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Can someone help me evaluate this: $$\int \frac{\cos^2(x)}{1 + \text{e}^x}dx\;?$$

I need it for determining whether the improper integral $\int_0^\infty {\frac{{\cos^2{{(x)}}}}{{1 + {{\text{e}}^x}}}}$ is convergent or not.

Using the software Maple is not possible to determine symbolically, but it is possible to evaluate it numerically:

$$\int_0^\infty {\frac{{\cos^2{{(x)}}}}{{1 + {{\text{e}}^x}}}}=0.3831765832$$

So apparently it converges, but how would show whether such an integral converges or diverges?

Using Wolfram Alpha I get symbolic result in terms of hypergeometric functions, but I want to know if it is possible to calculate in terms of elementary functions.

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A nondecreasing, bounded function always converges. So let $f(x) = \frac{\cos^2x}{1+e^x}$. Clearly $f(x) > 0\, \forall x$, so $\int_0^t f(x) dx$ is nondecreasing in $t$. We need to show it's bounded. $\int _0^t f(x)dx < \int_0^\infty \frac{dx}{1+e^x} < \int_0^\infty \frac{dx}{e^x} = 1$. Done.

It's unlikely to be expressible as an elementary function. This can be determined using Risch's algorithm.

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  • $\begingroup$ great response thanks !, $\endgroup$ – mathsalomon May 13 '15 at 9:15
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Since $$\mathcal{L}\left(\cos^2 x\right)=\frac{s^2+2}{s(s^2+4)}\tag{1}$$ and: $$\int_{0}^{+\infty}\frac{\cos^2 x}{1+x^2}\,dx = \sum_{n\geq 1}(-1)^{n+1}\int_{0}^{+\infty}\cos^2(x)\,e^{-nx}\,dx \tag{2}$$ we have: $$\int_{0}^{+\infty}\frac{\cos^2 x}{1+x^2}\,dx = \sum_{n\geq 1}(-1)^{n+1}\frac{2+n^2}{n^3+4n}=\frac{\log 2}{2}+\frac{1}{2}\sum_{n\geq 1}(-1)^{n+1}\frac{n}{n^2+4}.\tag{3}$$ The last series can be evaluated in terms of values of the digamma function: $$\psi(x)=\frac{d}{dx}\log\Gamma(x).$$

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  • $\begingroup$ Thanks, but I do not study series. $\endgroup$ – mathsalomon May 13 '15 at 9:14
  • $\begingroup$ @mathsalomon: in such a case, how have you defined improper integrals? $\endgroup$ – Jack D'Aurizio May 13 '15 at 10:02
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The given integral has a positive integrand less than $$\frac{1}{(1+x^2)}$$ whose integral from $0$ to $\infty$ is $\frac{\pi}{2}$ so it converges. But to evaluate the integral is a different story.

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  • $\begingroup$ Thanks, this answer is more simple and useful. $\endgroup$ – mathsalomon May 13 '15 at 9:13

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