# Homework Help: Help with ingetral

1. Sep 11, 2008

### phyguy321

So i dont really know where to paste this.

I'm reading a example in my modern physics book and i dont understand what they did.

$$\int$$ (2(pi)hc2)/($$\lambda$$5(ehc/$$\lambda$$kT-1))d$$\lambda$$

if we make the change of variable x=hc/$$\lambda$$kT

(2(pi)k4T4)/(c2h3)$$\int$$((x3)/(ex-1))dx

so how did they get to this:

(2(pi)k4T4)/(c2h3)$$\int$$((x3)/(ex-1))dx

when I solve for $$\lambda$$ in x=hc/$$\lambda$$kT and plug it into the first ingetral I get:

(2(pi)x5k5T5)/(h4c3)

basically, i'm off by 1 power for everything except for x (that i'm off by 2).

can someone explain this?

2. Sep 11, 2008

### HallsofIvy

That's
$$\int \frac{2\pi hc^2}{\lambda^5 e^{\frac{hc}{\lambda}kT^{-1}-1} d\lambda$$?

Did you forget to replace "$d\lambda$" as well?
$$x= \frac{hc}{kT}\lambda^{-1}$$
so $$dx= -\frac{hc}{kT}\lambda^{-2}d\lambda$$
Solving that for $d\lambda$,
$$d\lambda= -\frac{kT}{hc}\lambda^2 dx[/itex] and since $\lambda= (hc)/(kT) x^{-1}$, $\lambda^2= (h^2c^2)/(k^2T^2) x^{-2}$ giving [tex]d\lambda= \frac{hc}{kT}x^{-2}dx$$.