Is $\int\limits_0^\infty\frac{\sin y}{y^{s+1}}dy=-\Gamma(-s)\sin(\frac{\pi s}{2})$ for $\operatorname{Re}(s)\in (-1,0)$ obvious?

Question

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.

Answer 1

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.

Answer 2

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.

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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

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

Answer 3

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.

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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!

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See THIS ANSWER for reference.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Answer 5

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 $\Re(s)\in(0,1)$.