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I need to show that


but I don't really know why I'm not getting the result using contour integration (I'm not supposed to use the residue theorem).

Can't I use Cauchy's integral formula this way:

$$\int_{\gamma_r} \dfrac{\frac{\sin^2(z)}{z+i}}{z-i}dz =2\pi i\frac{\sin^2(i)}{2i}$$

along boundary of the upper semicircle of radius r? The other path would be 0 (the boundary without the path along the real axis). What's the problem with this approach?

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A related problem. –  Mhenni Benghorbal Mar 11 '13 at 14:31
Why do you think "the other path" would be zero? –  DonAntonio Mar 11 '13 at 14:52

2 Answers 2


$$\sin^2{x} = \frac{1}{2} (1-\cos{2 x})$$

Then the above integral is equal to

$$\frac{1}{2} \int_{-\infty}^{\infty} \frac{dx}{1+x^2} - \frac{1}{2} \int_{-\infty}^{\infty} dx\frac{\cos{2 x}}{1+x^2} = \frac{\pi}{2} - \frac{1}{2} \int_{-\infty}^{\infty} dx\frac{\cos{2 x}}{1+x^2}$$


$$\oint_C dz \frac{e^{i 2 z}}{1+z^2}$$

where $C$ is a semicircle in the upper half-plane of radius $R$. By the residue theorem, this integral is equal to $i 2 \pi$ time the sum of the residues of the poles inside $C$. In this case, the only pole inside $C$ is at $z=i$, at which the residue is $e^{-2}/(2 i)$.

Meanwhile, the contour integral vanishes along the semicircular contour in the limit as $R \rightarrow \infty$ (why?), so we are left with the integral along the real line:

$$\int_{-\infty}^{\infty} dx\frac{e^{i 2 x}}{1+x^2} = \int_{-\infty}^{\infty} dx\frac{\cos{2 x}}{1+x^2} = i 2 \pi \frac{e^{-2}}{2 i} = \frac{\pi}{e^2}$$

The result follows.

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Very nice ! Yet the OP mentions he's supposed not to use the residue theorem... –  DonAntonio Mar 11 '13 at 19:20
Sigh...too many late nights at work. But as it's not exactly wrong, maybe it'll be of use to someone else. –  Ron Gordon Mar 11 '13 at 19:22
By all means! Leave it here, it is a very nice answer. In fact, +1 –  DonAntonio Mar 11 '13 at 19:24

the poles of $ x^{2}+1 $ are $ i $ and $-i $ expand the sine function and apply residue theorem to

$$ \int_{-\infty}^{\infty}dx \frac{e^{2ix}}{x^{2}+1} $$

remmeber $ sin^{2}(x)= \frac{e^{2ix}+e^{-2ix}-2}{-4} $ and $ \int_{-\infty}^{\infty}dx \frac{1}{x^{2}+1}= \frac{\pi}{2} $

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Me parece que no leíste la parte en la que el consultante dice que no puede utilizar el teorema del residuo, @Jose Garcia –  DonAntonio Mar 11 '13 at 14:35
Gracias. Sí, no lo puedo usar (aunque el problema sea de una sección donde se explica). Lo que en realidad me gustaría saber es la razón por la que lo que escribo arriba parece no funcionar. –  safd Mar 11 '13 at 14:42
Ya te escribí allí un comentario. Me parece que la integral sobre el semicírculo superior no se anula cuando $\,R\to\infty\,$...Yo obtengo un resultado similar al tuyo pero yo tengo un factor de $\,1/4\,$ que tú no tienes. Por cierto, qué casualidad: yo le estaba escribiendo en español a José...;) –  DonAntonio Mar 11 '13 at 14:55

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