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Issue with $int_0^inftyfraccos(x^2)x^2dx$
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I have stumbled into a issue with gamma function, I will show the approach first:
$$Gamma(x)=int_0^inftyt^x-1e^-tdt$$ subtituting $t=iu^2$ gives:
$$Gamma(x)=2int_0^infty(iu^2)^x-1e^-iu^2iudurightarrowfracGamma(x)2i^x=int_0^inftyu^2x-1e^-iu^2du$$ Doing the same thing using $t=-iu^2,$results in$$fracGamma(x)-2i^x=int_0^inftyu^2x-1e^iu^2du$$ Now, summing those two and using that $i^x=e^fracipi2x ,$gives
$$fracGamma(x)2(e^fracipi2x+e^frac-ipi2x)=int_0^inftyu^2x-1(e^iu^2+e^-iu^2)du$$ which is just $$fracGamma(x)2cos(fracpi2x)=int_0^inftyu^2x-1cos(u^2)du$$ plugging $x=-frac12$ we get that$$int_0^inftyfraccos(x^2)x^2dx=-sqrtfracpi2$$ Well, obviously this integral diverges... But if Instead of summing we subtract we get that $$fracGamma(x)2sin(fracpi2x)=int_0^inftyu^2x-1sin(u^2)du$$ simmilarly with $$x=-frac12 rightarrow int_0^inftyfracsin(x^2)x^2dx=sqrtfracpi2$$ So its not that completely garbage. Now my question is, what goes wrong when I use the first substitution? And how do I prove that I am allowed to use this substitution for the sine integral?
integration complex-analysis substitution
edited Jul 24 at 20:06
Chappers
55k74190
asked Jul 24 at 19:51
Zacky
2,1771326
add a comment |Â
up vote
5
down vote
favorite
I have stumbled into a issue with gamma function, I will show the approach first:
$$Gamma(x)=int_0^inftyt^x-1e^-tdt$$ subtituting $t=iu^2$ gives:
$$Gamma(x)=2int_0^infty(iu^2)^x-1e^-iu^2iudurightarrowfracGamma(x)2i^x=int_0^inftyu^2x-1e^-iu^2du$$ Doing the same thing using $t=-iu^2,$results in$$fracGamma(x)-2i^x=int_0^inftyu^2x-1e^iu^2du$$ Now, summing those two and using that $i^x=e^fracipi2x ,$gives
$$fracGamma(x)2(e^fracipi2x+e^frac-ipi2x)=int_0^inftyu^2x-1(e^iu^2+e^-iu^2)du$$ which is just $$fracGamma(x)2cos(fracpi2x)=int_0^inftyu^2x-1cos(u^2)du$$ plugging $x=-frac12$ we get that$$int_0^inftyfraccos(x^2)x^2dx=-sqrtfracpi2$$ Well, obviously this integral diverges... But if Instead of summing we subtract we get that $$fracGamma(x)2sin(fracpi2x)=int_0^inftyu^2x-1sin(u^2)du$$ simmilarly with $$x=-frac12 rightarrow int_0^inftyfracsin(x^2)x^2dx=sqrtfracpi2$$ So its not that completely garbage. Now my question is, what goes wrong when I use the first substitution? And how do I prove that I am allowed to use this substitution for the sine integral?
integration complex-analysis substitution
edited Jul 24 at 20:06
Chappers
55k74190
asked Jul 24 at 19:51
Zacky
2,1771326
1
Your upper integration limit is incorrect after the change of variables. The correct intrgration limit are $u=0$ and $u=e^-ipi/4infty$. Attempt to deform the contour onto the real line fails.
– Mark Viola
Jul 24 at 20:24
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I have stumbled into a issue with gamma function, I will show the approach first:
$$Gamma(x)=int_0^inftyt^x-1e^-tdt$$ subtituting $t=iu^2$ gives:
$$Gamma(x)=2int_0^infty(iu^2)^x-1e^-iu^2iudurightarrowfracGamma(x)2i^x=int_0^inftyu^2x-1e^-iu^2du$$ Doing the same thing using $t=-iu^2,$results in$$fracGamma(x)-2i^x=int_0^inftyu^2x-1e^iu^2du$$ Now, summing those two and using that $i^x=e^fracipi2x ,$gives
$$fracGamma(x)2(e^fracipi2x+e^frac-ipi2x)=int_0^inftyu^2x-1(e^iu^2+e^-iu^2)du$$ which is just $$fracGamma(x)2cos(fracpi2x)=int_0^inftyu^2x-1cos(u^2)du$$ plugging $x=-frac12$ we get that$$int_0^inftyfraccos(x^2)x^2dx=-sqrtfracpi2$$ Well, obviously this integral diverges... But if Instead of summing we subtract we get that $$fracGamma(x)2sin(fracpi2x)=int_0^inftyu^2x-1sin(u^2)du$$ simmilarly with $$x=-frac12 rightarrow int_0^inftyfracsin(x^2)x^2dx=sqrtfracpi2$$ So its not that completely garbage. Now my question is, what goes wrong when I use the first substitution? And how do I prove that I am allowed to use this substitution for the sine integral?
integration complex-analysis substitution
edited Jul 24 at 20:06
Chappers
55k74190
asked Jul 24 at 19:51
Zacky
2,1771326
I have stumbled into a issue with gamma function, I will show the approach first:
$$Gamma(x)=int_0^inftyt^x-1e^-tdt$$ subtituting $t=iu^2$ gives:
$$Gamma(x)=2int_0^infty(iu^2)^x-1e^-iu^2iudurightarrowfracGamma(x)2i^x=int_0^inftyu^2x-1e^-iu^2du$$ Doing the same thing using $t=-iu^2,$results in$$fracGamma(x)-2i^x=int_0^inftyu^2x-1e^iu^2du$$ Now, summing those two and using that $i^x=e^fracipi2x ,$gives
$$fracGamma(x)2(e^fracipi2x+e^frac-ipi2x)=int_0^inftyu^2x-1(e^iu^2+e^-iu^2)du$$ which is just $$fracGamma(x)2cos(fracpi2x)=int_0^inftyu^2x-1cos(u^2)du$$ plugging $x=-frac12$ we get that$$int_0^inftyfraccos(x^2)x^2dx=-sqrtfracpi2$$ Well, obviously this integral diverges... But if Instead of summing we subtract we get that $$fracGamma(x)2sin(fracpi2x)=int_0^inftyu^2x-1sin(u^2)du$$ simmilarly with $$x=-frac12 rightarrow int_0^inftyfracsin(x^2)x^2dx=sqrtfracpi2$$ So its not that completely garbage. Now my question is, what goes wrong when I use the first substitution? And how do I prove that I am allowed to use this substitution for the sine integral?
integration complex-analysis substitution
edited Jul 24 at 20:06
Chappers
55k74190
asked Jul 24 at 19:51
Zacky
2,1771326
edited Jul 24 at 20:06
Chappers
55k74190
edited Jul 24 at 20:06
Chappers
55k74190
edited Jul 24 at 20:06
Chappers
55k74190
55k74190
asked Jul 24 at 19:51
Zacky
2,1771326
asked Jul 24 at 19:51
Zacky
2,1771326
asked Jul 24 at 19:51
Zacky
2,1771326
2,1771326
1
Your upper integration limit is incorrect after the change of variables. The correct intrgration limit are $u=0$ and $u=e^-ipi/4infty$. Attempt to deform the contour onto the real line fails.
– Mark Viola
Jul 24 at 20:24
add a comment |Â
1
Your upper integration limit is incorrect after the change of variables. The correct intrgration limit are $u=0$ and $u=e^-ipi/4infty$. Attempt to deform the contour onto the real line fails.
– Mark Viola
Jul 24 at 20:24
1
1
Your upper integration limit is incorrect after the change of variables. The correct intrgration limit are $u=0$ and $u=e^-ipi/4infty$. Attempt to deform the contour onto the real line fails.
– Mark Viola
Jul 24 at 20:24
Your upper integration limit is incorrect after the change of variables. The correct intrgration limit are $u=0$ and $u=e^-ipi/4infty$. Attempt to deform the contour onto the real line fails.
– Mark Viola
Jul 24 at 20:24
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
When you make a change of variables in a definite integral, the integration limits change as well.
The question now is why your formulas are correct. The original integral converges for $operatornameRe x > 0.$ When additionally $operatornameRe x < 1$, the integral of $u^2x-1 exp(-i u^2)$ over the arc of a large circle between $arg u = -pi/4$ and $arg u = 0$ is negligible and we don't have to take any singularities into account, therefore the integral over $[0, e^-i pi/4 infty)$ is the same as the integral over $[0, infty)$.
Similar reasoning for the second integral $-$ this time starting from $[0, e^i pi/4 infty)$ $-$ shows that both the sine and the cosine formulas are correct for $0 < operatornameRe x < 1$.
The integral of $u^2x-1 sin(u^2)$ is an analytic function on $-1 < operatornameRe x < 1$, and the reason why the sine formula is correct for $-1 < operatornameRe x < 1$ is the uniqueness of analytic continuation.
The cosine formula will be correct for $x = -1/2$ if you choose the regularization of the divergent integral that coincides with the analytic continuation. For $-2 < operatornameRe x < 0$, that regularization is the integral of $u^2x-1 (cos(u^2) - 1)$.
answered Jul 28 at 17:46
Maxim
2,045113
Thanks. Well I changed the limits but I took $e^-ipi/4infty$ as $infty$. My contour integration knowledge is low as I studied only from wikipedia. Is there a proof for the uniqueness of analytic continuation?
– Zacky
Jul 30 at 8:09
It's easier to see how the upper limit changes if you write the original integral as $lim_R to infty int_0^R$. There is a proof for the identity theorem here.
– Maxim
Jul 30 at 13:10
add a comment |Â
up vote
1
down vote
$smallunderlinetextFor,,Resgt0$ :
$$ Gamma(s)=int_0^inftyfracx^s-1e^x,dx $$
Substitute $,x=+it^2quadtext&quad x=-it^2,$, subtract, add, and simplify to get:
$$
beginalign
fracGamma(s)2,sinleft(fracpi2sright) &=int_0^inftyfracsinleft(x^2right)x^1-2s,dx \[2mm]
fracGamma(s)2,cosleft(fracpi2sright) &=int_0^inftyfraccosleft(x^2right)x^1-2s,dx
endalign
$$
$smallunderlinetextFor,,-1lt Reslt0$ :
$$ Gamma(s)=int_0^inftyx^s-1left(frac1e^xcolorred-1right),dx $$
Substitute $,x=+it^2quadtext&quad x=-it^2,$, subtract, add, and simplify to get:
$$
beginalign
fracGamma(s)2,sinleft(fracpi2sright) &=int_0^inftyfracsinleft(x^2right)x^1-2s,dx \[2mm]
fracGamma(s)2,cosleft(fracpi2sright) &=int_0^inftyfraccosleft(x^2right)colorred-1x^1-2s,dx
endalign
$$
$,colorred-1,$ cancelled each other in sine, and added to each other in cosine.
Nevertheless, do not forget to re-calculate the integration limits whenever you change the integration variable.
$$ smallGamma(s-N)=int_0^infty x^s-1-N,left[,frac1e^x-sum_n=0^N (-1)^n,fracx^nn!,right],dx quadcolon -1lt Reslt0,,,Nin0,,1,,2,,dots, $$
answered Jul 29 at 23:48
Hazem Orabi
2,2582427
Thank you, but how did you see that its valid for $-1lt Reslt0$?
– Zacky
Jul 30 at 8:08
@HazemOrabi Note that your second and third formulas are not valid for all $s$ in the right half-plane, the integrals diverge for $operatornameRe s geq 1$.
– Maxim
Jul 30 at 13:16
add a comment |Â
up vote
0
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What you get when you substitute, say, $t=iu$ into an integral
$$ int_a^b f(t); dt$$
where $a, b in mathbb R$ is an integral over a path in the complex plane
$$ i int_C f(iu); du$$
where $C$ is a path consisting of points mapped into the interval $[a,b]$ by
the mapping $u mapsto iu$.
answered Jul 24 at 20:06
Robert Israel
304k22201441
I didnt learn about mapping. So basically like an integral transform? I looked here en.wikipedia.org/wiki/Map_(mathematics)#Maps_as_functions
– Zacky
Jul 24 at 20:12
Simply the straight line from $a/i = -ai$ to $b/i = -bi$, i.e. all $u$ such that $iu in [a,b]$.
– Robert Israel
Jul 24 at 23:35
Sorry, but I dont see how does this help me.
– Zacky
Jul 26 at 18:09
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
When you make a change of variables in a definite integral, the integration limits change as well.
The question now is why your formulas are correct. The original integral converges for $operatornameRe x > 0.$ When additionally $operatornameRe x < 1$, the integral of $u^2x-1 exp(-i u^2)$ over the arc of a large circle between $arg u = -pi/4$ and $arg u = 0$ is negligible and we don't have to take any singularities into account, therefore the integral over $[0, e^-i pi/4 infty)$ is the same as the integral over $[0, infty)$.
Similar reasoning for the second integral $-$ this time starting from $[0, e^i pi/4 infty)$ $-$ shows that both the sine and the cosine formulas are correct for $0 < operatornameRe x < 1$.
The integral of $u^2x-1 sin(u^2)$ is an analytic function on $-1 < operatornameRe x < 1$, and the reason why the sine formula is correct for $-1 < operatornameRe x < 1$ is the uniqueness of analytic continuation.
The cosine formula will be correct for $x = -1/2$ if you choose the regularization of the divergent integral that coincides with the analytic continuation. For $-2 < operatornameRe x < 0$, that regularization is the integral of $u^2x-1 (cos(u^2) - 1)$.
answered Jul 28 at 17:46
Maxim
2,045113
Thanks. Well I changed the limits but I took $e^-ipi/4infty$ as $infty$. My contour integration knowledge is low as I studied only from wikipedia. Is there a proof for the uniqueness of analytic continuation?
– Zacky
Jul 30 at 8:09
It's easier to see how the upper limit changes if you write the original integral as $lim_R to infty int_0^R$. There is a proof for the identity theorem here.
– Maxim
Jul 30 at 13:10
add a comment |Â
up vote
3
down vote
accepted
When you make a change of variables in a definite integral, the integration limits change as well.
The question now is why your formulas are correct. The original integral converges for $operatornameRe x > 0.$ When additionally $operatornameRe x < 1$, the integral of $u^2x-1 exp(-i u^2)$ over the arc of a large circle between $arg u = -pi/4$ and $arg u = 0$ is negligible and we don't have to take any singularities into account, therefore the integral over $[0, e^-i pi/4 infty)$ is the same as the integral over $[0, infty)$.
Similar reasoning for the second integral $-$ this time starting from $[0, e^i pi/4 infty)$ $-$ shows that both the sine and the cosine formulas are correct for $0 < operatornameRe x < 1$.
The integral of $u^2x-1 sin(u^2)$ is an analytic function on $-1 < operatornameRe x < 1$, and the reason why the sine formula is correct for $-1 < operatornameRe x < 1$ is the uniqueness of analytic continuation.
The cosine formula will be correct for $x = -1/2$ if you choose the regularization of the divergent integral that coincides with the analytic continuation. For $-2 < operatornameRe x < 0$, that regularization is the integral of $u^2x-1 (cos(u^2) - 1)$.
answered Jul 28 at 17:46
Maxim
2,045113
Thanks. Well I changed the limits but I took $e^-ipi/4infty$ as $infty$. My contour integration knowledge is low as I studied only from wikipedia. Is there a proof for the uniqueness of analytic continuation?
– Zacky
Jul 30 at 8:09
It's easier to see how the upper limit changes if you write the original integral as $lim_R to infty int_0^R$. There is a proof for the identity theorem here.
– Maxim
Jul 30 at 13:10
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
When you make a change of variables in a definite integral, the integration limits change as well.
The question now is why your formulas are correct. The original integral converges for $operatornameRe x > 0.$ When additionally $operatornameRe x < 1$, the integral of $u^2x-1 exp(-i u^2)$ over the arc of a large circle between $arg u = -pi/4$ and $arg u = 0$ is negligible and we don't have to take any singularities into account, therefore the integral over $[0, e^-i pi/4 infty)$ is the same as the integral over $[0, infty)$.
Similar reasoning for the second integral $-$ this time starting from $[0, e^i pi/4 infty)$ $-$ shows that both the sine and the cosine formulas are correct for $0 < operatornameRe x < 1$.
The integral of $u^2x-1 sin(u^2)$ is an analytic function on $-1 < operatornameRe x < 1$, and the reason why the sine formula is correct for $-1 < operatornameRe x < 1$ is the uniqueness of analytic continuation.
The cosine formula will be correct for $x = -1/2$ if you choose the regularization of the divergent integral that coincides with the analytic continuation. For $-2 < operatornameRe x < 0$, that regularization is the integral of $u^2x-1 (cos(u^2) - 1)$.
answered Jul 28 at 17:46
Maxim
2,045113
When you make a change of variables in a definite integral, the integration limits change as well.
The question now is why your formulas are correct. The original integral converges for $operatornameRe x > 0.$ When additionally $operatornameRe x < 1$, the integral of $u^2x-1 exp(-i u^2)$ over the arc of a large circle between $arg u = -pi/4$ and $arg u = 0$ is negligible and we don't have to take any singularities into account, therefore the integral over $[0, e^-i pi/4 infty)$ is the same as the integral over $[0, infty)$.
Similar reasoning for the second integral $-$ this time starting from $[0, e^i pi/4 infty)$ $-$ shows that both the sine and the cosine formulas are correct for $0 < operatornameRe x < 1$.
The integral of $u^2x-1 sin(u^2)$ is an analytic function on $-1 < operatornameRe x < 1$, and the reason why the sine formula is correct for $-1 < operatornameRe x < 1$ is the uniqueness of analytic continuation.
The cosine formula will be correct for $x = -1/2$ if you choose the regularization of the divergent integral that coincides with the analytic continuation. For $-2 < operatornameRe x < 0$, that regularization is the integral of $u^2x-1 (cos(u^2) - 1)$.
answered Jul 28 at 17:46
Maxim
2,045113
edited Jul 28 at 18:23
answered Jul 28 at 17:46
Maxim
2,045113
answered Jul 28 at 17:46
Maxim
2,045113
answered Jul 28 at 17:46
Maxim
2,045113
2,045113
Thanks. Well I changed the limits but I took $e^-ipi/4infty$ as $infty$. My contour integration knowledge is low as I studied only from wikipedia. Is there a proof for the uniqueness of analytic continuation?
– Zacky
Jul 30 at 8:09
It's easier to see how the upper limit changes if you write the original integral as $lim_R to infty int_0^R$. There is a proof for the identity theorem here.
– Maxim
Jul 30 at 13:10
add a comment |Â
Thanks. Well I changed the limits but I took $e^-ipi/4infty$ as $infty$. My contour integration knowledge is low as I studied only from wikipedia. Is there a proof for the uniqueness of analytic continuation?
– Zacky
Jul 30 at 8:09
It's easier to see how the upper limit changes if you write the original integral as $lim_R to infty int_0^R$. There is a proof for the identity theorem here.
– Maxim
Jul 30 at 13:10
Thanks. Well I changed the limits but I took $e^-ipi/4infty$ as $infty$. My contour integration knowledge is low as I studied only from wikipedia. Is there a proof for the uniqueness of analytic continuation?
– Zacky
Jul 30 at 8:09
Thanks. Well I changed the limits but I took $e^-ipi/4infty$ as $infty$. My contour integration knowledge is low as I studied only from wikipedia. Is there a proof for the uniqueness of analytic continuation?
– Zacky
Jul 30 at 8:09
It's easier to see how the upper limit changes if you write the original integral as $lim_R to infty int_0^R$. There is a proof for the identity theorem here.
– Maxim
Jul 30 at 13:10
It's easier to see how the upper limit changes if you write the original integral as $lim_R to infty int_0^R$. There is a proof for the identity theorem here.
– Maxim
Jul 30 at 13:10
add a comment |Â
up vote
1
down vote
$smallunderlinetextFor,,Resgt0$ :
$$ Gamma(s)=int_0^inftyfracx^s-1e^x,dx $$
Substitute $,x=+it^2quadtext&quad x=-it^2,$, subtract, add, and simplify to get:
$$
beginalign
fracGamma(s)2,sinleft(fracpi2sright) &=int_0^inftyfracsinleft(x^2right)x^1-2s,dx \[2mm]
fracGamma(s)2,cosleft(fracpi2sright) &=int_0^inftyfraccosleft(x^2right)x^1-2s,dx
endalign
$$
$smallunderlinetextFor,,-1lt Reslt0$ :
$$ Gamma(s)=int_0^inftyx^s-1left(frac1e^xcolorred-1right),dx $$
Substitute $,x=+it^2quadtext&quad x=-it^2,$, subtract, add, and simplify to get:
$$
beginalign
fracGamma(s)2,sinleft(fracpi2sright) &=int_0^inftyfracsinleft(x^2right)x^1-2s,dx \[2mm]
fracGamma(s)2,cosleft(fracpi2sright) &=int_0^inftyfraccosleft(x^2right)colorred-1x^1-2s,dx
endalign
$$
$,colorred-1,$ cancelled each other in sine, and added to each other in cosine.
Nevertheless, do not forget to re-calculate the integration limits whenever you change the integration variable.
$$ smallGamma(s-N)=int_0^infty x^s-1-N,left[,frac1e^x-sum_n=0^N (-1)^n,fracx^nn!,right],dx quadcolon -1lt Reslt0,,,Nin0,,1,,2,,dots, $$
answered Jul 29 at 23:48
Hazem Orabi
2,2582427
Thank you, but how did you see that its valid for $-1lt Reslt0$?
– Zacky
Jul 30 at 8:08
@HazemOrabi Note that your second and third formulas are not valid for all $s$ in the right half-plane, the integrals diverge for $operatornameRe s geq 1$.
– Maxim
Jul 30 at 13:16
add a comment |Â
up vote
1
down vote
$smallunderlinetextFor,,Resgt0$ :
$$ Gamma(s)=int_0^inftyfracx^s-1e^x,dx $$
Substitute $,x=+it^2quadtext&quad x=-it^2,$, subtract, add, and simplify to get:
$$
beginalign
fracGamma(s)2,sinleft(fracpi2sright) &=int_0^inftyfracsinleft(x^2right)x^1-2s,dx \[2mm]
fracGamma(s)2,cosleft(fracpi2sright) &=int_0^inftyfraccosleft(x^2right)x^1-2s,dx
endalign
$$
$smallunderlinetextFor,,-1lt Reslt0$ :
$$ Gamma(s)=int_0^inftyx^s-1left(frac1e^xcolorred-1right),dx $$
Substitute $,x=+it^2quadtext&quad x=-it^2,$, subtract, add, and simplify to get:
$$
beginalign
fracGamma(s)2,sinleft(fracpi2sright) &=int_0^inftyfracsinleft(x^2right)x^1-2s,dx \[2mm]
fracGamma(s)2,cosleft(fracpi2sright) &=int_0^inftyfraccosleft(x^2right)colorred-1x^1-2s,dx
endalign
$$
$,colorred-1,$ cancelled each other in sine, and added to each other in cosine.
Nevertheless, do not forget to re-calculate the integration limits whenever you change the integration variable.
$$ smallGamma(s-N)=int_0^infty x^s-1-N,left[,frac1e^x-sum_n=0^N (-1)^n,fracx^nn!,right],dx quadcolon -1lt Reslt0,,,Nin0,,1,,2,,dots, $$
answered Jul 29 at 23:48
Hazem Orabi
2,2582427
Thank you, but how did you see that its valid for $-1lt Reslt0$?
– Zacky
Jul 30 at 8:08
@HazemOrabi Note that your second and third formulas are not valid for all $s$ in the right half-plane, the integrals diverge for $operatornameRe s geq 1$.
– Maxim
Jul 30 at 13:16
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$smallunderlinetextFor,,Resgt0$ :
$$ Gamma(s)=int_0^inftyfracx^s-1e^x,dx $$
Substitute $,x=+it^2quadtext&quad x=-it^2,$, subtract, add, and simplify to get:
$$
beginalign
fracGamma(s)2,sinleft(fracpi2sright) &=int_0^inftyfracsinleft(x^2right)x^1-2s,dx \[2mm]
fracGamma(s)2,cosleft(fracpi2sright) &=int_0^inftyfraccosleft(x^2right)x^1-2s,dx
endalign
$$
$smallunderlinetextFor,,-1lt Reslt0$ :
$$ Gamma(s)=int_0^inftyx^s-1left(frac1e^xcolorred-1right),dx $$
Substitute $,x=+it^2quadtext&quad x=-it^2,$, subtract, add, and simplify to get:
$$
beginalign
fracGamma(s)2,sinleft(fracpi2sright) &=int_0^inftyfracsinleft(x^2right)x^1-2s,dx \[2mm]
fracGamma(s)2,cosleft(fracpi2sright) &=int_0^inftyfraccosleft(x^2right)colorred-1x^1-2s,dx
endalign
$$
$,colorred-1,$ cancelled each other in sine, and added to each other in cosine.
Nevertheless, do not forget to re-calculate the integration limits whenever you change the integration variable.
$$ smallGamma(s-N)=int_0^infty x^s-1-N,left[,frac1e^x-sum_n=0^N (-1)^n,fracx^nn!,right],dx quadcolon -1lt Reslt0,,,Nin0,,1,,2,,dots, $$
answered Jul 29 at 23:48
Hazem Orabi
2,2582427
$smallunderlinetextFor,,Resgt0$ :
$$ Gamma(s)=int_0^inftyfracx^s-1e^x,dx $$
Substitute $,x=+it^2quadtext&quad x=-it^2,$, subtract, add, and simplify to get:
$$
beginalign
fracGamma(s)2,sinleft(fracpi2sright) &=int_0^inftyfracsinleft(x^2right)x^1-2s,dx \[2mm]
fracGamma(s)2,cosleft(fracpi2sright) &=int_0^inftyfraccosleft(x^2right)x^1-2s,dx
endalign
$$
$smallunderlinetextFor,,-1lt Reslt0$ :
$$ Gamma(s)=int_0^inftyx^s-1left(frac1e^xcolorred-1right),dx $$
Substitute $,x=+it^2quadtext&quad x=-it^2,$, subtract, add, and simplify to get:
$$
beginalign
fracGamma(s)2,sinleft(fracpi2sright) &=int_0^inftyfracsinleft(x^2right)x^1-2s,dx \[2mm]
fracGamma(s)2,cosleft(fracpi2sright) &=int_0^inftyfraccosleft(x^2right)colorred-1x^1-2s,dx
endalign
$$
$,colorred-1,$ cancelled each other in sine, and added to each other in cosine.
Nevertheless, do not forget to re-calculate the integration limits whenever you change the integration variable.
$$ smallGamma(s-N)=int_0^infty x^s-1-N,left[,frac1e^x-sum_n=0^N (-1)^n,fracx^nn!,right],dx quadcolon -1lt Reslt0,,,Nin0,,1,,2,,dots, $$
answered Jul 29 at 23:48
Hazem Orabi
2,2582427
edited Jul 30 at 7:57
answered Jul 29 at 23:48
Hazem Orabi
2,2582427
answered Jul 29 at 23:48
Hazem Orabi
2,2582427
answered Jul 29 at 23:48
Hazem Orabi
2,2582427
2,2582427
Thank you, but how did you see that its valid for $-1lt Reslt0$?
– Zacky
Jul 30 at 8:08
@HazemOrabi Note that your second and third formulas are not valid for all $s$ in the right half-plane, the integrals diverge for $operatornameRe s geq 1$.
– Maxim
Jul 30 at 13:16
add a comment |Â
Thank you, but how did you see that its valid for $-1lt Reslt0$?
– Zacky
Jul 30 at 8:08
@HazemOrabi Note that your second and third formulas are not valid for all $s$ in the right half-plane, the integrals diverge for $operatornameRe s geq 1$.
– Maxim
Jul 30 at 13:16
Thank you, but how did you see that its valid for $-1lt Reslt0$?
– Zacky
Jul 30 at 8:08
Thank you, but how did you see that its valid for $-1lt Reslt0$?
– Zacky
Jul 30 at 8:08
@HazemOrabi Note that your second and third formulas are not valid for all $s$ in the right half-plane, the integrals diverge for $operatornameRe s geq 1$.
– Maxim
Jul 30 at 13:16
@HazemOrabi Note that your second and third formulas are not valid for all $s$ in the right half-plane, the integrals diverge for $operatornameRe s geq 1$.
– Maxim
Jul 30 at 13:16
add a comment |Â
up vote
0
down vote
What you get when you substitute, say, $t=iu$ into an integral
$$ int_a^b f(t); dt$$
where $a, b in mathbb R$ is an integral over a path in the complex plane
$$ i int_C f(iu); du$$
where $C$ is a path consisting of points mapped into the interval $[a,b]$ by
the mapping $u mapsto iu$.
answered Jul 24 at 20:06
Robert Israel
304k22201441
I didnt learn about mapping. So basically like an integral transform? I looked here en.wikipedia.org/wiki/Map_(mathematics)#Maps_as_functions
– Zacky
Jul 24 at 20:12
Simply the straight line from $a/i = -ai$ to $b/i = -bi$, i.e. all $u$ such that $iu in [a,b]$.
– Robert Israel
Jul 24 at 23:35
Sorry, but I dont see how does this help me.
– Zacky
Jul 26 at 18:09
add a comment |Â
up vote
0
down vote
What you get when you substitute, say, $t=iu$ into an integral
$$ int_a^b f(t); dt$$
where $a, b in mathbb R$ is an integral over a path in the complex plane
$$ i int_C f(iu); du$$
where $C$ is a path consisting of points mapped into the interval $[a,b]$ by
the mapping $u mapsto iu$.
answered Jul 24 at 20:06
Robert Israel
304k22201441
I didnt learn about mapping. So basically like an integral transform? I looked here en.wikipedia.org/wiki/Map_(mathematics)#Maps_as_functions
– Zacky
Jul 24 at 20:12
Simply the straight line from $a/i = -ai$ to $b/i = -bi$, i.e. all $u$ such that $iu in [a,b]$.
– Robert Israel
Jul 24 at 23:35
Sorry, but I dont see how does this help me.
– Zacky
Jul 26 at 18:09
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What you get when you substitute, say, $t=iu$ into an integral
$$ int_a^b f(t); dt$$
where $a, b in mathbb R$ is an integral over a path in the complex plane
$$ i int_C f(iu); du$$
where $C$ is a path consisting of points mapped into the interval $[a,b]$ by
the mapping $u mapsto iu$.
answered Jul 24 at 20:06
Robert Israel
304k22201441
What you get when you substitute, say, $t=iu$ into an integral
$$ int_a^b f(t); dt$$
where $a, b in mathbb R$ is an integral over a path in the complex plane
$$ i int_C f(iu); du$$
where $C$ is a path consisting of points mapped into the interval $[a,b]$ by
the mapping $u mapsto iu$.
answered Jul 24 at 20:06
Robert Israel
304k22201441
answered Jul 24 at 20:06
Robert Israel
304k22201441
answered Jul 24 at 20:06
Robert Israel
304k22201441
answered Jul 24 at 20:06
Robert Israel
304k22201441
304k22201441
I didnt learn about mapping. So basically like an integral transform? I looked here en.wikipedia.org/wiki/Map_(mathematics)#Maps_as_functions
– Zacky
Jul 24 at 20:12
Simply the straight line from $a/i = -ai$ to $b/i = -bi$, i.e. all $u$ such that $iu in [a,b]$.
– Robert Israel
Jul 24 at 23:35
Sorry, but I dont see how does this help me.
– Zacky
Jul 26 at 18:09
add a comment |Â
I didnt learn about mapping. So basically like an integral transform? I looked here en.wikipedia.org/wiki/Map_(mathematics)#Maps_as_functions
– Zacky
Jul 24 at 20:12
Simply the straight line from $a/i = -ai$ to $b/i = -bi$, i.e. all $u$ such that $iu in [a,b]$.
– Robert Israel
Jul 24 at 23:35
Sorry, but I dont see how does this help me.
– Zacky
Jul 26 at 18:09
I didnt learn about mapping. So basically like an integral transform? I looked here en.wikipedia.org/wiki/Map_(mathematics)#Maps_as_functions
– Zacky
Jul 24 at 20:12
I didnt learn about mapping. So basically like an integral transform? I looked here en.wikipedia.org/wiki/Map_(mathematics)#Maps_as_functions
– Zacky
Jul 24 at 20:12
Simply the straight line from $a/i = -ai$ to $b/i = -bi$, i.e. all $u$ such that $iu in [a,b]$.
– Robert Israel
Jul 24 at 23:35
Simply the straight line from $a/i = -ai$ to $b/i = -bi$, i.e. all $u$ such that $iu in [a,b]$.
– Robert Israel
Jul 24 at 23:35
Sorry, but I dont see how does this help me.
– Zacky
Jul 26 at 18:09
Sorry, but I dont see how does this help me.
– Zacky
Jul 26 at 18:09
add a comment |Â
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1
Your upper integration limit is incorrect after the change of variables. The correct intrgration limit are $u=0$ and $u=e^-ipi/4infty$. Attempt to deform the contour onto the real line fails.
– Mark Viola
Jul 24 at 20:24