Pesky integral involving exponentials

[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<br /> = \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} <br /> = - \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<br /> = \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!

 

[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

 

[T-7]

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

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

Cheers.