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If $f$ is a continuous even function from $[-a,a]$ to $\Bbb R$ then calculate $\int_{-a}^a \frac{f(x)}{1+e^x}dx$

The answer that is given is $\int _{-a}^af(x)dx$

My attempt

$I=\int_{-a}^a \frac{f(x)}{1+e^x}dx=-\int_{-a}^a \frac{f(x)e^x}{1+e^x}dx+\int_{-a}^a {f(x)}dx$

Now for the first integral, we can use integration by parts but the equation is getting complicated and for the second integral would be $2\int_{0}^a {f(x)}dx$ as $f$ is even. Please help...

Ri-Li
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    Note that the correct answer is $\frac 12 \int _{-a}^af(x)dx = \int _{0}^af(x)dx$ – Martin R Dec 22 '18 at 17:52
  • Yes @MartinR when I searched it, I didn't get it. Strange!! Where do you search by the way? – Ri-Li Dec 22 '18 at 18:01
  • @Gimgim: With Approach0 – See also https://math.meta.stackexchange.com/questions/24978/announcing-a-third-party-search-engine-for-math-stackexchange. – Martin R Dec 22 '18 at 18:03
  • More generally, if $f$ and $g$ are integrable, $f$ is even, and $g(x)+g(-x)=1$, then $\int_{-a}^a f(x)g(x),dx=\int_0^a f(x),dx$. – Mark Viola Dec 22 '18 at 18:04

1 Answers1

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Hint:

$$I=\int^a_{-a} \frac {f(x)}{1+e^x} $$ Using King property:

Replace $x$ with $a+(-a)-x=-x$
$$I =\int^a_{-a} \frac {f(-x)}{1+e^{-x}}$$ Add both equations: $$2I= \int^a_{-a} \frac {f(x)}{1+e^{x}}+ \int^a_{-a} \frac {f(-x)}{1+e^{-x}}$$ $f(x)=f(-x)$ why? $$2I =\int^a_{-a} \frac {f(x)}{1+e^{x}}+\int^a_{-a} \frac {f(x)\cdot e^x}{1+e^{x}}=\int_{-a}^af(x)$$

$$I=\frac12 \int^a_{-a} {f(x)}=2\cdot \frac12 \int^a_{0} {f(x)}= \int^a_{0} {f(x)}$$

pooja somani
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