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I have a problem with my solution. $$\int_{0}^{\infty}\frac{2x}{e^{x}-e^{-x}}dx=\int_{0}^{a}\frac{2x}{e^{x}-e^{-x}}dx+\int_{a}^{\infty}\frac{2x}{e^{x}-e^{-x}}dx$$

So in first integral, if I compare it with : $ \frac{1}{\sqrt x}$ I have: $$\lim_{x \to 0^+ }\frac{\frac{2x}{e^{x}-e^{-x}}}{\frac{1}{\sqrt x}} =0$$ and I also know that $\int_{0}^{a}\frac{1}{\sqrt x}$ converges. And I can conclude that my integral also converges. But if I choose to compare it with $\frac{1}{x^2}$ , again I get $$\lim_{x \to 0^+ }\frac{\frac{2x}{e^{x}-e^{-x}}}{\frac{1}{x^2}} =0$$ but now I conclude that my integral diverges together with $\int_{0}^{a}\frac{1}{x^2}$. What the right solution and what is wrong with my solution??! Thanks a lot!

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Expand the denominator as follows:

$$\begin{align}\int_0^{\infty} dx \frac{2 x}{e^x-e^{-x}} &= 2 \int_0^{\infty} dx \frac{x \, e^{-x}}{1-e^{-2 x}} \\ &= 2 \sum_{k=0}^{\infty} \int_0^{\infty} dx\, x \, e^{-(2 k+1) x}\\ &= 2 \sum_{k=0}^{\infty} \frac1{(2 k+1)^2}\\ &= 2 \left [\sum_{k=1}^{\infty} \frac1{k^2} - \sum_{k=1}^{\infty}\frac1{(2 k)^2} \right ] \\ &= \frac{3}{2} \sum_{k=1}^{\infty} \frac1{k^2} \\ &= \frac{\pi^2}{4}\end{align}$$

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  • $\begingroup$ Thank you, but what is wrong with my solution? $\endgroup$ – Laura Feb 10 '14 at 7:08
  • $\begingroup$ $$e^x - e^{-x} \sim 2 x \quad (x \to 0)$$ $\endgroup$ – Ron Gordon Feb 10 '14 at 7:10
  • $\begingroup$ Thanks again, can you please explain why? $\endgroup$ – Laura Feb 10 '14 at 7:20
  • $\begingroup$ @Laura: from the Taylor expansion of an exponential: $e^x = 1+x+x^2/2+\cdots$ $\endgroup$ – Ron Gordon Feb 10 '14 at 7:42

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