I need to find radiance, and I've found out that the integral of Planck's fuction allows me to do this. However, I don't understand the integration between 0 and infinity.
(from http://www.spectralcalc.com/blackbody/images/eq0036M.gif)
I don't understand why they've put $2k^3T^3/ h^2c^2$ before the integral symbol, as in the first line there was nothing before it. If anyone could explain this integration for me it would be appreciated. Thanks.
On
There is a ways ways smarter way to evaluate this.
$$\frac{2k^3T^3}{h^2c^2}\cdot \frac{h^3}{k^3T^3}\int_0^{+\infty} \frac{v^3\ \text{d}v}{e^{av} - 1}$$
Where I set $a = h/kT$.
Little algebra and we get
$$\alpha\int_0^{+\infty} \frac{v^3\ \text{d}v}{e^{av} - 1}$$
Where $\alpha = 2h/c^2$
Now, let's collect the exponential at the denominator
$$\alpha\int_0^{+\infty} \frac{v^3\ \text{d}v}{e^{av}(1 - e^{-av})}$$
Make use of the geometric Series
$$\frac{1}{1 - e^{-av}} = \sum_{k = 0}^{+\infty} (e^{-av})^k$$
To get
$$\alpha\int_0^{+\infty} v^3 e^{-av} \sum_{k = 0}^{+\infty} e^{-avk}\ \text{d}v$$
$$\alpha\sum_{k = 0}^{+\infty}\int_0^{+\infty} v^3 e^{-av(1 + k)}\ \text{d}v$$
Let's call $\beta = a(1+k)$ and the integral is trivial:
$$\int_0^{+\infty} v^3 e^{-\beta v}\ \text{d}v = \frac{\Gamma(4)}{\beta^4}$$
Now, $\Gamma(4) = 3! = 6$ and $b^4 = a^4(1+k)^4$ hence:
$$\alpha\sum_{k = 0}^{+\infty}\frac{6}{a^4(1+k)^4} = \frac{6\alpha}{a^4}\sum_{k = 0}^{+\infty}\frac{1}{(1+k)^4}$$
The constants are $a = h/kT$, $\alpha = 2h/c^2$ hence in front of the sum we have
$$\frac{6\cdot 2h}{c^2}\frac{k^4T^4}{h^4} = \frac{12k^4T^4}{c^2h^3}$$
Finally the sum is nothing but:
$$\sum_{k = 0}^{+\infty}\frac{1}{(1+k)^4} = \sum_{k = 1}^{+\infty}\frac{1}{(k)^4} = \zeta(4) = \frac{\pi^4}{90}$$
Hence at the end you have:
$$\frac{12k^4T^4}{c^2h^3}\frac{\pi^4}{90} = \boxed{\frac{2\pi^4}{15}\frac{k^4T^4}{c^2h^3}}$$
To end up with the $x$ as integration variable, which showed up before in the exponent of the exponential function, where it had the $h \nu /(kT)$ value.