How to Compute a Definite Integral with Symmetry: The Case of $f(-x)=f(x)$

[MarkFL]

Suppose $f(-x)=f(x)$, then compute the following definite integral:

\(\displaystyle \int_{-a}^{a}\frac{1}{1+2^{f(x)}}\,dx\) where $0<a\in\mathbb{R}$.

 

[mathbalarka]

Spoiler

$$\int_{-a}^{a} \frac{d x}{1+2^{f(x)}} = \int_{0}^{a} \frac{dx}{1+2^{f(x)}} + \int_{-a}^{0} \frac{dx}{1+2^{f(x)}} = \int_{0}^{a} \frac{dx}{1+2^{f(x)}} + \int_{0}^{a} \frac{dx}{1+2^{f(x)}} = 2\int_{0}^{a} \frac{dx}{1+2^{f(x)}}$$

Note : Although not related to the original problem that was meant, it is not worthless to note that the calculations above works only if $\frac{1}{1+2^{f(x)}}$ has no vertical asymptotes.

 

[MarkFL]

Spoiler

$$\int_{-a}^{a} \frac{d x}{1+2^{f(x)}} = \int_{0}^{a} \frac{dx}{1+2^{f(x)}} + \int_{-a}^{0} \frac{dx}{1+2^{f(x)}} = \int_{0}^{a} \frac{dx}{1+2^{f(x)}} + \int_{0}^{a} \frac{dx}{1+2^{f(x)}} = 2\int_{0}^{a} \frac{dx}{1+2^{f(x)}}$$

Haha...that is quite correct (well done!)...but I messed up and actually meant for $f$ to be odd, i.e.:

\(\displaystyle f(-x)=-f(x)\)

(Bug)

 

[mathbalarka]

Spoiler

$$\int_{-a}^{a} \frac{dx}{1+2^{f(x)}} = \int_{0}^{a} \frac{dx}{1+2^{f(x)}} + \int_{-a}^{0} \frac{dx}{1+2^{f(x)}} = \int_{0}^{a} \frac{dx}{1+2^{f(x)}} - \int_{a}^{0} \frac{dx}{1+2^{f(-x)}} = \int_{0}^{a} \frac{dx}{1+2^{f(x)}} + \int_{0}^{a} \frac{dx}{1+2^{-f(x)}} \\ = \int_{0}^{a} \left ( \frac{1}{1+2^{f(x)}} + \frac{1}{1+2^{-f(x)}} \right ) dx = \int_{0}^{a} \frac{2 + 2^{f(x)} + 2^{-f(x)}}{2 + 2^{f(x)} + 2^{-f(x)}} dx = a$$

 

[MarkFL]

Spoiler

$$\int_{-a}^{a} \frac{dx}{1+2^{f(x)}} = \int_{0}^{a} \frac{dx}{1+2^{f(x)}} + \int_{-a}^{0} \frac{dx}{1+2^{f(x)}} = \int_{0}^{a} \frac{dx}{1+2^{f(x)}} - \int_{a}^{0} \frac{dx}{1+2^{f(-x)}} = \int_{0}^{a} \frac{dx}{1+2^{f(x)}} + \int_{0}^{a} \frac{dx}{1+2^{-f(x)}} \\ = \int_{0}^{a} \left ( \frac{1}{1+2^{f(x)}} + \frac{1}{1+2^{-f(x)}} \right ) dx = \int_{0}^{a} \frac{2 + 2^{f(x)} + 2^{-f(x)}}{2 + 2^{f(x)} + 2^{-f(x)}} dx = a$$

That's correct! :D

Here's my solution:

Spoiler

\(\displaystyle I=\int_{-a}^{a}\frac{1}{1+2^{f(x)}}\,dx\)

\(\displaystyle I=\int_{-a}^{a}\frac{1}{1+2^{f(x)}}-\frac{1}{2}+\frac{1}{2}\,dx\)

\(\displaystyle I=\frac{1}{2}\int_{-a}^{a}\frac{1-2^{f(x)}}{1+2^{f(x)}}+\frac{1}{2}\int_{-a}^{a}\,dx\)

The first integrand is odd, and the second even, hence:

\(\displaystyle I=0+\int_0^a\,dx=a\)

 

[DreamWeaver]

Great thread! (Heidy) :D