I got stuck at a homework question:
I shall proof that the improper integral exists for:
$\frac{x^3}{e^x-1}$ between 0 and infinity.
So the technique that I know is to split up integrals, integrate the function and let one boundary approach to the point that makes trouble.
But after I even tried to solve the indefinite integral of this by an online calculator - which didn't work - I don't have an idea how to continue.
So I am basically looking for the technique, the keyword or a first step that I could work with to solve the integral. I am not looking for a solution !
The usual intention after lots of these questions is that the two boundaries of the splitted integral will be [0,1] and [1,infinity]. But then I get stuck unable to solve the integral to analyse the limits as n approaches 0,1 or infinity Integral in symbolab
You can split the integral in the following way: $$ \int \limits_0^\infty \frac{x^3}{\mathrm{e}^x-1} \, \mathrm{d} x = \int \limits_0^1 \frac{x^3}{\mathrm{e}^x-1} \, \mathrm{d} x + \int \limits_1^\infty \frac{x^3}{\mathrm{e}^x-1} \, \mathrm{d} x \, . $$ For the first integral you can use the inequality $\mathrm{e}^x -1 \geq x$ for $x \in \mathbb{R}$ . For $x\geq0$ (which we need here) this is a direct consequence of the series representation of the exponential function: $$ \mathrm{e}^x - 1 = \sum \limits_{n=1}^\infty \frac{x^n}{n!} \geq \frac{x^1}{1!} = x \, . $$ Alternatively, the line $y = x$ is the tangent line to the graph of $x \mapsto \mathrm{e}^x - 1$ at $x=0$. Since this function is convex, the inequality follows.
For the second integral $ \mathrm{e}^{x} - 1 \geq \frac{1}{2} \mathrm{e}^x \, , \, x \geq \ln(2),$ should help. This can be seen by rearranging: $$ x \geq \ln(2) \Leftrightarrow \mathrm{e}^x \geq 2 \Leftrightarrow \frac{1}{2} \mathrm{e}^x \geq 1 \Leftrightarrow \mathrm{e}^x - 1 \geq \frac{1}{2}\mathrm{e}^x \, .$$
Since all integrals are non-negative, you only need to estimate the two parts from above. You do not have to evaluate them exactly. You can use the first inequality to find $$ \int \limits_0^1 \frac{x^3}{\mathrm{e}^x-1} \, \mathrm{d} x \leq \int \limits_0^1 \frac{x^3}{x} \, \mathrm{d} x = \int \limits_0^1 x^2 \, \mathrm{d} x = \frac{1}{3} < \infty $$ and proceed in the same manner with the second integral.