Can we evaluate the integral $ I(a)=\int_0^{\infty} \frac{\sin x}{x^a} e^{-x} d x, $ without Gamma functions?

Question

Encountering the integral in the post stating that $$ \int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x}\,dx=\frac\pi4, $$ I started to investigate the integral in a more general form as

$$ I(a)=\int_0^{\infty} \frac{\sin x}{x^a} e^{-x} d x, $$ where $1<a<2$.

Using the Euler’s identity: $e^{xi}=\cos x+i\sin x$, we have $$ \begin{aligned} I(a) & = \Im \int_0^{\infty} \frac{e^{xi} \cdot e^{-x}}{x^a} d x \\ & = \Im\int_0^{\infty} \frac{e^{-(1-i) x}}{x^a} d x \\ & = \Im \int_0^{\infty} x^{-a} e^{-(1-i) x} d x \\ & =\Im\left[\frac{\Gamma(1-a)}{(1-i)^{1-a}}\right] \end{aligned} $$ By expressing the denominator in polar form, we have $$ \frac{1}{(1-i)^{1-a}}=\left(\sqrt{2} e^{-\frac{\pi}{4} i}\right)^{a-1} =2^{\frac{a-1}{2}} e^{\frac{(1-a) \pi}{4}i} $$ Now we can conclude that $$ \boxed{I(a)=2^{\frac{a-1}{2}} \Gamma(1-a) \sin \frac{(1-a) \pi}{4}} $$ For example, $$ I\left(\frac{3}{2}\right) =2^{\frac{1}{4}} (-2 \sqrt{\pi})\left(-\sin \frac{\pi}{8}\right) =2\sqrt[4]{2}\cdot\frac{\sqrt{(2-\sqrt{2} )\pi}}{2}=\sqrt{2(\sqrt{2}-1)\pi} $$

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Do we have any other methods to evaluate the integral? Your comments and alternative methods are highly appreciated.

Answer 1

Without complex functions.

Using the series expansion of the sine function $$I(a)=\int_0^{\infty} \frac{\sin (x)}{x^a} e^{-x} d x=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\int_0^{\infty}e^{-x} x^{2 n+1-a}\,dx$$ $$\int e^{-x} x^{2 n+1-a}\,dx=-\Gamma (2 n+2-a,x)$$ $$\int_0^{\infty} e^{-x} x^{2 n+1-a}\,dx=\Gamma (2 n+2-a)$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\Gamma (2 n+2-a)=\frac{2^{\frac{a-1}{2}} \cos \left(\frac{a+1}{4} \pi \right)\,\, \Gamma (2-a)}{1-a}$$ $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\Gamma (2 n+2-a)=2^{\frac{a-1}{2}} \cos \left(\frac{a+1}{4} \pi \right)\,\,\Gamma (1-a)$$

Answer 2

I'm going to complete Claude Leibovici's alternative approach by showing that

$$ \begin{align} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!} \, \Gamma(2n+2-a) &= 2^{a/2-1/2} \, \Gamma(1-a) \sin \left( \tfrac{\pi}{4}(1-a) \right) \\ &= 2^{a/2-1/2} \, \Gamma(1-a) \cos \left(\tfrac{\pi}{4}(1+a) \right) \end{align}$$ for $1 <\Re(a) <2$.

We'll use the hypergeometric identity $$_{2}F_{1}\left(a, a+ \tfrac{1}{2}; \tfrac{3}{2} ;-\tan^{2}(z) \right) = \cos^{2a}(z) \, \frac{\sin \left((1-2a) z\right)}{(1-2a) \sin (z)}. $$

I'll prove this at the end.

The series representation of the above hypergeometric function converges absolutely at $z= \frac{\pi}{4}$ if $ \Re(1-2a) >0$. (See here.)

Using this identity and Legendre's duplication formula, we have

$$ \begin{align} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!} \, \Gamma(2n+2-a) &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{\Gamma(2n+2)} \, \Gamma(2n+2-a) \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n} \, 2^{2n+1-a} \, \Gamma(n+1-\tfrac{a}{2}) \Gamma(n+\tfrac{3}{2}-\tfrac{a}{2})}{2^{2n+1}\Gamma(n+1)\Gamma(n+ \tfrac{3}{2})} \\ &= 2^{-a} \, \frac{\Gamma(1- \tfrac{a}{2}) \Gamma(\tfrac{3}{2}-\tfrac{a}{2})}{\Gamma (\tfrac{3}{2})} \, _{2}F_{1} \left(1-\tfrac{a}{2}, \tfrac{3}{2}- \tfrac{a}{2}; \tfrac{3}{2};-1 \right) \\ &= \Gamma(2-a) \, _{2}F_{1} \left(1-\tfrac{a}{2}, \tfrac{3}{2}- \tfrac{a}{2}; \tfrac{3}{2};-1 \right) \\ &= \Gamma(2-a) \, \frac{\left(\cos(\tfrac{\pi}{4})\right)^{2-a} \sin \left(\tfrac{\pi}{4}(a-1) \right)}{(a-1) \sin (\tfrac{\pi}{4})} \\ &= \Gamma(2-a) \, \frac{\left(\tfrac{1}{\sqrt{2}}\right)^{2-a} \sin \left(\tfrac{\pi}{4}(1-a) \right)}{(1-a) \, \tfrac{1}{\sqrt{2}}} \\ &= 2^{a/2-1/2} \, \Gamma(1-a)\sin \left(\tfrac{\pi}{4}(1-a) \right). \end{align}$$

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Proof of hypergeometric identity:

We'll first prove the identity $$_{2}F_{1}\left(a, a+ \tfrac{1}{2}; \tfrac{3}{2};z^{2}\right) = \frac{(1-z)^{1-2a}-(1+z)^{1-2a} }{(4a-2)z}. $$

Using the generalized binomial theorem and Legendre's duplication formula, we have

$$\begin{align} \frac{(1-z)^{1-2a}-(1+z)^{1-2a} }{(4a-2)z} &= \frac{1}{(4a-2)z} \left(\sum_{n=0}^{\infty} \binom{n+2a-2}{n}z^{n} - \sum_{n=0}^{\infty} \binom{n+2a-2}{n} (-z)^{n} \right) \\ &= \frac{2}{(4a-2)z} \sum_{n=0}^{\infty} \binom{2n+2a-1}{2n +1}z^{2n+1} \\ &= \frac{1}{2a-1} \sum_{n=0}^{\infty} \frac{\Gamma(2n+2a)}{\Gamma(2n+2) \Gamma(2a-1)} \, z^{2n} \\ &= \sum_{n=0}^{\infty} \frac{\Gamma(2n+2a)}{\Gamma(2n+2) \Gamma(2a)} \, z^{2n} \\ &= \sum_{k=0}^{\infty} \frac{\Gamma(n+a) \Gamma(n+a+\tfrac{1}{2}) \Gamma (\tfrac{1}{2})}{2 \Gamma(a) \Gamma(a+\tfrac{1}{2})\Gamma(n+ \tfrac{3}{2})} \, \frac{z^{2n}}{n!} \\ &= \sum_{k=0}^{\infty} \frac{\Gamma(n+a) \Gamma(n+a+\tfrac{1}{2}) \Gamma (\tfrac{3}{2})}{\Gamma(a) \Gamma(a+\tfrac{1}{2})\Gamma(n+ \tfrac{3}{2})} \, \frac{z^{2n}}{n!} \\ &= \, _{2}F_{1}\left(a, a+ \tfrac{1}{2}; \tfrac{3}{2};z^{2}\right) . \end{align}$$

Replacing $z$ with $i \tan(z)$, we get

$$ \begin{align} \, _{2}F_{1}\left(a, a+ \tfrac{1}{2}; \tfrac{3}{2};-\tan^{2}(z)\right) &= \frac{(1-i \tan z)^{1-2a}-(1+i\tan z)^{1-2a}}{(4a-2) i \tan (z)} \\ &= \frac{(\cos z - i \sin z)^{1-2a}-(\cos z + i \sin z)^{1-2a} }{(4a-2)i \tan (z) \cos^{1-2a}(z)} \\ &= \cos^{2a}(z) \, \frac{e^{-i(1-2a)z}-e^{i(1-2a)z}}{2i(2a-1) \sin(z)} \\ &= \cos^{2a}(z) \, \frac{\sin \left((1-2a)z \right)}{(1-2a)\sin(z)}. \end{align}$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#44f}{\on{I}\pars{a}} & \equiv \color{#44f}{\left.\int_{0}^{\infty}{\sin\pars{x} \over x^{a}} \expo{-x}\dd x\,\right\vert_{a\ <\ 2}} \\[5mm] & = -\,\Im\int_{0}^{\infty}x^{\pars{1 - a} - 1}\,\,\bracks{\expo{-\pars{1 + \ic}x} - \expo{-x}}\dd x \\[5mm] & \mbox{Note that} \\ & \expo{-\pars{1 + \ic}x}\ -\ \expo{-x} = \sum_{n = 0}^{\infty} \bracks{\pars{1 + \ic}^{\,n}\ - 1} {\pars{-x}^{n} \over n!} \\ & \mbox{such that} \\[5mm] \color{#44f}{\on{I}\pars{a}} & \equiv \color{#44f}{\left.\int_{0}^{\infty}{\sin\pars{x} \over x^{a}} \expo{-x}\dd x\right\vert_{a\ <\ 2}} \\[5mm] & = -\Im\overbrace{\braces{\Gamma\pars{1 - a}\bracks{\pars{1 + \ic}^{\,-1 + a}\ - 1}}} ^{\ds{Ramanujan's\ Master\ Theorem}} \\[5mm] & = -\Gamma\pars{1 - a} \Im\bracks{\pars{1 + \ic}^{\,-1 + a}\,} \\[5mm] & = -\Gamma\pars{1 - a} \Im\bracks{2^{\pars{a - 1}/2}\,\expo{\pars{a - 1}\pi\ic/4}\,} \\[5mm] & = \bbx{\color{#44f}{2^{\pars{a - 1}/2}\,\,\,\Gamma\pars{1 - a}\sin\pars{\bracks{1 - a}\pi \over 4}}} \\[5mm] & \mbox{Note that the final result}\ \sr{{\rm as}\ a\ \to\ 1}{\to}\ {\pi \over 4}. \end{align}