Demonstration of $\int_{-a}^a \frac{f(x)}{1+e^x} \,dx= \int_0^a f(x) \,dx$

Question

Good morning,

Can you give me a help to demonstrate this proposition:

$f$ is an even and continuous function on the interval $[-a,a], a>0$. Demonstrate:

$$\int_{-a}^a \frac{f(x)}{1+e^x} \,dx= \int_0^a f(x) \,dx$$

The problem can be generalized: $\int_{-a}^a g(x)f(x) dx = \int_0^a f(x)dx$ for all even $f$ and all (integrable) functions $g$ satisfying $g(x) + g(-x) = 1$. — Winther, Dec 25 '14 at 22:02

score 6 · Accepted Answer · answered Dec 25 '14 at 20:31

$$\begin{align}\int_{-a}^a dx \frac{f(x)}{1+e^x} &= \int_{-a}^0 dx \frac{f(x)}{1+e^x}+\int_{0}^a dx \frac{f(x)}{1+e^x}\\ &= \int_0^a dx \frac{f(-x)}{1+e^{-x}}+\int_{0}^a dx \frac{f(x)}{1+e^x} \\ &= \int_0^a dx \left (\frac1{1+e^{-x}}+\frac1{1+e^x} \right ) f(x)\\ &= \int_0^a dx \frac{2+e^x+e^{-x}}{(1+e^x)(1+e^{-x})}f(x) \\ &= \int_0^a dx \, f(x)\end{align} $$

score 2 · Answer 2 · answered Dec 25 '14 at 20:29

2

As $f(x)=f(-x)$: $$\int_{-a}^a \frac{f(x)}{1+e^x} \,dx=\underbrace{\int_{0}^a \frac{f(x)}{1+e^x} \,dx}_{x\ge0}+\underbrace{\int_{0}^a \frac{f(-x)}{1+e^{-x}}}_{\substack{x\le0\\x\mapsto-x}} \,dx=\int_0^a f(x) \,dx$$

answered Dec 25 '14 at 20:29

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Demonstration of $\int_{-a}^a \frac{f(x)}{1+e^x} \,dx= \int_0^a f(x) \,dx$

2 Answers2