ? By the way, do you agree with me that splitting the integral does not seem to be a good idea . Cheers.
– Claude Leibovici
Jul 20 at 10:16
@Bonjour. I did not try since I have to go now. Please, could you let me know if you get an answer ? By the way, do you agree with me that splitting the integral does not seem to be a good idea . Cheers.
– Claude Leibovici
Jul 20 at 10:16
ok, thank you, I agree. See you.
– user565198
Jul 20 at 10:40
ok, thank you, I agree. See you.
– user565198
Jul 20 at 10:40
add a comment |Â
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I have a problem with $K$ which can not integrated if the lower bound is $0$
– Claude Leibovici
Jul 20 at 8:47
sorry for the mistake, the limit is $int_0^infty$
– user565198
Jul 20 at 8:53
The problem remains with $K$ around $x=0$. I do not see the trick but I suppose that we must not split as $I=J-K$.
– Claude Leibovici
Jul 20 at 9:00
When I read answers such as the ones you received, I realize how bad I am !
– Claude Leibovici
Jul 21 at 2:07