Find
$$\int_{-\infty}^{\infty} \dfrac{\cos(x)}{x^2 + 2x + 4}\,dx$$
and
$$\int_{-\infty}^{\infty} \dfrac{\sin(x)}{x^2 + 2x + 4}\,dx$$
I find it really difficult.
Much appreciate it if anyone can solve this for me thanks.
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3 
- $\begingroup$ Yes I know the basic of residues theorem and other complex analysis but I am stuck how to start this. $\endgroup$Mike– Mike2015-05-18 21:39:38 +00:00Commented May 18, 2015 at 21:39
- $\begingroup$ Hint use $e^{ix} =i sin(x)+cos(x)$ to do both integrals at once. $\endgroup$tired– tired2015-05-18 21:46:56 +00:00Commented May 18, 2015 at 21:46
- $\begingroup$ @Mike: please unaccept my answer and, if you wish, accept Aaron Maroja's answer. It seems that I have included too much detail while trying to show that Jan Eerland's result-only answer has an error. $\endgroup$robjohn– robjohn ♦2015-05-19 22:14:44 +00:00Commented May 19, 2015 at 22:14
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3 Answers
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6 Hint: Consider the function $$f(z) = \frac{e^{iz}}{z^2 +2z +4}$$ Notice that $f(z)$ has poles at $-1 \pm i \sqrt 3$. Choose one root and take the path $C_R$ to be a semi-circunference of radius $R > 2$.
Use the Residue Theorem, to find the real and imaginary parts of your function $f(z)$ as $R \to \infty$.
- $\begingroup$ I don't get why is radius R >2? $\endgroup$Mike– Mike2015-05-18 22:54:23 +00:00Commented May 18, 2015 at 22:54
- $\begingroup$ To use The Residue Theorem, you need a path that has a pole inside its region. What is $\sqrt {1^2 + (\sqrt 3)^2}$? $\endgroup$Aaron Maroja– Aaron Maroja2015-05-18 23:33:01 +00:00Commented May 18, 2015 at 23:33
- $\begingroup$ alright, thanks allots! $\endgroup$Mike– Mike2015-05-18 23:39:38 +00:00Commented May 18, 2015 at 23:39
- $\begingroup$ You're most welcome! I'm glad you could make it. $\endgroup$Aaron Maroja– Aaron Maroja2015-05-18 23:40:21 +00:00Commented May 18, 2015 at 23:40
- $\begingroup$ Hi Aaron, I have one last question, shouldn't cos(x)=[exp(ix)+exp(-ix)]/2? according to Euler formula? $\endgroup$Mike– Mike2015-05-19 00:21:33 +00:00Commented May 19, 2015 at 0:21
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5 To add a bit of detail:
Let $\gamma_1$ be the contour $[-R,R]$ and $\gamma_2$ be the contour $Re^{i[0,\pi]}$ as $R\to\infty$. Then $$ \begin{align} &\int_{-\infty}^\infty\frac{\cos(x)\,\mathrm{d}x}{x^2+2x+4} +i\int_{-\infty}^\infty\frac{\sin(x)\,\mathrm{d}x}{x^2+2x+4}\\[6pt] &=\int_{-\infty}^\infty\frac{e^{ix}\,\mathrm{d}x}{(x+1)^2+3}\tag{1}\\[6pt] &=\frac1{2i\sqrt3}\int_{\gamma_1\cup\gamma_2}\left(\frac1{z+1-i\sqrt3}-\frac1{z+1+i\sqrt3}\right)e^{iz}\,\mathrm{d}z\\ &-\int_{\gamma_2}\frac{e^{iz}\,\mathrm{d}z}{(z+1)^2+3}\tag{2}\\[6pt] &=\frac{2\pi i}{2i\sqrt3}\operatorname*{Res}_{z=-1+i\sqrt3}\left(\frac{e^{iz}}{z-(-1+i\sqrt3)}\right)\tag{3}\\[6pt] &=\frac\pi{\sqrt3}e^{i(-1+i\sqrt3)}\tag{4}\\[6pt] &=\frac\pi{\sqrt3}e^{-\sqrt3}(\cos(1)-i\sin(1))\tag{5} \end{align} $$ Explanation:
$(1)$: Euler's Formula
$(2)$: the integral over the line is the limit of the difference of two contour integrals
$(3)$: Residue Theorem and $$\small\left|\int_{\gamma_2}\frac{e^{iz}\,\mathrm{d}z}{(z+1)^2+3}\right| \le2\int_0^{\pi/2}\frac{e^{-R\sin(\theta)}}{(R-1)^2}R\,\mathrm{d}\theta \le2\int_0^{\pi/2}\frac{e^{-2R\theta/\pi}}{(R-1)^2}R\,\mathrm{d}\theta\le\frac{\pi}{(R-1)^2}$$ $(4)$: evaluate the residue
$(5)$: Euler's Formula
Equating the real and imaginary parts above, we get $$ \int_{-\infty}^\infty\frac{\cos(x)\,\mathrm{d}x}{x^2+2x+4}=\frac\pi{\sqrt3}e^{-\sqrt3}\cos(1)\tag{6} $$ and $$ \int_{-\infty}^\infty\frac{\sin(x)\,\mathrm{d}x}{x^2+2x+4}=-\frac\pi{\sqrt3}e^{-\sqrt3}\sin(1)\tag{7} $$
- $\begingroup$ I have worked it out and got the same answer as yours thanks $\endgroup$Mike– Mike2015-05-19 20:20:39 +00:00Commented May 19, 2015 at 20:20
- $\begingroup$ A bit of detail? $\endgroup$Aaron Maroja– Aaron Maroja2015-05-19 20:23:54 +00:00Commented May 19, 2015 at 20:23
- $\begingroup$ @AaronMaroja: Since I was getting a different answer than Jan Eerland, I put the whole computation. I'm sorry if it seems like too much detail. $\endgroup$2015-05-19 21:47:42 +00:00Commented May 19, 2015 at 21:47
- $\begingroup$ @robjohn It's just that you as a moderator should know that this question is not "qualified" according to the MSE terms, that is, it's off-topic. And as far as I was explained these types of question should receive answers, if at all received, according to what has been presented. I don't know, sometimes I don't get the terms here. The OP didn't know how to begin and apparently managed to get the hint right. I don't see why you did this. But it's alright. $\endgroup$Aaron Maroja– Aaron Maroja2015-05-19 21:54:21 +00:00Commented May 19, 2015 at 21:54
- $\begingroup$ @AaronMaroja: If the OP will unaccept my answer, I will delete it. I've had things go funny when an accepted answer is deleted. $\endgroup$2015-05-19 22:13:25 +00:00Commented May 19, 2015 at 22:13
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3 It gave us:
$$\int_{-\infty}^{\infty} \dfrac{\cos(x)}{x^2 + 2x + 4}\,dx=\frac{e^{-\sqrt{3}}\pi \cos(1)}{\sqrt{3}}$$
And
$$\int_{-\infty}^{\infty} \dfrac{\sin(x)}{x^2 + 2x + 4}\,dx=\frac{e^{-\sqrt{3}}\pi \sin(1)}{\sqrt{3}}$$
- $\begingroup$ What gave those? $\endgroup$2015-05-19 18:04:26 +00:00Commented May 19, 2015 at 18:04
- $\begingroup$ @robjohn the answers on the integral! $\endgroup$Jan Eerland– Jan Eerland2015-05-19 18:05:18 +00:00Commented May 19, 2015 at 18:05
- $\begingroup$ I believe the sign is wrong on the integral with the sine. $\endgroup$2015-05-19 19:12:38 +00:00Commented May 19, 2015 at 19:12