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My goal is to calculate the integral

$\int_{0}^{\infty} x^{a-1} \cos(x) dx = \Gamma(a) \cos (\pi a/2)$,

where $0<a<1$, and my textbook provides the hint: integrate $z^{a-1} e^{iz}$ around the boundary of a quarter disk.

However, I couldn't figure out how to control the integral over the quarter arc. Any hints?

a12345
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  • This question was studied in detail here at this MSE link by some of the very same people who participated in this (the current) thread but this being a duplicate was apparently overlooked. – Marko Riedel Nov 27 '13 at 23:49

3 Answers3

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Consider the contour integral

$$\oint_C dz \, z^{a-1} \, e^{i z}$$

where $C$ is a quarter circle of radius $R$ in the 1st quadrant (real and imaginary $> 0$), with a small quarter circle of radius $\epsilon$ about the origin cut out (to avoid the branch point at the origin).

This integral is equal to

$$\int_{\epsilon}^R dx \, x^{a-1} \, e^{i x} + i R^a \int_0^{\pi/2} d\theta \, e^{i a \theta} e^{i R \cos{\theta}} \, e^{-R \sin{\theta}}\\+ i \int_R^{\epsilon} dy \, e^{i \pi (a-1)/2} y^{a-1} e^{-y} + i \epsilon^a \int_{\pi/2}^0 d\phi \, e^{i a \phi} \, e^{i \epsilon e^{i \phi}}$$

We note that the second integral vanishes as $R\to\infty$ because $\sin{\theta} \gt 2 \theta/\pi$, so that the magnitude of that integral is bounded by

$$R^a \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le R^a \int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi} \le \frac{2}{\pi R^{1-a}}$$

We also note that the fourth integral vanishes as $\epsilon^a$ as $\epsilon \to 0$. In the third integral, we write $i=e^{i \pi/2}$ to make a simplification.

The contour integral is zero by Cauchy's Theorem (no poles in the interior of $C$). Ths we have (+)

$$\int_{0}^{\infty} dx \, x^{a-1} \, e^{i x} - e^{i \pi a/2} \int_0^{\infty} dy \, y^{a-1} \, e^{-y}=0$$

We use the definition of the gamma function:

$$\Gamma(a) = \int_0^{\infty} dy \, y^{a-1} \, e^{-y}$$

and take real parts of (+) to obtain the sought-after result.

Ron Gordon
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  • Was waiting for you to show up :p – Bennett Gardiner Nov 27 '13 at 05:26
  • This is crystal clear. Thank you. – a12345 Nov 27 '13 at 15:42
  • @Ron Gordon, I was wondering about one thing. What if we take instead of a quarter circle a random arc. Wouldn't the integral involving $R\to \infty$ and $\epsilon \to 0$ still be $0$ and that makes gamma integrand something like symmetric from all rays emanating from origin? – Someone Mar 07 '16 at 16:03
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{\infty} x^{a - 1}\cos\pars{x}\,\dd x = \Gamma\pars{a}\cos\pars{\pi a \over 2}}:\ {\large ?}}$.

Note that $\ds{\int_{0}^{\infty}x^{a - 1} \cos\pars{x}\,\dd x = \Re\int_{0}^{\infty}x^{a - 1}\expo{\ic x}\,\dd x}$.
With the change of variable $x \equiv \expo{\ic\pi/2}\,t$, we'll have: \begin{align} &\int_{0}^{\infty}x^{a - 1}\cos\pars{x}\,\dd x = \Re\int_{0}^{-\ic\infty}\pars{\expo{\ic\pi/2}t}^{a - 1} \expo{-t}\,\ic\,\dd x \\[3mm] = &\ \Re\bracks{\expo{\ic\pi a/2} \int_{0}^{-\ic\infty}t^{a - 1} \expo{-t}\,\dd x} \\[3mm] = &\ \Re\braces{\expo{\ic\pi a/2}\bracks{-\int_{\infty}^{0}t^{a - 1}\expo{-t}\,\dd x}} \\[3mm] = &\ \Re\bracks{\expo{\ic\pi a/2}\ \overbrace{\int_{0}^{\infty}t^{a - 1}\expo{-t}\,\dd x\ }^{\ds{=\ \Gamma\pars{a}}}} \\[5mm] = &\ \underbrace{\ \Re\bracks{\expo{\ic\pi a/2}}\ } _{\ds{=\ \cos\pars{\pi a/2}}} \Gamma\pars{a} \end{align}
\begin{align} &\mbox{} \\ &\bbox[10px,border:1px groove navy]{% \int_{0}^{\infty}x^{a - 1}\cos\pars{x}\,\dd x = \Gamma\pars{a}\cos\pars{\pi a \over 2}} \\ & \end{align}
Felix Marin
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2

I recently read a short, yet interesting, article by Boas and Friedman where the authors calculated this same integral by using a different kind of contour:

a triangle with vertices $-p_{1}, \;\ p_{2}, \;\ (p_{1}+p_{2})i$.

Consider $R(z)e^{iz}$, where R(z) is a rational function that vanishes at $\infty$

Then, $p_{1}, \;\ p_{2}$ are taken to be sufficiently large as to enclose the poles in the UHP.

Instead of using the arc of a circle in the first quadrant, they use a straight line segment.

By letting $p_{1}, \;\ p_{2} \to \infty$ the integral over the real axis is equal to $2\pi i$ times the sum of the residues in the UHP.

This computation is claimed to be simpler because the slope of the triangular contour is bounded away from 0 in the UHP.

Connect the points $z=p, \;\ z=pi$ with a line rather than the arc of a circle.

On this line, $|dz|=\sqrt{2}dy$ and $\frac{1}{|z|}\leq \frac{\sqrt{2}}{p}$.

It follows immediately that the integral over the line is bounded by a constant times

$p^{s-1}\int_{0}^{p}e^{-y}dy\leq p^{s-1}$ which tends to 0 as $p\to \infty$ since $s<1$.

I have not included all of the intricacies, so if any one is interested in looking over this article it is in JSTOR. Look for "a simplification in certain contour integrals".

Cody
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