What is one way to show that $$\int_{-\infty}^\infty \frac{x^4 e^{x/2}}{e^x+1}\, dx$$ converges? I see it is an even function so it is enough to show that it converges for $[0,1]$. Moreover, it is nonnegative and decreasing after some point so I can use the integral test. But then it reduces to showing that $$\sum_0^\infty \frac{x^4 e^{x/2}}{e^x+1} < \infty,$$ but that is difficult.
Alternatively we can use integration by substitution. Letting $u = e^x+1, du = e^x dx$, we have $[\log(u-1)]^4 = x^4$ and $u \to \pm \infty$ as $x \to \pm \infty$. Therefore we can rewrite the integral as \begin{align*} \int_{-\infty}^\infty \frac{x^4 (e^x)^{1/2}}{e^x+1} \frac{e^x}{e^x}\, dx &= \int_{-\infty}^\infty \frac{[\log(u-1)]^4u^{1/2}}{u(u-1)}\, du \\ &= \int_{-\infty}^\infty \frac{[\log(u-1)]^4}{u^{3/2} - u^{1/2}}\, du. \end{align*} How can I proceed from here? I was thinking to use the $p$-series test but am unsure. Thanks. The other alternative is the ratio test but I think that would be inconclusive.
Split the integral up into two: $x>0$ and $x<0$. In the first case, the integrand is less than $x^4 e^{-x/2}$ which has a finite value
$$\int_0^{\infty} dx \, x^4 \, e^{-x/2} = 2^5 \cdot 4!$$
The the second case, sub $-x \leftarrow x$ and you get a similar bound.