For what values of $s$ will the integral $\int_{-\infty}^{\infty} \frac{x^se^x}{e^x+1} \ dx$ converge? For what values of $s$ will the integral $\displaystyle\int_{-\infty}^{\infty} \frac{x^se^x}{e^x+1} \,dx$ converge?
I'm having a hard time approaching the problem. It would seem that I need to simplify the fraction to get something I can work with, but how I cannot seem to do anything to get it into some nicer form. If I could find an upper bound for it and find $s$ such that the upper bound converges then this would also converge for the same $s$? I tried looking at the series expansion for $e^x$, but that didn't seem very helpful.
Edit: Managed to get $$\frac{x^se^x}{e^x+1} = \frac{x^s}{1+\frac{1}{e^x}}.$$
 A: The key thing here is to think about where potential "interesting" behavior happens.
As $x\to\infty$, you know that $\frac{e^x}{e^x+1}\to1$. So, the behavior is essentially the same as $x^s$.
As $x\to-\infty$, $e^x\to0$ and so this integrand behaves like $x^se^x$.
Finally, as $x\to0$, $\frac{e^x}{e^x+1}\to\frac{1}{2}$. So, again, the behavior here is like $x^s$.
Can you figure out what values of $s$ make these proxy integrands converge or diverge as $x$ escapes to $\pm\infty$ or tends toward $0$?
A: Lets look at two similar integrals:
$$I_1(s)=\int_\epsilon^\infty\frac{x^se^x}{e^x+1}\,dx$$
the problem is that:
$$\lim_{x\to\infty^+}\frac{e^x}{e^x+1}=1$$
so the converge of this integral is the same as the convergence of:
$$\int_\epsilon^\infty x^s\,dx$$
We could do the same thing for the negative part of your integral and notice that:
$$\lim_{x\to-\infty}\frac{x^se^x}{e^x+1}\sim x^s$$
We may also look around the axis of this function, and notice that:
$$\lim_{x\to\pm0}\frac{x^se^x}{e^x+1}=\left(\lim_{x\to\pm0}x^s\right)\left(\lim_{x\to\pm0}\frac{e^x}{e^x+1}\right)=\frac12\lim_{x\to\pm0}x^s$$
So overall the convergence of our integral is dependent on the convergence of the integral:
$$\int_{-\infty}^\infty x^s\,dx$$
However notice that (if we ignore $x\to0$ for a minute) this will only converge is $s<-1$. Now if we draw our attention to around zero we see that the function is not symmetric, put a principle value may exist.

It makes sense to break our integral up into the following:
$$\int_{-\infty}^\infty=\int_{-\infty}^{-1}+\int_{-1}^{-\epsilon}+\int_{\epsilon}^1+\int_1^\infty$$
where this is an equality for $\epsilon\to0$. Now in the domains involving infinity approximate the function as $x^s$ and for the ones involving epsilon break the integral up further into:
$$\int_{-1}^{-\epsilon}\frac{x^se^x}{e^x+1}dx=\int_{-1}^{-\epsilon}x^s\,dx-\int_{-1}^{-\epsilon}\frac{x^s}{e^x+1}dx$$
and doing the same for the other side, try to get a closed form expression then combine them and take the limit
