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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

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  • $\begingroup$ Do you just want to prove the integral exists, or do you want to find the value? If you just want to prove it exists you can bound it with nicer functions in the area of $0$ and $\infty$ instead of integrating it directly. $\endgroup$ – Ross Millikan Jul 20 '18 at 16:16
  • $\begingroup$ Just the existence $\endgroup$ – LurioTabasco Jul 20 '18 at 16:17
  • $\begingroup$ Then ComplexYetTrivial has it. $\endgroup$ – Ross Millikan Jul 20 '18 at 16:18
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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.

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  • $\begingroup$ Basically I do not have a clue yet, where your inequalities come from and how they are useful - I guess I am lacking some kind of knowledge to use your answer. What is this technique called? $\endgroup$ – LurioTabasco Jul 20 '18 at 16:21

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