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$$\int_{0}^{\infty}\frac{x\sin 2x}{9+x^{2}} \, dx$$

Some rearranging eventually gives $$\int_{0}^{\infty}\frac{x\sin 2x}{9+x^{2}} \, dx = \frac{-i}{2}\int_{-\infty}^{\infty} \frac{xe^{2ix}}{9+x^{2}}$$ Consider $f(z) = \frac{ze^{2iz}}{9+z^{2}}$ and the contour $\gamma$ of the semicircle laying in the upper half of the plane: contour

Let $\gamma_{R}$ denote the circular part with radius $R$ and $\gamma_{L}$ denote the part lying on the real axis with length $2R$.

Computing the residue at the only pole, $z = 3i$, we have that $$\oint_{\gamma}f(z) \, dz = \frac{i\pi}{e^{6}} $$ On the other hand, \begin{align*} \oint_{\gamma}f(z) \, dz &= \oint_{\gamma_{R}}f(z) \, dz + \oint_{\gamma_{L}} f(z) \, dz \\ &= \oint_{\gamma_{R}} f(z) \, dz + \int_{-R}^{R} \frac{xe^{2ix}}{9+x^{2}} \, dx \end{align*} We may evaluate the first integral in the usual way by parameterizing the contour and taking $z = Re^{i\theta}$. \begin{align*} \oint_{\gamma_{R}} f(z) \, dz = \int_{0}^{\pi} \, \frac{R^{2}e^{2i\theta}e^{2i\cos\theta}e^{-2R\sin\theta}}{(9+R^{2}e^{2i\theta})} d\theta \end{align*}

I'm aiming to show that this integral goes to $0$ as $R$ goes to infinity. This gives the desired result as the factor of $\frac{-i}{2}$ is all that is missing according to WolframAlpha. I'm not sure how to finish it though.

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  • $\begingroup$ Factor $R^2$ from the denominator to cancel the one in the numerator and then take the limit as R approaches infinity which will show it goes to zero $\endgroup$ – Triatticus Mar 4 '17 at 22:58
  • $\begingroup$ Shouldn't that be $9+R^2e^{2i\theta}$ in the denominator of the last integral? $\endgroup$ – Simply Beautiful Art Mar 4 '17 at 23:06
  • $\begingroup$ You're right. edited. $\endgroup$ – user367387 Mar 4 '17 at 23:18
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Before addressing the question regarding the limit of the integral over $\gamma_R$, it is important to first understand that

$$\int_0^\infty \frac{x\sin(2x)}{9+x^2}\,dx=\frac12\text{Im}\left(\int_{-\infty}^\infty \frac{xe^{i2x}}{9+x^2}\,dx\right)$$

Next, using the residue theorem we find that

$$\oint_\gamma \frac{ze^{i2z}}{9+z^2}\,dz=2\pi i \frac{3i(e^{-6})}{6i}=\pi ie^{-6}$$


Now, to address the question regarding the limit of the integral over $\gamma_R$, we first note that

$$\begin{align} \left|\int_0^\pi \frac{Re^{i\phi }e^{i2Re^{i\phi}}}{9+R^2e^{i2\phi}}\,iRe^{i\phi}\,d\phi\right|&\le \int_0^\pi \frac{R^2e^{-2R\sin(\phi)}}{|9+R^2e^{i2\phi}|}\\\\ &\le \frac{2R^2}{|R^2-9|}\int_0^{\pi/2} e^{-2R\sin(\phi)}\,d\phi\tag 1 \end{align}$$

Next, we use the fact that $\sin(\phi)\ge 2\phi/\pi$ for $\phi \in[0,\pi/2]$. Hence, we can write

$$\int_0^{\pi/2} e^{-2R\sin(\phi)}\,d\phi\le \int_0^{\pi/2} e^{-4R\phi/\pi}\,d\phi=\frac{1-e^{-2R}}{4R/\pi}\tag 2$$

Using $(2)$ in $(1)$ we find that

$$\begin{align} \left|\int_0^\pi \frac{Re^{i\phi }e^{i2Re^{i\phi}}}{9+R^2e^{i2\phi}}\,iRe^{i\phi}\,d\phi\right|\le \frac{2R^2}{|R^2-9|}\frac{1-e^{-2R}}{4R/\pi}\to 0\,\,\text{as}\,\,R\to \infty \end{align}$$


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  • $\begingroup$ Lol, don't doubt yourself so much XD $\endgroup$ – Simply Beautiful Art Mar 4 '17 at 23:43
  • $\begingroup$ @SimplyBeautifulArt I'm not sure what you mean? $\endgroup$ – Mark Viola Mar 4 '17 at 23:44
  • $\begingroup$ :P You deleted your own answer, though I thought it was good. $\endgroup$ – Simply Beautiful Art Mar 4 '17 at 23:45
  • $\begingroup$ @SimplyBeautifulArt There was a flaw in the first post that I needed to edit. But thank you for your comment! $\endgroup$ – Mark Viola Mar 4 '17 at 23:46
  • $\begingroup$ Safe and Sound. (+1) $\endgroup$ – Hazem Orabi Mar 5 '17 at 0:02
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I'll use the duality of the convolution theorem to find the spectrum $\mathcal{F}\left(\frac{x}{9 + x^2}\right)$ as the convolution of the two spectra $\mathcal{F}\left(\frac{1}{9 + x^2}\right)$ and $\mathcal{F}\left(x\right)$. The spectrum of the first (Cauchy) function is well-known, being a symmetric mono-exponential, and the second spectrum is $\left(-i \sqrt{2\pi} δ'(ω)\right)$, being the differential operator in the Fourier domain times $i$.

$$\begin{align} I &= \int_{0}^{\infty}\frac{x\sin 2x}{9+x^{2}} \, dx \\ &=\frac{1}{2}\Im\left(\sqrt{2\pi}\mathcal{F}\left(\frac{x}{9 + x^2}\right)_{\omega = 2}\right) \\ &=\frac{1}{2}\Im\left(\frac{\sqrt{2\pi}\mathcal{F}\left(\frac{1}{9 + x^2}\right) \otimes \sqrt{2\pi}\mathcal{F}\left(x\right)}{2\pi}\right)_{\omega = 2} \\ &=\frac{1}{2}\Im\left(\frac 1 3 \sqrt{\frac\pi 2} e^{(-3|ω|)} \otimes -i \sqrt{2\pi} δ'(ω)\right)_{\omega = 2} \\ &=\frac{\pi}{2}\Im\left(i\left[\frac 1 3 e^{(-3|ω|)}\right]' \right)_{\omega = 2}\\ &=\frac{\pi}{2}\left[\frac 1 3 e^{(-3|ω|)}\right]'_{\omega = 2} \\ &=\frac{\pi}{2 e^{6}} \\ &\approx 0.003893... \end{align}$$

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    $\begingroup$ I don't know anything about the machinery you're using, but there must be an error somewhere. Thank you though. $\endgroup$ – user367387 Mar 5 '17 at 0:42
  • $\begingroup$ @JPelter : thanks, I have corrected my solution. $\endgroup$ – Job Bouwman Mar 5 '17 at 1:03
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We may also use the Laplace transform to evaluate the integral. Using this useful identity of the Laplace transform we have $$I=\int_{0}^{\infty}\frac{x\sin\left(2x\right)}{x^{2}+9}dx\stackrel{2x\rightarrow x}{=}\int_{0}^{\infty}\frac{x\sin\left(x\right)}{x^{2}+36}dx=\int_{0}^{\infty}\frac{\cos\left(6x\right)}{x^{2}+1}dx.$$ Let us consider $$I\left(a\right)=\int_{0}^{\infty}\frac{\cos\left(ax\right)}{x^{2}+1}dx,\,a>0.$$ Then if we apply again the Laplace transform we have $$\mathfrak{L}\left(I\left(a\right)\right)\left(s\right)=\int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos\left(ax\right)}{x^{2}+1}e^{-as}dadx=\int_{0}^{\infty}\frac{s}{\left(x^{2}+s^{2}\right)\left(1+x^{2}\right)}dx$$ hence using partial fractions $$\mathfrak{L}\left(I\left(a\right)\right)\left(s\right)=\frac{\pi}{2\left(s+1\right)}$$ so inverting $$I\left(a\right)=\color{red}{\frac{\pi}{2e^{a}}}.$$ Taking $a=6$ we have the result.

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