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The complete problem says: Let $f(x) \in C^1 [-a,a]$ and even, calculate

$\displaystyle \int_{-a}^{a} \left[\frac{f(x)}{1+e^x} + f'(x)\log(1+e^x) \right]dx$

The first part is easy to see.

I don't know how to use the hypothesis that $f'(x)$ is continuous. I used that if $f(x)$ is even $f'(x)$ is odd, but this lead me that $\int_{-a}^{a}f'(x) = 0$...

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David
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1 Answers1

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What about a little integration by parts?

$$u=\log(1+e^x)\;,\;\;\;u'=\frac{e^x}{1+e^x}=1-\frac{1}{1+e^x}\\v'=f'(x)\;,\;\;v=f(x)$$

$$\left.\int\limits_{-a}^af'(x)\log(1+e^x)\,dx=f(x)\log(1+e^x)\right|_{-a}^a-\int_{-a}^a\left(f(x)-\frac{f(x)}{1+e^x}\right)dx=$$

$$=f(a)\overbrace{\left(\log\frac{1+e^a}{1+e^{-a}}\right)}^{\color{red}{\text{this equals}\; a}}+\int\limits_0^af(x)\,dx$$

using the result you linked to and, of course, the fact that the integral of an even (integrable) function on a symmetric interval is zero

DonAntonio
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  • Ach! Was just in the middle of typing this! The $\log$ at the end simplifies to $a$ of course. – L. F. Mar 10 '13 at 03:25
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    Good catch on that log thing, L.F. I was so concentrated on the other part of the equation that I didn't even pay attention to that. Thanks. – DonAntonio Mar 10 '13 at 03:27
  • Thanks!. I still having problems with integrations by parts. – David Mar 10 '13 at 03:39
  • @David, do you mean you have problems with that method or with the particular integration by parts shown in the answer? By parts is one of the most basic and useful integration techniques, and I assumed you already know it well if you're dealing with this kind of problems. – DonAntonio Mar 10 '13 at 03:40
  • Yes, I know how integrate by parts, but when use it?... is my problem. – David Mar 10 '13 at 03:44
  • Then, the integral is $a f(a) + \int_{-a}^{a}f(x)dx$ – David Mar 10 '13 at 03:45
  • @David, that is everybody's problem. I guess we all must develop an intuition on this. There is no pre-cooked formula that tells us when to use this or that method on integrating in each and every case, though there are many cases that beg this or that method. Practice, practice and more practice. – DonAntonio Mar 10 '13 at 03:46
  • The integral from the beginning in fact is $$af(a)+2\int\limits_0^a f(x),dx$$ – DonAntonio Mar 10 '13 at 03:48
  • Well, but $2\int_{0}^{a}f(x)dx = \int_{-a}^{a}f(x)dx$ when $f(x)$ is even? – David Mar 10 '13 at 03:49
  • Of course, @David, you're right. – DonAntonio Mar 10 '13 at 03:51