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I am trying to see how, in my textbook,

[tex]\frac{1}{3}\int_{0}^{\infty}x^3 ln(1-e^{-x}) dx

= \frac{1}{3}\left[ x^3 ln(1-e^{-x}) \right]_{0}^{\infty} - \frac{1}{3}\int_{0}^{\infty} \frac{x^3 dx}{e^x - 1}

= - \frac{\pi^4}{45}[/tex]

The integration by parts is clear enough, it's clear that first term of the middle expression evaluates at zero, and I recognise the standard integral that has been used to resolve the remaining integral... but how has exponent of e in the second integral suddenly gone from negative to being positive, without any change in the limits etc?

I would have written:

[tex]\frac{1}{3}\int_{0}^{\infty}x^3 ln(1-e^{-x}) dx

= \frac{1}{3}\left[ x^3 ln(1-e^{-x}) \right]_{0}^{\infty} - \frac{1}{3}\int_{0}^{\infty} \frac{x^3 dx}{1 - e^{-x}} = \frac{1}{3}\int_{0}^{\infty} \frac{x^3 dx}{e^{-x} - 1}[/tex]

I'm sure it's obvious, but I'm not seeing it just now.

Cheers!

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# Homework Help: Pesky integral involving exponentials

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