13
$\begingroup$

This is a part of computation in Titchmash, Theories of Zeta Functions which I do not find obvious but there is no explanation. I did figure out the computation.

$$\int_0^\infty\frac{\sin(y)}{y^{s+1}}dy=-\Gamma(-s)\sin\left(\frac{\pi s}{2}\right)$$

Q: There is no explanation in the book for this step. Why is this obvious without explanation? My recipe goes as the following. It suffices to restrict to real axis part with $s\in (-1,0)$ region. Now integral is real valued in this region. Here I need $\Gamma(-s)=\frac{\Gamma(-s+1)}{s}$ extension to obtain real valuedness. Consider the integral as the imaginary part of $\int_0^{i\infty} \frac{e^{z}}{i^s z^{s+1}}dz$ where I have already rotated axis by $i$ multiplication. Now to obtain $\Gamma$ function, close contour from $(+\infty,0)$ axis portion and connect to $(0,i\infty)$ portion. Then close the contour by arc. The arc contour contribution is $0$ via exponential suppresion. Then apply residue theorem easily as the whole thing is holomorphic by $s\in (-1,0)$ region. Hence equality follows. This is not $1-2$ line naive computation though not hard. However, it did take me a while to figure out.

$\endgroup$
  • 2
    $\begingroup$ If you are interested in an attempt not relying on complex analysis I can offer a method using Ramanujan's Master Theorem which is quite simple too. $\endgroup$ – mrtaurho Dec 30 '18 at 20:04
  • 1
    $\begingroup$ My guess: the authors thought it not worth going into, that anyone who cared enough to verify could do so. And if, as you say, the computation isn't particularly difficult, it's probably (to them) not worth the trouble to go into, Personally I hate that philosophy myself (albeit depending on the context) but some people seem to like it. $\endgroup$ – Eevee Trainer Dec 30 '18 at 20:05
  • $\begingroup$ @mrtaurho Would you care to demonstrate it? The book did cover ramanujan sums in the first chapter. However the book did not define it. It might be correlated to that part of the book. $\endgroup$ – user45765 Dec 30 '18 at 20:07
  • $\begingroup$ Possible duplicate of Finding the value of improper integral given some other integral's value $\endgroup$ – mrtaurho Dec 30 '18 at 20:38
  • 1
    $\begingroup$ I think it is trivial from the Mellin transform point of view $M(\sin(t))(s)=\int^{\infty}_{0}\sin(t)t^{s-1}dt=\Gamma(s)\sin\left(\frac{s\pi}{2}\right)$, $-1<Re(s)<1$ $\endgroup$ – Nikos Bagis Dec 31 '18 at 21:47
12
$\begingroup$

I realised this question has been asked before as you can see here. Anyway I will write down my solution here again. First of all consider Ramanuajan's Master Theorem.

Ramanujan's 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 $$\int_0^{\infty}x^{p-1}f(x)dx=\Gamma(p)\phi(-p)$$

In order to use this Theorem we may expand the sine function as a series followed by the substitution $y^2=t$ which yields to

$$\begin{align*} \mathfrak{I}=\int_0^{\infty}y^{-s-1}\sin(y)dy&=\int_0^{\infty}y^{-s-1}\sum_{n=0}^\infty (-1)^n \frac{y^{2n+1}}{(2n+1)!}dy\\ &=\frac12\int_0^{\infty}y^{-s-1}\sum_{n=0}^\infty (-1)^n \frac{n!/(2n+1)!}{n!}(-y^2)^n[2ydy]\\ &=\frac12\int_0^\infty t^{-(s+1)/2}\sum_{n=0}^\infty\frac{n!/(2n+1)!}{n!}(-t)^ndt \end{align*}$$

Now we can use Ramanuajan's Master Theorem by setting $p=-\frac{s-1}2$ and $\phi(n)=\frac{n!}{(2n+1)!}=\frac{\Gamma(n+1)}{\Gamma(2(n+1))}$ and so we get

$$\begin{align*} \mathfrak{I}=\frac12\int_0^\infty t^{-(s+1)/2}\sum_{n=0}^\infty\frac{n!/(2n+1)!}{n!}(-t)^ndt&=\frac12\Gamma\left(-\frac{s-1}2\right)\frac{\Gamma\left(1+\frac{s-1}2\right)}{\Gamma\left(2\left(\frac{s-1}2+1\right)\right)}\\ &=\frac1{2\Gamma(s+1)}\Gamma\left(\frac{s+1}2\right)\Gamma\left(-\frac{s-1}2\right)\tag1\\ &=\frac1{2\Gamma(s+1)}\frac{\pi}{\sin\left(\pi\frac{s+1}2\right)}\\ &=\frac1{2\Gamma(s+1)}\frac{\pi}{\cos\left(\frac{\pi s}2\right)}\\ &=\frac{\pi}{\Gamma(s+1)}\frac{\sin\left(\frac{\pi s}2\right)}{2\sin\left(\frac{\pi s}2\right)\cos\left(\frac{\pi s}2\right)}\\ &=-\sin\left(\frac{\pi s}2\right)\frac{\pi}{\Gamma(s+1)\sin(\pi(s+1))}\tag2\\ &=-\sin\left(\frac{\pi s}2\right)\Gamma(-s) \end{align*}$$

$$\therefore~\mathfrak{I}=\int_0^{\infty}y^{-s-1}\sin(y)dy~=~-\Gamma(-s)\sin\left(\frac{\pi s}2\right)$$

For the simplification of the final solution we excessively used Euler's Reflection Formula which is a key property of the Gamma Function. Within line $(1)$ we applied the formula for $z=\frac{s+1}2$ and within line $(2)$ for $z=s+1$. The trigonometric reshaping utilized the double-angle formula as well as the periodic property of the sine function.

$\endgroup$
7
$\begingroup$

I'd say most of the book relies heavily on the same kind of derivation : complex analysis, change of variable, change of contour, recognizing famous integrals, restricting to domains where everything is easier then extending by continuity/analyticity.


For $\Re(s) < 0$ and $\Re(e^a) >0$ then $$\int_0^\infty t^{-s-1} e^{-e^a t}dt = \int_0^{e^{\overline{a}}\infty} (e^{-a }u)^{-s-1} e^{-u}d(e^{-a}u) =e^{a s}\int_0^{e^{\overline{a}}\infty}+\int_{e^{\overline{a}}\infty}^\infty u^{-s-1} e^{-u}du= e^{a s} \Gamma(-s)$$

For $Re(s) \in (-1,0)$ and $a =b+ i\pi/2$ then $$2i\int_0^\infty t^{-s-1} \sin(t) dt = \lim_{b \to 0^+} \int_0^\infty t^{-s-1} (e^{-e^{b+i\pi/2} t}-e^{-e^{b-i\pi/2} t})dt = \lim_{b \to 0^+}(e^{s(b+i\pi/2) }-e^{s(b-i\pi/2)}) \Gamma(-s)= 2i \sin(\pi s/2)\Gamma(-s)$$

And $\int_0^\infty t^{-s-1} \sin(t) dt =\sin(\pi s/2)\Gamma(-s)$ stays true for $\Re(s) \in(-1,1)$ by analytic continuation


Note a similar derivation with $\int_0^\infty t^{s-1} \log(1-e^{-t})dt$ yields the functional equation for $\zeta(s)$, as $Im(\log(1-e^{4i \pi t})) = 2i\pi t - 2i\pi\lfloor t \rfloor$

$\endgroup$
  • 1
    $\begingroup$ Yes, then, and now, such computations are eminently "standard" in certain circles... even if completely unknown to a more general mathematical community. $\endgroup$ – paul garrett Dec 30 '18 at 23:01
5
$\begingroup$

I thought it might be instructive to present an approach that uses Laplace Transforms, an integral representation of the Beta Function, the relationship between the Beta Function and Gamma Function, and Euler's Reflection Formula for the Gamma Function. To that end we now proceed.


Let $f(x)=\sin(x)$ and $g(x)=\frac{1}{x^{s+1}}$. Then, the Laplace Transform of $f$ is

$$\mathscr{L}\{f\}(x)=\frac{1}{x^2+1}\tag1$$

and for $\text{Re}(s)\in(-1,0)$, the inverse Laplace Transform of $g$ is

$$\mathscr{L}^{-1}\{g\}(x)=\frac{x^s}{\Gamma(s+1)}\tag2$$

Using $(1)$ and $(2)$ we see that

$$\begin{align} \int_0^\infty \frac{\sin(y)}{y^{s+1}}\,dy&=\frac1{\Gamma(s+1)}\int_0^\infty \frac{x^s}{x^2+1}\,dx\\\\ &=\frac{1}{2\Gamma(s+1)}\int_0^\infty \frac{x^{(s-1)/2}}{1+x}\,dx\\\\ &=\frac1{2\Gamma(s+1)}B\left(\frac{1+s}{2},\frac{1-s}{2}\right)\\\ &=\frac{\Gamma\left(\frac{1+s}{2}\right)\Gamma\left(\frac{1-s}{2}\right)}{2\Gamma(s+1)}\\\\ &=\frac{\frac{\pi}{\cos(\pi s/2)}}{2\frac{\pi}{\Gamma(-s)\sin(\pi(s+1))}}\\\\ &=-\Gamma(-s)\sin(\pi s/2) \end{align}$$

as expected!


See THIS ANSWER for reference.

$\endgroup$
4
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\left.\int_{0}^{\infty} {\sin\pars{y} \over y^{s + 1}}\,\dd y \,\right\vert_{\ -1\ <\ \Re\pars{s}\ <\ 1}} = \int_{0}^{\infty}\!\!\!\!\!\!\sin\pars{y}\ \overbrace{\bracks{{1 \over \Gamma\pars{s + 1}} \int_{0}^{\infty}t^{s}\expo{-yt}\,\dd t}} ^{\ds{\,\,\,\,\,\,=\ {1 \over y^{s + 1}}}}\ \,\dd y \\[5mm] = &\ {1 \over \Gamma\pars{s + 1}}\int_{0}^{\infty}t^{s} \int_{0}^{\infty}\sin\pars{y}\expo{-ty}\dd y\,\dd t \\[5mm] = &\ {1 \over \pi/\braces{\sin\pars{\pi\bracks{-s}}\Gamma\pars{-s}}} \int_{0}^{\infty}t^{s} \bracks{\Im\int_{0}^{\infty}\expo{-\pars{t - \ic}y} \dd y}\dd t \\[5mm] = &\ -\,{\sin\pars{\pi s}\Gamma\pars{-s} \over \pi} \int_{0}^{\infty}t^{s} \pars{1 \over t^{2} + 1}\dd t = -\,{\sin\pars{\pi s}\Gamma\pars{-s} \over \pi} \int_{0}^{\infty}{t^{s} \over t^{2} + 1}\,\dd t \\[5mm] = &\ -\,{\sin\pars{\pi s}\Gamma\pars{-s} \over \pi}\,{1 \over 2} \int_{0}^{\infty}{t^{s/2 - 1/2} \over t + 1}\,\dd t = -\,{\sin\pars{\pi s}\Gamma\pars{-s} \over 2\pi} \int_{1}^{\infty}{\pars{t - 1}^{s/2 - 1/2} \over t}\,\dd t \\[5mm] = &\ -\,{\sin\pars{\pi s}\Gamma\pars{-s} \over 2\pi} \int_{1}^{0}{\pars{1/t - 1}^{s/2 - 1/2} \over 1/t}\,\pars{-\,{\dd t \over t^{2}}} \\[5mm] = &\ -\,{\sin\pars{\pi s}\Gamma\pars{-s} \over 2\pi} \int_{0}^{1}t^{-s/2 - 1/2}\pars{1 - t}^{s/2 - 1/2}\,\dd t \\[5mm] = &\ -\,{\sin\pars{\pi s}\Gamma\pars{-s} \over 2\pi}\, {\Gamma\pars{-s/2 + 1/2}\Gamma\pars{s/2 + 1/2} \over \Gamma\pars{1}} \\[5mm] = &\ -\,{\sin\pars{\pi s}\Gamma\pars{-s} \over 2\pi}\, {\pi \over \sin\pars{\pi\bracks{s/2 + 1/2}}} \\[5mm] = &\ -\,{\bracks{2\sin\pars{\pi s/2} \cos\pars{\pi s/2}}\Gamma\pars{-s} \over 2} \,{1 \over \cos\pars{\pi s/2}} = \bbx{-\Gamma\pars{-s}\sin\pars{\pi s \over 2}} \end{align}

$\endgroup$
  • $\begingroup$ It seems like you missed a minus sign in between since your answers differs from the given one exactly by a minus sign. $\endgroup$ – mrtaurho Dec 31 '18 at 12:32
  • $\begingroup$ @mrtaurho Yes, it's true. I already fixed it. Thanks. $\endgroup$ – Felix Marin Dec 31 '18 at 20:26
2
$\begingroup$

For $0<\Re(s)<1$, we have $$ \int^{\infty}_{0}\sin(t)t^{s-1}dt=-\operatorname{Im}\left(\int^{\infty}_{0}e^{-it}t^{s-1}dt\right)=-\operatorname{Im}\left((-i)^{s}\int^{\infty i}_{0 i}e^{-z}z^{s-1}dz\right)= $$ $$ -\operatorname{Im}\left(e^{-i\pi s/2}\Gamma(s)\right)=\sin\left(\frac{\pi s}{2}\right)\Gamma(s). $$ About the question below, it have to be proved that $$ \int^{i\infty}_{0}e^{-z}z^{s-1}dz=\Gamma(s), $$ when $\Re(s)\in(0,1)$.

$\endgroup$
  • $\begingroup$ Out of curiosity: How does the given integral equals the Gamma Function? So far I only has seen the definition $$\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}dt$$ but I do not know how one could justify that $$\Gamma(s)\stackrel{?}{=}\int_0^{\color{red}{i}\infty} t^{s-1}e^{-t}dt$$ Has it something to do with the fact that you are only taking the imaginary part? Could you explain this detail to me? $\endgroup$ – mrtaurho Jan 1 at 14:12
  • $\begingroup$ Wihtin the last line: should the borders not be $0$ and $\infty$ after the substitution $z\to it$? Hence otherwise it does not make sense to my mind ^^ $\endgroup$ – mrtaurho Jan 9 at 19:12
  • $\begingroup$ Yes. I have been trying to resolve the problem of $"i"$ unsuccessfully a few hours here. $\endgroup$ – Nikos Bagis Jan 9 at 19:44
  • $\begingroup$ Mathematica program finds the result symbolicaly, but I don't have proof. $\endgroup$ – Nikos Bagis Jan 9 at 19:46
  • $\begingroup$ I guess it is right like this. Consider the substitution $z=it$. Hence $t$ is the new variable we have to change the borders aswell and thus we get $t_b=\frac zi=\frac{i0}i=0$ and $t_t=\frac zi=\frac{i\infty}i=\infty$ if I am not mistaken. However, I cannot tell how the factor $e^{-i\pi s/2}$ is gained by Mathematica. $\endgroup$ – mrtaurho Jan 9 at 19:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.