Proving that two integral are equal (changing variable of integration)

Question

I want to prove that $$ \frac{1}{\beta^2}\int_0^\beta d\tilde{\beta} \frac{n^2 \tilde{\beta}}{e^{\tilde{\beta} n}+1}=-\int_{n}^\infty dz \frac{z}{e^{\beta z}+1} $$ how do we change the variable of integration?

Answer 1

This is clearly false as you stated it, since (for $\beta>0$) the left-hand side is positive and the right-hand side is negative.

However, substituting $z = \left(\dfrac n\beta\right)\tilde\beta$ gives $$\frac{n^2}{\beta^2}\int_0^\beta \frac{\tilde\beta\,d\tilde\beta}{e^{\tilde\beta n}+1} = \int_0^n \frac{z\,dz}{e^{\beta z}+1}.$$