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- #1

- Jan 31, 2012

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How do you show that $\displaystyle \int_{0}^{\infty} \Big( \frac{x^{a-1}}{\sinh x} - x^{a-2} \Big) \ dx $ converges for $0 \le a <1$?

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- Thread starter Random Variable
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- Thread starter
- #1

- Jan 31, 2012

- 253

Last edited:

- Feb 13, 2012

- 1,704

Remembering that ...

$\displaystyle \frac{1}{\sinh x} = \frac{1}{x} - \frac{x}{6} + \frac{7}{360}\ x^{2} - ...$ (1)

... You derive that is...

$\displaystyle \frac{x^{a-1}}{\sinh x} - x^{a-2} = x^{a-2}\ (1 - \frac{x^{2}}{6} + \frac{7}{360}\ x^{4} + ... -1) = - x^{a}\ (\frac{1}{6} - \frac{7}{360}\ x^{2} + ...)$ (2)

... so that the function around x=0 is $\displaystyle \sim - \frac{x^{a}}{6}$. For 'large x' the function is $\displaystyle \sim - x^{a-2}$ so that, combining the two conditions, the integral should converge for $-1 < a < 1$...

Kind regards

$\chi$ $\sigma$