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For $t\in [ 0, 1 )$ is $$ \frac{xe^{tx}}{e^{x}-1}$$ integrable over $x\in (0 , \infty )$? I.e., $$ \int_{0}^{\infty} \frac{xe^{tx}}{e^{x}-1} dx < \infty?$$ How do I show this?

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4 Answers 4

As $x\to0$, $x/(e^x-1)$ approaches a finite limit. As $x\to\infty$, do a limit-comparison of the integrand to $xe^{tx}/e^x$.

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As $\frac x{e^x-1}$ as a limit when $x\to 0$ (namely $1$), the only problem is when $x\to\infty$. We have $e^x-1\sim e^x$ at $+\infty$, so $\dfrac{xe^{tx}}{e^x-1}\sim xe^{(t-1)x}$. Using Taylor's series, $$e^{(t-1)x}\leq \frac 1{1+(1-t)x+x^2(1-t)^2/2+x^3(1-t)^3/6},$$ the integral is convergent for $t\in[0,1)$.

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Doesn't the first limit go to $1$ instead of $e^{–1}$? –  Pedro Tamaroff Oct 23 '12 at 13:05
    
@PeterTamaroff Right. Fixed now. –  Davide Giraudo Oct 23 '12 at 13:14

What matters in the improper integral of a nice function (e.g. elementary function) is the existence of singularities. In a broad sense, there are two kinds of singularities that counts.

  1. A point where the integrand does not behave well. For example, the function can explode to infinite or oscillate infinitely.

  2. A point at infinity. That is, $\pm \infty$.

Away from singularities, the behavior of the function is quite under control, allowing us to concentrate our attention on those singularities.

There is a basic method to establish the convergence (or divergence) of the integral near each singularity point. In many cases, except for the oscillatory case, you can find a dominating function that determines the order of magnitude of the function near the point. If the dominating function is easy to integrate, then you can make a comparison with this dominating function to conclude the convergence behavior.

For example, let us consider

$$ \int_{0}^{\frac{\pi}{2}} \tan^2 x \, dx \quad \text{and} \quad \int_{0}^{\infty} \frac{x^2 e^{-x}}{1+x^2} \, dx. $$

We can easily check that $\tan^2 x$ is bounded below by $(x-\frac{\pi}{2})^{-2}$ near the singularity $x = \frac{\pi}{2}$ and $x^2 e^{-x} / (1 + x^2)$ is bounded above by $e^{-x}$ near the singularity $x = \infty$. Then

$$ \int_{\frac{\pi}{2}-\delta}^{\frac{\pi}{2}} \tan^2 x \, dx \geq \int_{\frac{\pi}{2}-\delta}^{\frac{\pi}{2}} \left(x - \frac{\pi}{2}\right)^{2} \, dx = \infty$$

for sufficiently small $\delta > 0$ and

$$ \int_{R}^{\infty} \frac{x^2 e^{-x}}{1+x^2}\,dx \leq \int_{R}^{\infty} e^{-x} \, dx < \infty$$

for sufficiently large $R > 0$. Thus we find that the former diverges to $\infty$ and the latter converges.

In our example, there are two seemingly singular points, namely $x = 0$ and $x = \infty$. At $x = 0$, we find that

$$\lim_{x \to 0} \frac{x e^{tx}}{e^x - 1} = 1.$$

This means that this singularity is removable, in the sense that the function can be extended in a continuous manner to this point. Thus we need not count this point and we can move our attention to the point at infinity.

To establish the convergence (or possibly divergence) of the integral near $x = \infty$, we write

$$ \frac{x e^{tx}}{e^x - 1} = \frac{x}{1 - e^{-x}} e^{-(1-t)x}. $$

It is clear that for sufficiently large $x$, the term $\frac{x}{1 - e^{-x}}$ is bounded above by some constant $C > 0$. Thus the dominating function is $e^{-(1-t)x}$ and

$$ \int_{R}^{\infty} \frac{x e^{tx}}{e^x - 1} \, dx \leq \int_{R}^{\infty} C e^{-(1-t)x} \, dx < \infty$$

for large $R$. Therefore the improper integral converges.

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Thanks for your answers and especially for this limit method. But in this way, I actually found a simpler bound, namely, the following: observe $$ \frac{xe^{tx}}{e^{x}-1}=\frac{xe^{(1/2)(t-1)x}}{1-e^{-x}}e^{(1/2)(t-1)x} $$ $$\frac{xe^{(1/2)(t-1)x}}{1-e^{-x}}<M$$ for a constant $M>0$ as $$ \frac{xe^{(1/2)(t-1)x}}{1-e^{-x}} $$ is continuous and the limits for $x\to 0$ and $x \to \infty$ are finite. And this can be directly used for the integrability of the function.

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