eRealIntegral

Answer To solve $\int_{0}^{\infty}\frac{x\sin{mx}}{x^2 + 1}dx$ , we consider the following contour $C$

: The line $C_1$

which connects $-R$

and

, The curve $C_R$

which connects $R$

and

．Then．　

$\displaystyle \int_{C_1}\frac{-ixe^{imx}}{x^2 + 1}dx + \int_{C_R}\frac{-iz e^{imz}}{z^2 + 1}dz = \oint_{C}\frac{-i z e^{imz}}{z^2 + 1}dz$

**図 1:** Curve
$\includegraphics[width=6cm]{COMPFIG/Fig-counter.eps}$

Using Residue theorem, we evaluate $\oint_{C}\frac{-i z e^{imz}}{z^2 + 1}dz$ ． $z = \pm i$ are singularities. But $z = -i$

is outside of the curve $C$

．Then we find the residue at $z = i$

is the pole of th 1st-order.

$\displaystyle Res[i] = \lim_{z \to i}(z-i)\frac{-iz e^{imz}}{(z+i)(z-i)} =　\frac{e^{-m}}{2i}$

$\displaystyle \oint_{C}\frac{-ize^{imz}}{z^2 + 1}dz = 2\pi i\frac{e^{-m}}{2i} = \pi e^{-m}$

$\displaystyle \int_{C_1}\frac{-ixe^{imx}}{x^2 + 1}dx$	$\displaystyle =$	$\displaystyle \int_{-R}^{R}\frac{-ixe^{imx}}{x^2 + 1}dx$
	$\displaystyle =$	$\displaystyle \int_{-R}^{0}\frac{-ixe^{imx}}{x^2 + 1}dx + \int_{0}^{R}\frac{-ixe^{imx}}{x^2 + 1}dx$
	$\displaystyle =$	$\displaystyle \int_{0}^{R}\frac{i x e^{-imx}}{x^2 + 1}dx + \int_{0}^{R}\frac{-ixe^{imx}}{x^2 + 1}dx$
	$\displaystyle =$	$\displaystyle \int_{0}^{R}\frac{-i x(e^{imx}- e^{-imx})}{x^2 + 1}dx = 2 \int_{0}^{R}\frac{x \sin{mx}}{x^2 + 1}dx$

Consider the integral on $C_R$

． We show $\int_{C_R}\frac{-ize^{imz}}{z^2 + 1}dz$ converges to 0 as $R \to \infty$ ．On the $C_R$

, $z = Re^{i\theta}$ . Thus $dz = Rie^{i\theta}$

$\displaystyle \int_{C_R}\vert\frac{-ize^{imz}}{z^2 + 1}dz$	$\displaystyle =$	$\displaystyle \int_{0}^{\pi}\frac{-iRe^{i\theta}e^{iRme^{i\theta}}}{R^2 e^{2i\theta} + 1}R i e^{i\theta}\vert d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{\pi}\vert e^{(iRme^{i\theta})}\vert d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{\pi}\vert e^{iRm(\cos{\theta}+i\sin{\theta})}\vert d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{\pi}\vert e^{iRm (\cos{\theta})}\cdot e^{iRm (i\sin{\theta})}\vert d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{\pi}e^{-Rm \sin{\theta}}d\theta$

$\displaystyle \int_{0}^{\pi}e^{-Rm \sin{\theta}}d\theta = 2\int_{0}^{\frac{\pi}{2}}e^{-Rm \sin{\theta}}d\theta$

$\displaystyle 2\int_{0}^{\frac{\pi}{2}}e^{-Rm \sin{\theta}}d\theta$	$\displaystyle \leq$	$\displaystyle 2\int_{0}^{\frac{\pi}{2}}e^{-Rm \frac{2}{\pi}}\theta d\theta$
	$\displaystyle =$	$\displaystyle 2\frac{-\pi}{2Rm}e^{-Rm\frac{2}{\pi}}\theta\mid_{0}^{\frac{\pi}{2}}$
	$\displaystyle =$	$\displaystyle \frac{\pi}{Rm}(e^{-Rm} -1) \to 0 \ (R \to \infty)$