# Double Integral Evaluation

1. Aug 26, 2009

### Nabeshin

I have come across the following integral which I need to compute:

$$\int_0^{t_1} \int_{\nu_0}^{\infty} \left(\frac{h \nu ^3}{c^2}\right) \frac{1}{e^{\frac{h\nu}{k T(t)}}-1} d\nu dt$$

The problem is, since the inner integral cannot be computed analytically, I have to resort to numerical methods. But, I don't think I can use numerical methods since the inner integral contains the variable t and as such is not purely a function of nu. Any idea how to evaluate something like this?

One thought I had was writing a taylor series for the inner integral treating the time dependent factor as a constant, and just taking the first few terms and proceeding with the integration. This seems rather crude though, so I wonder: is there a better way?

2. Aug 26, 2009

### Marin

3. Aug 26, 2009

### Cyosis

I am not so sure if the usual tricks are going to work seeing as the lower bound is not zero. We could attempt it by using power series.

Substitute $u=\frac{h \nu}{kT}$

\begin{align*} I=\int_{\nu_0}^{\infty} \left(\frac{h \nu ^3}{c^2}\right) \frac{1}{e^{\frac{h\nu}{k T(t)}}-1} d\nu & = \frac{h}{c^2}\left(\frac{kT}{h}\right)^4 \int_{\frac{h \nu_0}{kT}}^\infty \frac{u^3}{e^u-1}du \end{align*}

Ignoring the constant in front of the integral, using the geometric series and using the substitution $z=nu$ we get:

\begin{align*} \int_{\frac{h \nu_0}{kT}}^\infty \frac{u^3}{e^u-1}du & = \int_{\frac{h \nu_0}{kT}}^\infty u^3 e^{-u} \frac{1}{1-e^{-u}} du \\ & = \int_{\frac{h \nu_0}{kT}}^\infty u^3 e^{-u} \sum_{n=0}^\infty e^{-nu} du \\ & = \int_{\frac{h \nu_0}{kT}}^\infty u^3 \sum_{n=1}^\infty e^{-nu} du \\ & = \sum_{n=1}^\infty \int_{\frac{h \nu_0}{kT}}^\infty u^3 e^{-nu} du \\ & = \sum_{n=1}^\infty \left(\frac{1}{n^4}\int_{na}^\infty z^3 e^{-z} dz \right) \\ & = \sum_{n=1}^\infty \frac{e^{-an}(a^3n^3+3a^2n^2+6an+6)}{n^4} \end{align*}

With $a=\frac{h \nu_0}{k T}$.

Plugging everything back into the original integral:
\begin{align*} \int_{\nu_0}^{\infty} \left(\frac{h \nu ^3}{c^2}\right) \frac{1}{e^{\frac{h\nu}{k T(t)}}-1} d\nu & =\frac{h}{c^2}\left(\frac{k T}{h}\right)^4 \sum_{n=1}^\infty \frac{e^{-\frac{n h \nu_0}{k T}}}{n^4}\left[ \left(\frac{h \nu_0}{k T}\right)^3 n^3+3\left(\frac{h \nu_0}{k T}\right)^2 n^2+\frac{6 nh \nu_0}{k T}+6\right] \\ & =\frac{k^4}{c^3 h^3} \sum_{n=1}^\infty \frac{e^{-\frac{n h \nu_0}{k T}}}{n^4}\left[ \left(\frac{h \nu_0}{k}\right)^3 T n^3+3\left(\frac{h \nu_0}{k}\right)^2 T^2 n^2+\frac{6 nh \nu_0}{k}T^3+6T^4\right] \\ & = \frac{k \nu_0^3 T}{c^3} Li_1(e^{-\frac{h \nu_0}{k T}})+\frac{3 k^2 \nu_0^2T^2}{h c^3} Li_2(e^{-\frac{h \nu_0}{k T}})+\frac{6 \nu_0 k^3 T^3}{c^3 h^2}Li_3(e^{-\frac{h \nu_0}{k T}})+\frac{6k^4 T^4}{c^3 h^3}Li_4(e^{-\frac{h \nu_0}{k T}}) \end{align*}

Here I used the definition of the polylogarithm, $Li_k(z)=\sum_{n=1}^\infty \frac{z^n}{n^k}$.

Using $Li_1(z)=-\log(1-z)$ we get:

I= \frac{\nu_0^4h}{c^3}-\frac{k \nu_0^3T}{c^3}\log\left(e^{\frac{h \nu_0}{kT}}-1\right)+\frac{3 k^2 \nu_0^2T^2}{h c^3} Li_2(e^{-\frac{h \nu_0}{k T}})+\frac{6 \nu_0 k^3 T^3}{c^3 h^2}Li_3(e^{-\frac{h \nu_0}{k T}})+\frac{6k^4 T^4}{c^3 h^3}Li_4(e^{-\frac{h \nu_0}{k T}}) \end{align*}

If T(t) is simple enough it may be possible to evaluate the second integral.

ps. I wouldn't be surprised if I made some errors regarding constants.

Last edited: Aug 26, 2009
4. Aug 26, 2009

### Nabeshin

Thank you both for the replies,

Marin: I did not think it could be computed analytically because mathematica got nowhere with it, and its basic form didn't lend itself to any methods I am familiar with (I am not familiar with contour integrals, such as those referenced in your post).

Cyosis: This is more what I had in mind for evaluating it, although I am again not familiar with the polylogarithms you used to arrive at an (apparently) closed form for a solution. I will study this a little more and see if I can figure it out using this method.

Also, T(t) does not have a very simple form, in fact it is:
$$T(t)=T_0 \frac{ \left(1-Tanh\left[\frac{a_0 t}{c}\right] \cos{\theta}\right) }{\sqrt{1-\frac{\left(c Tanh\left[\frac{a_0 t}{c}\right]\right)^2}{c^2}}}$$

I really don't need an analytical solution though, so long as I get a number to a reasonable amount of accuracy I'm satisfied. Certainly, I should think, numerical methods will be in order for the outer integral.