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How would I prove that$$\int_{0}^\infty \frac{\cos (3x)}{x^2+4}dx= \frac{\pi}{4e^6}$$ I changed it to $$\int_{0}^\infty \frac{\cos (3z)}{(z+2i)(z-2i)}dz$$, and so the two singularities are $2i$ and $-2i$, but how would I go on from there?

$f(z)=\dfrac{e^{i3z}}{(z+2i)(z-2i)},$

$\lim\limits_{z\rightarrow\ 2i}(z-2i)*f(z) = \frac{e^-6}{4i}$

Using Residue theorem: $$2\pi i*\frac{1}{4ie^6}=\frac{\pi}{2e^6}$$

Américo Tavares
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Thomas
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2 Answers2

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You can consider $f(z)=\displaystyle\frac{e^{3zi}}{(z+2i)(z-2i)}$. For $R>0$, consider the counter $\gamma=\gamma_1+\gamma_2$ where $\gamma_1(t)=t$ where $-R\leq t\leq R$, and $\gamma_2(t)=Re^{it}$ where $0\leq t\leq \pi$. Then $$\int_{\gamma_1}f(z)dz=\int_{-R}^R\frac{e^{i3t}}{t^2+4}dt=\int_{-R}^R\frac{\cos(3t)}{t^2+4}dt+i\int_{-R}^R\frac{\sin(3t)}{t^2+4}dt$$ $$\rightarrow\int_{-\infty}^\infty\frac{\cos(3t)}{t^2+4}dt+i\int_{-\infty}^\infty\frac{\sin(3t)}{t^2+4}dt\mbox{ as }R\rightarrow\infty,$$ and $$\left|\int_{\gamma_2}f(z)dz\right|=\left|\int_{0}^{\pi}\frac{e^{i3Re^{it}}}{R^2e^{2it}+4}Rie^{it}dt\right|$$ $$\leq\int_{0}^{\pi}\frac{R|e^{i3Re^{it}}|}{R^2-4}dt= \int_{0}^{\pi}\frac{Re^{-3R\sin t}}{R^2-4}dt\leq\int_{0}^{\pi}\frac{R}{R^2-4}dt\rightarrow 0\mbox{ as }R\rightarrow\infty.$$

On the other hand, by Residue Therorem, we have $$\int_{\gamma} f(z)dx=2\pi iRes(f(z),2i)$$ since $2i$ is in the interior of $\gamma$ and $-2i$ is not. Note that $$Res(f(z),2i)=Res(\frac{e^{3zi}}{(z+2i)(z-2i)},2i)=\frac{e^{3zi}}{(z+2i)}\Big|_{z=2i}=\frac{e^{-6}}{4i}.$$ Combining all these, we get $$\int_{-\infty}^\infty\frac{\cos(3t)}{t^2+4}dt+i\int_{-\infty}^\infty\frac{\sin(3t)}{t^2+4}dt=2\pi i\cdot\frac{e^{-6}}{4i}=\frac{\pi}{2e^6}.$$ Equating real and imaginary parts, we get $$\int_{-\infty}^\infty\frac{\cos(3t)}{t^2+4}dt=\frac{\pi}{2e^6}\mbox{ and } \int_{-\infty}^\infty\frac{\sin(3t)}{t^2+4}dt=0.$$

Now note that $\frac{\cos(3t)}{t^2+4}$ is an even function, i.e. it is symmetric about the $y$-axis: $\frac{\cos(-3t)}{(-t)^2+4}=\frac{\cos(3t)}{t^2+4}$. We have $\int_{-\infty}^\infty\frac{\cos(3t)}{t^2+4}dt=2\int_{0}^\infty\frac{\cos(3t)}{t^2+4}dt$, which implies that $$\int_{0}^\infty\frac{\cos(3t)}{t^2+4}dt=\frac{1}{2}\int_{-\infty}^\infty\frac{\cos(3t)}{t^2+4}dt=\frac{\pi}{4e^6},$$ as required.

Paul
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  • Thanks, I just posted my working also, from working from Américo Tavares's earlier answer, I seem to get $\frac{\pi}{2e^6}$ also :s – Thomas Jan 20 '12 at 00:04
  • What I did is the same as the method that Americo suggested. I think what we got is right. You may want to go back and look at your book once more to see if the integral is from $\int_0^\infty$ instead of $\int_{-\infty}^\infty$. – Paul Jan 20 '12 at 00:07
  • sorry..Typo, yes it is indeed $\int_0^\infty$. From what it seems, do we need to divide by 2 to obtain the answer? If so, why? Im slightly confused.. – Thomas Jan 20 '12 at 00:10
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    Since $\frac{\cos(3t)}{t^2+4}$ is an even function, i.e. it is symmetric about the $y$-axis: $\frac{\cos(-3t)}{(-t)^2+4}=\frac{\cos(3t)}{t^2+4}$. We have $\int_{-\infty}^\infty\frac{\cos(3t)}{t^2+4}dt=2\int_{0}^\infty\frac{\cos(3t)}{t^2+4}dt$. Therefore, what we got is right. – Paul Jan 20 '12 at 00:14
  • Thanks so much, yeah I can imagine the last part requires a bit more intuition in order to complete it! – Thomas Jan 20 '12 at 00:18
  • @Thomas: Good that we solved the problem. You may want to accept the answer. – Paul Jan 20 '12 at 00:24
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    Yeah, glad we solved it, now at least I know this trick when doing future integrals! – Thomas Jan 20 '12 at 00:45
  • I am reading your answer, and you wrote $\leq\int_{0}^{\pi}\frac{R|e^{i3Re^{it}}|}{R^2-4}dt$, I think we have $\vert R^2-4\vert$, isn't it ? –  Jun 24 '15 at 20:41
  • Since we let $R\to\infty$ eventually, we can just consider $R$ sufficiently large so that $R^2-4>0$. – Paul Jun 25 '15 at 07:30
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Similarly to this answer to the question Verify integrals with residue theorem, the following should work. Choose $$f(z)=\dfrac{e^{i3z}}{z^{2}+4},$$ use a contour $\gamma _{R}$ consisting of the boundary of the upper half of the disk $|z|=R$ and the segment $[-R,R]$ described counterclockwise and apply the Jordan's lemma.

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Américo Tavares
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