how to integrate $\int_{-\infty}^\infty\frac{dx}{x^2(1+e^x)}$? How can I integrate this difficult form of integration, I tried many times to solve it by parts or by substitution but I didn't get it.
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See this and this. There's no simple closed form. – Rushabh Mehta Dec 21 '19 at 18:18
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Do you have any specific bounds? Are you trying to solve this with something else? Given more context, this question might be reasonable to answer, but as is, the only answer is you can't, as far as closed forms are concerned. – Simply Beautiful Art Dec 21 '19 at 18:25
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the bounds are from negative infinity to infinity – Mario Dec 21 '19 at 18:28
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If the bounds are from $-\infty$ to $\infty$, I think you better edit your question. – Ali Shadhar Dec 21 '19 at 19:57
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1Please correct it. I don't know how to write the boundaries. – Mario Dec 21 '19 at 20:32
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I'm glad to see someone added the integral's limits for you, after you said you didn't know how. Here is a guide. – J.G. Dec 21 '19 at 20:49
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Using $\int_{\Bbb R}f(x)dx=\int_0^\infty[f(x)+f(-x)]dx$,$$\int_{\Bbb R}\frac{1}{x^2}\frac{1}{1+e^x}dx=\int_0^\infty\frac{1}{x^2}\underbrace{\left[\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right]}_1dx=\int_0^\infty\frac{dx}{x^2}=\infty.$$
J.G.
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I don't understand this step $\int_0^\infty\frac{1}{x^2}\underbrace{\left[\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right]}_1dx$ – Mario Dec 21 '19 at 20:52
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Let u = -x => Integral from -infinity to infinity of du/(u^2*(1+e-u)) = integral of e^u/(u^2(1+e^u)) du.
Take the original integral and add it to this one to get(switch u to x): 2I = integral (e^x+1)/x^2 *1/(e^x+1) du=integral of 1/x^2 dx.
Integral from -infinity to infinity of 1/x^2 dx = 2*int from 0 to infinity of 1/x^2 dx = -1/x from 0 to infinity which diverges.
So the original integral equals divergent/2 which still diverges.
