Show that for all $ a \in \mathbb{R} $ applies: $ \int \limits_{-r}^{r} \frac{f(x)}{1+\mathrm{e}^{a x}} d x=\int \limits_{0}^{r} f(x) d x$

Question

Let $ r>0 $ and $ f:[-r, r] \rightarrow \mathbb{R} $ be even, i. e. $ f(x)=f(-x) \ (x \in[-r, r]) . $

Show that for all $ a \in \mathbb{R} $ applies: $ \int \limits_{-r}^{r} \frac{f(x)}{1+\mathrm{e}^{a x}} d x=\int \limits_{0}^{r} f(x) d x $

Hint: Show that $ \frac{1}{1+\mu}+\frac{1}{1+\frac{1}{\mu}}=1$ $(\mu>0) $ and use the substitution rule.

My solution approach so far

$\frac{1}{1+\mu}+\frac{1}{1+\frac{1}{\mu}}=\frac{1}{1+\mu}+\frac{1}{\frac{\mu}{\mu}+\frac{1}{\mu}}=\frac{1}{1+\mu}+\frac{1}{\frac{\mu+1}{\mu}}=\frac{1}{1+\mu}+\frac{\mu}{\mu+1}=\frac{1+\mu}{1+\mu}=1$.

Should I substitute $\mu=e^{ax}$ now, or how do I continue? Thanks in advance!

Answer 1

First of all, because $f$ is even, we have:

$$\int_{-r}^rf(x)dx = \int_{-r}^0f(x)dx+\int_{0}^rf(x)dx=\int_{r}^0f(-y)(-dy)+\int_{0}^rf(x)dx $$ $$= \int_{0}^rf(-y)dy+\int_{0}^rf(x)dx=\int_{0}^rf(y)dy+\int_{0}^rf(x)dx=2\int_{0}^rf(x)dx$$

So, using your hint, we get

$$\int_{-r}^r\frac{f(x)}{1+e^{ax}}dx+\int_{-r}^r\frac{f(x)}{1+e^{-ax}}dx=\int_{-r}^rf(x)dx=2\int_{0}^rf(x)dx \ \ \ \ \ \ \ (*)$$

Now substituting $y=-x$, we have:

$$\int_{-r}^r\frac{f(x)}{1+e^{-ax}}dx = \int_{r}^{-r}\frac{f(-y)}{1+e^{ay}}(-dy)=\int_{-r}^{r}\frac{f(-y)}{1+e^{ay}}dy$$

$$=\int_{-r}^{r}\frac{f(y)}{1+e^{ay}}dy = \int_{-r}^{r}\frac{f(x)}{1+e^{ax}}dx$$

Substitute this back into $(*)$ to get:

$$ 2\int_{-r}^r\frac{f(x)}{1+e^{ax}}dx=2\int_{0}^rf(x)dx $$

and divide by $2$ to complete it.