Mixing
Show that $int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx=-frac1{alpha^s}frac{pi}{sin(pi alpha)}$
$begingroup$
Recently I have come across the following integral while going over this list (Problem $35$)
$$int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx=-frac1{alpha^s}frac{pi}{sin(pi alpha)}~~~~s>0, alphain(0,1)$$
where $operatorname{Li}_s(x)$ denotes the Polylogarithm Function.
I tried to use the series expansion of $operatorname{Li}_s(-x)$ which yields to
$$begin{align}
int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx&=int_0^{infty}frac1{x^{alpha+1}}left[sum_{n=1}^{infty}frac{(-x)^n}{n^s}right]dx\
&=sum_{n=1}^{infty}frac{(-1)^n}{n^s}int_0^{infty}frac{x^n}{x^{alpha+1}}dx\
&=sum_{n=1}^{infty}frac{(-1)^n}{n^s}int_0^{infty}x^{n-alpha-1}dx
end{align}$$
But one can easily see the problems concerning the convergence of the last integral. Furthermore I am not sure whether it is possible to interchange the order of summation and integration in this case or not.
Another approach would be to use the an integral represantation of $operatorname{Li}_s(-x)$ so that the given integral becomes
$$begin{align}
int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx&=int_0^{infty}frac1{x^{alpha+1}}left[frac1{Gamma(s)}int_0^{infty}frac{t^{s-1}}{e^t/(-x)-1}dtright]dx\
&=frac{-1}{Gamma(s)}int_0^{infty}int_0^{infty}frac{t^{s-1}}{x^{alpha}(e^t+x)}dxdt\
end{align}$$
but from hereon I have no clue how to proceed. Since the solution reminds me of Euler's Reflection Formula it is maybe possible somehow to reshaphe the integral in terms of the Gamma Function.
Therefore I am asking for a whole evaluation of the given integral. I did not found anything closely connected to this question but you can correct me if I have overseen something.
Thanks in advance!
integration definite-integrals closed-form polylogarithm
edited Sep 29 '18 at 15:17
pisco
11.6k21742
asked Sep 29 '18 at 14:37
mrtaurhomrtaurho
4,25421234
$endgroup$
add a comment |
$begingroup$
Recently I have come across the following integral while going over this list (Problem $35$)
$$int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx=-frac1{alpha^s}frac{pi}{sin(pi alpha)}~~~~s>0, alphain(0,1)$$
where $operatorname{Li}_s(x)$ denotes the Polylogarithm Function.
I tried to use the series expansion of $operatorname{Li}_s(-x)$ which yields to
$$begin{align}
int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx&=int_0^{infty}frac1{x^{alpha+1}}left[sum_{n=1}^{infty}frac{(-x)^n}{n^s}right]dx\
&=sum_{n=1}^{infty}frac{(-1)^n}{n^s}int_0^{infty}frac{x^n}{x^{alpha+1}}dx\
&=sum_{n=1}^{infty}frac{(-1)^n}{n^s}int_0^{infty}x^{n-alpha-1}dx
end{align}$$
But one can easily see the problems concerning the convergence of the last integral. Furthermore I am not sure whether it is possible to interchange the order of summation and integration in this case or not.
Another approach would be to use the an integral represantation of $operatorname{Li}_s(-x)$ so that the given integral becomes
$$begin{align}
int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx&=int_0^{infty}frac1{x^{alpha+1}}left[frac1{Gamma(s)}int_0^{infty}frac{t^{s-1}}{e^t/(-x)-1}dtright]dx\
&=frac{-1}{Gamma(s)}int_0^{infty}int_0^{infty}frac{t^{s-1}}{x^{alpha}(e^t+x)}dxdt\
end{align}$$
but from hereon I have no clue how to proceed. Since the solution reminds me of Euler's Reflection Formula it is maybe possible somehow to reshaphe the integral in terms of the Gamma Function.
Therefore I am asking for a whole evaluation of the given integral. I did not found anything closely connected to this question but you can correct me if I have overseen something.
Thanks in advance!
integration definite-integrals closed-form polylogarithm
edited Sep 29 '18 at 15:17
pisco
11.6k21742
asked Sep 29 '18 at 14:37
mrtaurhomrtaurho
4,25421234
$endgroup$
1
$begingroup$
Looks like an application of Ramanujan's Master Theorem, coupled with the reflection formula for the Gamma function.
$endgroup$
– nospoon
Sep 29 '18 at 14:52
$begingroup$
@nospoon Is it possible to use this Theorem here? I also thought about a potential connection to the Mellin Transform but hence the $x$-term has the form $x^{-alpha-1}$ instead of $x^{alpha-1}$ I do not know whether this possible or not.
$endgroup$
– mrtaurho
Sep 29 '18 at 14:58
add a comment |
$begingroup$
Recently I have come across the following integral while going over this list (Problem $35$)
$$int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx=-frac1{alpha^s}frac{pi}{sin(pi alpha)}~~~~s>0, alphain(0,1)$$
where $operatorname{Li}_s(x)$ denotes the Polylogarithm Function.
I tried to use the series expansion of $operatorname{Li}_s(-x)$ which yields to
$$begin{align}
int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx&=int_0^{infty}frac1{x^{alpha+1}}left[sum_{n=1}^{infty}frac{(-x)^n}{n^s}right]dx\
&=sum_{n=1}^{infty}frac{(-1)^n}{n^s}int_0^{infty}frac{x^n}{x^{alpha+1}}dx\
&=sum_{n=1}^{infty}frac{(-1)^n}{n^s}int_0^{infty}x^{n-alpha-1}dx
end{align}$$
But one can easily see the problems concerning the convergence of the last integral. Furthermore I am not sure whether it is possible to interchange the order of summation and integration in this case or not.
Another approach would be to use the an integral represantation of $operatorname{Li}_s(-x)$ so that the given integral becomes
$$begin{align}
int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx&=int_0^{infty}frac1{x^{alpha+1}}left[frac1{Gamma(s)}int_0^{infty}frac{t^{s-1}}{e^t/(-x)-1}dtright]dx\
&=frac{-1}{Gamma(s)}int_0^{infty}int_0^{infty}frac{t^{s-1}}{x^{alpha}(e^t+x)}dxdt\
end{align}$$
but from hereon I have no clue how to proceed. Since the solution reminds me of Euler's Reflection Formula it is maybe possible somehow to reshaphe the integral in terms of the Gamma Function.
Therefore I am asking for a whole evaluation of the given integral. I did not found anything closely connected to this question but you can correct me if I have overseen something.
Thanks in advance!
integration definite-integrals closed-form polylogarithm
edited Sep 29 '18 at 15:17
pisco
11.6k21742
asked Sep 29 '18 at 14:37
mrtaurhomrtaurho
4,25421234
$endgroup$
Recently I have come across the following integral while going over this list (Problem $35$)
$$int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx=-frac1{alpha^s}frac{pi}{sin(pi alpha)}~~~~s>0, alphain(0,1)$$
where $operatorname{Li}_s(x)$ denotes the Polylogarithm Function.
I tried to use the series expansion of $operatorname{Li}_s(-x)$ which yields to
$$begin{align}
int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx&=int_0^{infty}frac1{x^{alpha+1}}left[sum_{n=1}^{infty}frac{(-x)^n}{n^s}right]dx\
&=sum_{n=1}^{infty}frac{(-1)^n}{n^s}int_0^{infty}frac{x^n}{x^{alpha+1}}dx\
&=sum_{n=1}^{infty}frac{(-1)^n}{n^s}int_0^{infty}x^{n-alpha-1}dx
end{align}$$
But one can easily see the problems concerning the convergence of the last integral. Furthermore I am not sure whether it is possible to interchange the order of summation and integration in this case or not.
Another approach would be to use the an integral represantation of $operatorname{Li}_s(-x)$ so that the given integral becomes
$$begin{align}
int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx&=int_0^{infty}frac1{x^{alpha+1}}left[frac1{Gamma(s)}int_0^{infty}frac{t^{s-1}}{e^t/(-x)-1}dtright]dx\
&=frac{-1}{Gamma(s)}int_0^{infty}int_0^{infty}frac{t^{s-1}}{x^{alpha}(e^t+x)}dxdt\
end{align}$$
but from hereon I have no clue how to proceed. Since the solution reminds me of Euler's Reflection Formula it is maybe possible somehow to reshaphe the integral in terms of the Gamma Function.
Therefore I am asking for a whole evaluation of the given integral. I did not found anything closely connected to this question but you can correct me if I have overseen something.
Thanks in advance!
integration definite-integrals closed-form polylogarithm
integration definite-integrals closed-form polylogarithm
edited Sep 29 '18 at 15:17
pisco
11.6k21742
asked Sep 29 '18 at 14:37
mrtaurhomrtaurho
4,25421234
edited Sep 29 '18 at 15:17
pisco
11.6k21742
asked Sep 29 '18 at 14:37
mrtaurhomrtaurho
4,25421234
edited Sep 29 '18 at 15:17
pisco
11.6k21742
edited Sep 29 '18 at 15:17
pisco
11.6k21742
edited Sep 29 '18 at 15:17
pisco
11.6k21742
11.6k21742
asked Sep 29 '18 at 14:37
mrtaurhomrtaurho
4,25421234
asked Sep 29 '18 at 14:37
mrtaurhomrtaurho
4,25421234
asked Sep 29 '18 at 14:37
mrtaurhomrtaurho
4,25421234
4,25421234
1
$begingroup$
Looks like an application of Ramanujan's Master Theorem, coupled with the reflection formula for the Gamma function.
$endgroup$
– nospoon
Sep 29 '18 at 14:52
$begingroup$
@nospoon Is it possible to use this Theorem here? I also thought about a potential connection to the Mellin Transform but hence the $x$-term has the form $x^{-alpha-1}$ instead of $x^{alpha-1}$ I do not know whether this possible or not.
$endgroup$
– mrtaurho
Sep 29 '18 at 14:58
add a comment |
1
$begingroup$
Looks like an application of Ramanujan's Master Theorem, coupled with the reflection formula for the Gamma function.
$endgroup$
– nospoon
Sep 29 '18 at 14:52
$begingroup$
@nospoon Is it possible to use this Theorem here? I also thought about a potential connection to the Mellin Transform but hence the $x$-term has the form $x^{-alpha-1}$ instead of $x^{alpha-1}$ I do not know whether this possible or not.
$endgroup$
– mrtaurho
Sep 29 '18 at 14:58
1
1
$begingroup$
Looks like an application of Ramanujan's Master Theorem, coupled with the reflection formula for the Gamma function.
$endgroup$
– nospoon
Sep 29 '18 at 14:52
$begingroup$
Looks like an application of Ramanujan's Master Theorem, coupled with the reflection formula for the Gamma function.
$endgroup$
– nospoon
Sep 29 '18 at 14:52
$begingroup$
@nospoon Is it possible to use this Theorem here? I also thought about a potential connection to the Mellin Transform but hence the $x$-term has the form $x^{-alpha-1}$ instead of $x^{alpha-1}$ I do not know whether this possible or not.
$endgroup$
– mrtaurho
Sep 29 '18 at 14:58
$begingroup$
@nospoon Is it possible to use this Theorem here? I also thought about a potential connection to the Mellin Transform but hence the $x$-term has the form $x^{-alpha-1}$ instead of $x^{alpha-1}$ I do not know whether this possible or not.
$endgroup$
– mrtaurho
Sep 29 '18 at 14:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You're just one step away: for $b>0$, we have $$int_0^infty {frac{1}{{{x^alpha }(b + x)}}dx} = {b^{ - alpha }}int_0^infty {frac{{{x^{ - alpha }}}}{{1 + x}}dx} = frac{{{b^{ - alpha }}pi }}{{sin alpha pi }}$$
so
$$int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx=-frac{1}{Gamma(s)}int_0^{infty}int_0^{infty}frac{t^{s-1}}{x^{alpha}(e^t+x)}dxdt = - frac{pi }{{Gamma (s)sin alpha pi }}int_0^infty {{t^{s - 1}}{e^{ - alpha t}}dt} $$
The series expansion of $text{Li}_s(-x)$ only converges for $|x| < 1$, so your first method is illegitimate.
answered Sep 29 '18 at 15:06
piscopisco
11.6k21742
$endgroup$
$begingroup$
First of all thank you for your answer. I guessed for myself that I went wrong somewhere concerning the convergence of the series expansion. Furhermore I was not aware of this integral identity at the moment. I think it can be shown using the Beta Function hence I remember some kind of identity similiar to this? But nevertheless I am glad you provided an asnwer :)
$endgroup$
– mrtaurho
Sep 29 '18 at 15:17
add a comment |
$begingroup$
I have finally figured out how to use Ramanujans Master Theorem in this case.
Ramanujans Master Theorem
Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form
$$f(x)=sum_{k=0}^{infty}frac{phi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by
$$int_0^{infty}x^{s-1}f(x)dx=Gamma(s)phi(-s)$$
Lets get back to the given integral. The series representation of $operatorname{Li}_s(-x)$ is given by $displaystylesum_{k=1}^{infty}frac{(-x)^k}{k^s}$ and therefore, by using the Gamma Function as extension of the factorial, we obtain $displaystylephi(k)=frac{Gamma(k+1)}{k^s}$. Now applying the Mellin Transform with $s=-alpha$ yields to
$$int_0^{infty}x^{-alpha-1}operatorname{Li}_s(-x)dx=Gamma(-alpha)phi(alpha)=Gamma(-alpha)frac{Gamma(alpha+1)}{alpha^s}$$
This term can be simplified by using Eulers Reflection Formula with $z=alpha+1$ which finally gives
$$frac1{alpha^s}Gamma(alpha+1)Gamma(-alpha)=frac1{alpha^s}frac{pi}{sin(pi(alpha+1))}=-frac1{alpha^s}frac{pi}{sin(pialpha)}$$
answered Sep 30 '18 at 20:23
mrtaurhomrtaurho
4,25421234
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
2
active
oldest
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votes
$begingroup$
You're just one step away: for $b>0$, we have $$int_0^infty {frac{1}{{{x^alpha }(b + x)}}dx} = {b^{ - alpha }}int_0^infty {frac{{{x^{ - alpha }}}}{{1 + x}}dx} = frac{{{b^{ - alpha }}pi }}{{sin alpha pi }}$$
so
$$int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx=-frac{1}{Gamma(s)}int_0^{infty}int_0^{infty}frac{t^{s-1}}{x^{alpha}(e^t+x)}dxdt = - frac{pi }{{Gamma (s)sin alpha pi }}int_0^infty {{t^{s - 1}}{e^{ - alpha t}}dt} $$
The series expansion of $text{Li}_s(-x)$ only converges for $|x| < 1$, so your first method is illegitimate.
answered Sep 29 '18 at 15:06
piscopisco
11.6k21742
$endgroup$
$begingroup$
First of all thank you for your answer. I guessed for myself that I went wrong somewhere concerning the convergence of the series expansion. Furhermore I was not aware of this integral identity at the moment. I think it can be shown using the Beta Function hence I remember some kind of identity similiar to this? But nevertheless I am glad you provided an asnwer :)
$endgroup$
– mrtaurho
Sep 29 '18 at 15:17
add a comment |
$begingroup$
You're just one step away: for $b>0$, we have $$int_0^infty {frac{1}{{{x^alpha }(b + x)}}dx} = {b^{ - alpha }}int_0^infty {frac{{{x^{ - alpha }}}}{{1 + x}}dx} = frac{{{b^{ - alpha }}pi }}{{sin alpha pi }}$$
so
$$int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx=-frac{1}{Gamma(s)}int_0^{infty}int_0^{infty}frac{t^{s-1}}{x^{alpha}(e^t+x)}dxdt = - frac{pi }{{Gamma (s)sin alpha pi }}int_0^infty {{t^{s - 1}}{e^{ - alpha t}}dt} $$
The series expansion of $text{Li}_s(-x)$ only converges for $|x| < 1$, so your first method is illegitimate.
answered Sep 29 '18 at 15:06
piscopisco
11.6k21742
$endgroup$
$begingroup$
First of all thank you for your answer. I guessed for myself that I went wrong somewhere concerning the convergence of the series expansion. Furhermore I was not aware of this integral identity at the moment. I think it can be shown using the Beta Function hence I remember some kind of identity similiar to this? But nevertheless I am glad you provided an asnwer :)
$endgroup$
– mrtaurho
Sep 29 '18 at 15:17
add a comment |
$begingroup$
You're just one step away: for $b>0$, we have $$int_0^infty {frac{1}{{{x^alpha }(b + x)}}dx} = {b^{ - alpha }}int_0^infty {frac{{{x^{ - alpha }}}}{{1 + x}}dx} = frac{{{b^{ - alpha }}pi }}{{sin alpha pi }}$$
so
$$int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx=-frac{1}{Gamma(s)}int_0^{infty}int_0^{infty}frac{t^{s-1}}{x^{alpha}(e^t+x)}dxdt = - frac{pi }{{Gamma (s)sin alpha pi }}int_0^infty {{t^{s - 1}}{e^{ - alpha t}}dt} $$
The series expansion of $text{Li}_s(-x)$ only converges for $|x| < 1$, so your first method is illegitimate.
answered Sep 29 '18 at 15:06
piscopisco
11.6k21742
$endgroup$
You're just one step away: for $b>0$, we have $$int_0^infty {frac{1}{{{x^alpha }(b + x)}}dx} = {b^{ - alpha }}int_0^infty {frac{{{x^{ - alpha }}}}{{1 + x}}dx} = frac{{{b^{ - alpha }}pi }}{{sin alpha pi }}$$
so
$$int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}dx=-frac{1}{Gamma(s)}int_0^{infty}int_0^{infty}frac{t^{s-1}}{x^{alpha}(e^t+x)}dxdt = - frac{pi }{{Gamma (s)sin alpha pi }}int_0^infty {{t^{s - 1}}{e^{ - alpha t}}dt} $$
The series expansion of $text{Li}_s(-x)$ only converges for $|x| < 1$, so your first method is illegitimate.
answered Sep 29 '18 at 15:06
piscopisco
11.6k21742
answered Sep 29 '18 at 15:06
piscopisco
11.6k21742
answered Sep 29 '18 at 15:06
piscopisco
11.6k21742
answered Sep 29 '18 at 15:06
piscopisco
11.6k21742
11.6k21742
$begingroup$
First of all thank you for your answer. I guessed for myself that I went wrong somewhere concerning the convergence of the series expansion. Furhermore I was not aware of this integral identity at the moment. I think it can be shown using the Beta Function hence I remember some kind of identity similiar to this? But nevertheless I am glad you provided an asnwer :)
$endgroup$
– mrtaurho
Sep 29 '18 at 15:17
add a comment |
$begingroup$
First of all thank you for your answer. I guessed for myself that I went wrong somewhere concerning the convergence of the series expansion. Furhermore I was not aware of this integral identity at the moment. I think it can be shown using the Beta Function hence I remember some kind of identity similiar to this? But nevertheless I am glad you provided an asnwer :)
$endgroup$
– mrtaurho
Sep 29 '18 at 15:17
$begingroup$
First of all thank you for your answer. I guessed for myself that I went wrong somewhere concerning the convergence of the series expansion. Furhermore I was not aware of this integral identity at the moment. I think it can be shown using the Beta Function hence I remember some kind of identity similiar to this? But nevertheless I am glad you provided an asnwer :)
$endgroup$
– mrtaurho
Sep 29 '18 at 15:17
$begingroup$
First of all thank you for your answer. I guessed for myself that I went wrong somewhere concerning the convergence of the series expansion. Furhermore I was not aware of this integral identity at the moment. I think it can be shown using the Beta Function hence I remember some kind of identity similiar to this? But nevertheless I am glad you provided an asnwer :)
$endgroup$
– mrtaurho
Sep 29 '18 at 15:17
add a comment |
$begingroup$
I have finally figured out how to use Ramanujans Master Theorem in this case.
Ramanujans Master Theorem
Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form
$$f(x)=sum_{k=0}^{infty}frac{phi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by
$$int_0^{infty}x^{s-1}f(x)dx=Gamma(s)phi(-s)$$
Lets get back to the given integral. The series representation of $operatorname{Li}_s(-x)$ is given by $displaystylesum_{k=1}^{infty}frac{(-x)^k}{k^s}$ and therefore, by using the Gamma Function as extension of the factorial, we obtain $displaystylephi(k)=frac{Gamma(k+1)}{k^s}$. Now applying the Mellin Transform with $s=-alpha$ yields to
$$int_0^{infty}x^{-alpha-1}operatorname{Li}_s(-x)dx=Gamma(-alpha)phi(alpha)=Gamma(-alpha)frac{Gamma(alpha+1)}{alpha^s}$$
This term can be simplified by using Eulers Reflection Formula with $z=alpha+1$ which finally gives
$$frac1{alpha^s}Gamma(alpha+1)Gamma(-alpha)=frac1{alpha^s}frac{pi}{sin(pi(alpha+1))}=-frac1{alpha^s}frac{pi}{sin(pialpha)}$$
answered Sep 30 '18 at 20:23
mrtaurhomrtaurho
4,25421234
$endgroup$
add a comment |
$begingroup$
I have finally figured out how to use Ramanujans Master Theorem in this case.
Ramanujans Master Theorem
Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form
$$f(x)=sum_{k=0}^{infty}frac{phi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by
$$int_0^{infty}x^{s-1}f(x)dx=Gamma(s)phi(-s)$$
Lets get back to the given integral. The series representation of $operatorname{Li}_s(-x)$ is given by $displaystylesum_{k=1}^{infty}frac{(-x)^k}{k^s}$ and therefore, by using the Gamma Function as extension of the factorial, we obtain $displaystylephi(k)=frac{Gamma(k+1)}{k^s}$. Now applying the Mellin Transform with $s=-alpha$ yields to
$$int_0^{infty}x^{-alpha-1}operatorname{Li}_s(-x)dx=Gamma(-alpha)phi(alpha)=Gamma(-alpha)frac{Gamma(alpha+1)}{alpha^s}$$
This term can be simplified by using Eulers Reflection Formula with $z=alpha+1$ which finally gives
$$frac1{alpha^s}Gamma(alpha+1)Gamma(-alpha)=frac1{alpha^s}frac{pi}{sin(pi(alpha+1))}=-frac1{alpha^s}frac{pi}{sin(pialpha)}$$
answered Sep 30 '18 at 20:23
mrtaurhomrtaurho
4,25421234
$endgroup$
add a comment |
$begingroup$
I have finally figured out how to use Ramanujans Master Theorem in this case.
Ramanujans Master Theorem
Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form
$$f(x)=sum_{k=0}^{infty}frac{phi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by
$$int_0^{infty}x^{s-1}f(x)dx=Gamma(s)phi(-s)$$
Lets get back to the given integral. The series representation of $operatorname{Li}_s(-x)$ is given by $displaystylesum_{k=1}^{infty}frac{(-x)^k}{k^s}$ and therefore, by using the Gamma Function as extension of the factorial, we obtain $displaystylephi(k)=frac{Gamma(k+1)}{k^s}$. Now applying the Mellin Transform with $s=-alpha$ yields to
$$int_0^{infty}x^{-alpha-1}operatorname{Li}_s(-x)dx=Gamma(-alpha)phi(alpha)=Gamma(-alpha)frac{Gamma(alpha+1)}{alpha^s}$$
This term can be simplified by using Eulers Reflection Formula with $z=alpha+1$ which finally gives
$$frac1{alpha^s}Gamma(alpha+1)Gamma(-alpha)=frac1{alpha^s}frac{pi}{sin(pi(alpha+1))}=-frac1{alpha^s}frac{pi}{sin(pialpha)}$$
answered Sep 30 '18 at 20:23
mrtaurhomrtaurho
4,25421234
$endgroup$
I have finally figured out how to use Ramanujans Master Theorem in this case.
Ramanujans Master Theorem
Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form
$$f(x)=sum_{k=0}^{infty}frac{phi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by
$$int_0^{infty}x^{s-1}f(x)dx=Gamma(s)phi(-s)$$
Lets get back to the given integral. The series representation of $operatorname{Li}_s(-x)$ is given by $displaystylesum_{k=1}^{infty}frac{(-x)^k}{k^s}$ and therefore, by using the Gamma Function as extension of the factorial, we obtain $displaystylephi(k)=frac{Gamma(k+1)}{k^s}$. Now applying the Mellin Transform with $s=-alpha$ yields to
$$int_0^{infty}x^{-alpha-1}operatorname{Li}_s(-x)dx=Gamma(-alpha)phi(alpha)=Gamma(-alpha)frac{Gamma(alpha+1)}{alpha^s}$$
This term can be simplified by using Eulers Reflection Formula with $z=alpha+1$ which finally gives
$$frac1{alpha^s}Gamma(alpha+1)Gamma(-alpha)=frac1{alpha^s}frac{pi}{sin(pi(alpha+1))}=-frac1{alpha^s}frac{pi}{sin(pialpha)}$$
answered Sep 30 '18 at 20:23
mrtaurhomrtaurho
4,25421234
edited Jan 8 at 10:21
answered Sep 30 '18 at 20:23
mrtaurhomrtaurho
4,25421234
answered Sep 30 '18 at 20:23
mrtaurhomrtaurho
4,25421234
answered Sep 30 '18 at 20:23
mrtaurhomrtaurho
4,25421234
4,25421234
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$begingroup$
Looks like an application of Ramanujan's Master Theorem, coupled with the reflection formula for the Gamma function.
$endgroup$
– nospoon
Sep 29 '18 at 14:52
$begingroup$
@nospoon Is it possible to use this Theorem here? I also thought about a potential connection to the Mellin Transform but hence the $x$-term has the form $x^{-alpha-1}$ instead of $x^{alpha-1}$ I do not know whether this possible or not.
$endgroup$
– mrtaurho
Sep 29 '18 at 14:58