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I'm currently just starting integration of complex improper integrals, and I'm already struggling on the following integral:

$$I = \int\limits_{-\infty}^\infty \frac{(x+1)\sin(2x)}{x^2 + 2x + 2}dx = \int\limits_{-\infty}^\infty f(x)dx$$

So far I've thought about finding residues to apply Cauchy Residues Theorem after defining $I$ as $\lim\limits_{R\to\infty}\int\limits_{-R}^R f(x)dx$. However I'm not very comfortable with this integral, and I'm not sure my method can be applied.. could anyone tell me if that's a right way to start, or give me a clue if not?

That would be great!

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2 Answers 2

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Integrals of this type are treated using Jordan's lemma: $$I=\Im\int_{-\infty}^\infty f(z)\,dz=\Im\lim_{R\to\infty}\int_{C_R}f(z)\,dz,\quad f(z)=\frac{(z+1)e^{2iz}}{z^2+2z+2},$$ where $C_R$ is the boundary of $\{z\in\mathbb{C}:|z|<R,\Im z>0\}$, and the residue theorem: $$I=\Im\left(2\pi i\operatorname*{Res}\limits_{z=i-1}f(z)\right)=\pi e^{-2}\cos 2.$$

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  • $\begingroup$ I think you've mixed up the real & imaginary parts. It should be $\pi e^{-2}\color{blue}{\cos}2$. $\endgroup$
    – J.G.
    Apr 26, 2021 at 18:28
  • $\begingroup$ @J.G.: Indeed. Fixed. Thanks! $\endgroup$
    – metamorphy
    Apr 26, 2021 at 18:34
  • $\begingroup$ Indeed I've just seen Jordan lemma so I didn't think of it but it is a great idea to ! Thanks @metamorphy $\endgroup$
    – Rhaena
    Apr 27, 2021 at 8:47
  • $\begingroup$ I've tried your method @metamorphy, and also the method of Mostafa Ayaz in the answer below, but in any case the answer I find is different from yours. Indeed with your method I find $\pi i(e^{-2i-2})cos(2)$ because I find that the residue of $f$ at $-1+i$ is $\frac{e^{-2i-2}}{2}$; and with Mostafa's method I find a closer answer to yours but still not exactly the same because of the missing $i$: $\pi ie^{-2}cos(2)$ So could you either elaborate on your final result, or tell me what is wrong in mine ? $\endgroup$
    – Rhaena
    Apr 27, 2021 at 10:29
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    $\begingroup$ @Rhaena: Your evaluation of the residue is correct, but you have to multiply it by $2\pi i$, and take the imaginary part afterwards. [I.e., we're computing $\Im(\pi i e^{-2-2i})$.] $\endgroup$
    – metamorphy
    Apr 27, 2021 at 14:12
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Write the integral as following $$ I{=\int_{-\infty}^\infty \frac{(x+1)\sin(2x+2-2)}{(x+1)^2+1}dx \\ = \int_{-\infty}^\infty \frac{u\sin(2u-2)}{u^2+1}du \\ = \int_{-\infty}^\infty \frac{u(\sin(2u)\cos2+\cos(2u)\sin2)}{u^2+1}du } $$ and proceed the rest of the process through complex integration.

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  • $\begingroup$ That seems like a nice trick ! I guess by "proceed with complex integral" you just mean to integrate this considering $u$ as a variable of $\mathbb{C}$? $\endgroup$
    – Rhaena
    Apr 26, 2021 at 16:21
  • $\begingroup$ That is right and you will end up with two complex integrals. $\endgroup$
    – Mostafa Ayaz
    Apr 26, 2021 at 16:25
  • $\begingroup$ However after some reflexion I'm not sure how to compute those two integrals because the first one has $u\sin(2u)$ as a numerator and the second $u\cos(2u)$; I first thought it was a Fourier transform but it can't be because of the $1$ in the denominator and the $2u$ in $\sin$ and $\cos$. So how can we proceed ? If it's not too much to ask of you $\endgroup$
    – Rhaena
    Apr 26, 2021 at 16:54
  • $\begingroup$ @Rhaena The cosine part is odd, so vanishes; the sine part is easy with the residue theorem, using $\sin2u=\Im\exp2iu$. $\endgroup$
    – J.G.
    Apr 26, 2021 at 18:49
  • $\begingroup$ Okey I think I see for the sine part, but why does to cosine part vanish entirely if it is odd ? $\endgroup$
    – Rhaena
    Apr 27, 2021 at 8:46

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