How to Solve the Definite Integral for Deriving Saha's Equation?

[Ichimaru]

I'm trying to derive Saha's equation, and I've come up against this definite integral, which I can't seem to find anywhere and may not even be doable; I'm not sure.

\begin{equation}

\int\limits_0^\infty \frac{x^{2}}{e^{x}+1}dx

\end{equation}

Can anyone help? Thanks!

 

[maajdl]

Mathematica gives: 3/2 Zeta[3] = 1.80309 .

Where Zeta is the Riemann zeta function (http://en.wikipedia.org/wiki/Riemann_zeta_function)

 

[D H]

It's doable. That's essentially the Fermi-Dirac integral F₂(0).

 

[arildno]

Presumably, traipsing along a different curve in the complex plane, using Cauchy's theorem, might give you the answer.

 

[Saitama]

Here is a more elementary solution:

Rewrite the given integral as:
$$I=\int_0^{\infty} \frac{x^2e^{-x}}{1+e^{-x}}\,dx$$
Use the substitution $e^{-x}=t$ to obtain:
$$I=\int_0^1 \frac{\ln^2 t}{1+t}\,dt$$
Next, use integration by parts in the following way:
$$I=\left(\ln^2t \ln(1+t)\right|_0^1-2\int_0^1 \frac{2\ln t\ln(1+t)}{t}\,dt$$
Notice that the first term is zero, so we are left with:
$$I=-2\int_0^1 \frac{\ln t \ln(1+t)}{t}\,dt$$
Since,
$$\ln(1+t)=-\sum_{k=1}^{\infty} \frac{(-1)^kt^k}{k}$$
Hence,
$$I=2\sum_{k=1}^{\infty} \frac{(-1)^k}{k}\int_0^1 \frac{t^k\ln t}{t}\,dt$$
It is not difficult to show that
$$\int_0^1 \frac{t^k\ln t}{t}\,dt =-\frac{1}{k^2}$$
From above, we get:
$$I=-2\sum_{k=1}^{\infty} \frac{(-1)^k}{k^3}$$
It can be easily shown that
$$\sum_{k=1}^{\infty} \frac{(-1)^k}{k^3}=-\frac{3}{4}\zeta(3)$$
Hence, the final answer is:
$$I=\frac{3}{2}\zeta(3)$$
which agrees with post #2.

I hope that helps.

 

[arildno]

Cool, Pranav-Arora!

 

[Saitama]

Cool, Pranav-Arora!

Thank you! :)