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

AI Thread Summary
The discussion focuses on computing the definite integral $$\int_{-a}^{a}\frac{1}{1+2^{f(x)}}\,dx$$ under the condition that the function satisfies $f(-x)=f(x)$. It is noted that the calculations are valid only if the integrand $\frac{1}{1+2^{f(x)}}$ does not have vertical asymptotes. A clarification is made that the original intent was for $f$ to be an odd function, where $f(-x)=-f(x)$. Participants express enthusiasm for the problem-solving process and share solutions. The thread emphasizes the importance of function properties in evaluating definite integrals.
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Suppose $f(-x)=f(x)$, then compute the following definite integral:

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

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.
 
Last edited:
mathbalarka said:
$$\int_{-a}^{a} \frac{d x}{1+2^{f(x)}} = \int_{0}^{a} \frac{dx}{1+2^{f(x)}} + \int_{-a}^{0} \frac{dx}{1+2^{f(x)}} = \int_{0}^{a} \frac{dx}{1+2^{f(x)}} + \int_{0}^{a} \frac{dx}{1+2^{f(x)}} = 2\int_{0}^{a} \frac{dx}{1+2^{f(x)}}$$

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

$$f(-x)=-f(x)$$

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

That's correct! :D

Here's my solution:

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

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

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

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

$$I=0+\int_0^a\,dx=a$$
 
Great thread! (Heidy) :D
 
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