Help with Integral: Solve for Power Difference

[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)/(\lambda⁵(e^hc/\lambdakT-1))d\lambda

if we make the change of variable x=hc/\lambdakT

(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/\lambdakT 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)/(\lambda⁵(e^hc/\lambdakT-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/\lambdakT

(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/\lambdakT 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,
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.