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@Yigeng You have asked $9$ questions so far. It is time to start accepting some of the answers to your other questions. – user17762 Oct 21 '12 at 3:05
@Marvis: I agree with that. – Mhenni Benghorbal Oct 21 '12 at 3:07
up vote 6 down vote accepted

Showing that $\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$, when $f$ is even

From the above, we get that $$\int_{-1}^{1} \dfrac{dx}{(1+e^x)(1+x^2)} = \int_{0}^{1} \dfrac{dx}{(1+x^2)} = \arctan(x) \left. \right\vert_{x=0}^{1} = \dfrac{\pi}4$$

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@MhenniBenghorbal The existence of the integral is pretty obvious. The integrand is a nice function bounded above by $1$ – user17762 Oct 21 '12 at 3:13
@MhenniBenghorbal: What do you mean? – wj32 Oct 21 '12 at 3:14
@Marivis: Sorry. I was looking at the wrong integral $\int_{-1}^1\frac{dx}{(e^x-1)(x^2+1)}\,.$ – Mhenni Benghorbal Oct 21 '12 at 3:16
@Marvis: indeed; the even part of $\frac1{1+e^x}$ is $\frac12$. – robjohn Oct 21 '12 at 8:23
@MhenniBenghorbal: Since the even part of $\frac1{e^x-1}$ is $-\frac12$, the Cauchy Principal Value of your integral would be $-\frac\pi4$. – robjohn Oct 21 '12 at 8:31

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