# Pesky integral involving exponentials

1. Dec 16, 2007

### T-7

Hi,

I am trying to see how, in my textbook,

$$\frac{1}{3}\int_{0}^{\infty}x^3 ln(1-e^{-x}) dx = \frac{1}{3}\left[ x^3 ln(1-e^{-x}) \right]_{0}^{\infty} - \frac{1}{3}\int_{0}^{\infty} \frac{x^3 dx}{e^x - 1} = - \frac{\pi^4}{45}$$

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:

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

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

Cheers!

Last edited: Dec 16, 2007
2. Dec 16, 2007

### VeeEight

Hopefully this is what you are asking:

if you think of integration by parts as being f g' = f g - $$\int g f$$, then they are letting ln (1-e^-x) be the f in this case.

Taking the derivative gives $$1/1 - e^-x$$(-e^-x)(-1) by the chain rule. The two negatives cancel out to give

$$1/1 - e^-x$$(e^-x).

Bring e^-x = $$1/e^x$$ so your f' will be $$1/e^x (1 - e^-x)$$

Multiply through to get f' = $$1/e^x - 1$$

3. Dec 16, 2007

### T-7

Of course. I just wasn't handling the log properly when I differentiated it (!).

Cheers.