Change of variables in Planck's radiation integral

[phyguy321]

So i don't really know where to paste this.

I'm reading a example in my modern physics book and i don't understand what they did.

$$\int$$ (2(pi)hc²)/($$\lambda$$⁵(e^{hc/$$\lambda$$kT}-1))d$$\lambda$$

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

(2(pi)k⁴T⁴)/(c²h³)$$\int$$((x³)/(e^x-1))dx

so how did they get to this:

(2(pi)k⁴T⁴)/(c²h³)$$\int$$((x³)/(e^x-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?

 

[HallsofIvy]

So i don't really know where to paste this.

I'm reading a example in my modern physics book and i don't understand what they did.

$$\int$$ (2(pi)hc²)/($$\lambda$$⁵(e^{hc/$$\lambda$$kT}-1))d$$\lambda$$

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

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

(2(pi)k⁴T⁴)/(c²h³)$$\int$$((x³)/(e^x-1))dx

so how did they get to this:

(2(pi)k⁴T⁴)/(c²h³)$$\int$$((x³)/(e^x-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?

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 $d\lambda$,
[tex]d\lambda= -\frac{kT}{hc}\lambda^2 dx[/itex]<br /> and since $\lambda= (hc)/(kT) x^{-1}$, $\lambda^2= (h^2c^2)/(k^2T^2) x^{-2}$ giving<br /> $$d\lambda= \frac{hc}{kT}x^{-2}dx$$.[/tex]