How do I integrate this exponential function with a constant exponent value?

I'm trying to integrate $$(1)$$, regarding $$T$$:

$$\left( \frac{\partial f_s}{\partial T}\right)_{E/T}= -\frac{1}{HT} f_\alpha \tag{1}$$

where $$\displaystyle f_\alpha= \frac{1}{e^{E/T}+1}$$ .

• $$f_\alpha$$ and $$f_s$$ are distribution functions.
• $$E$$ represents energy/momentum;
• $$T$$ represents temperature
• The subscript $$E/T$$ represents a constant ration of these quantities.

Questions:

• Even though $$E/T$$ is constant, the individual quantities, $$E$$ and $$T$$ are still variables and therefore the $$T$$ in $$f_\alpha$$ must still be integrated. Is this statement correct? Or should I ignore the $$f_\alpha$$ term?
• If the statement is correct, how do I integrate the $$(1)$$? I've attempted integration by parts but I keep on getting extra $$T$$ terms, I might have done something wrong.

Attempt:

When trying to integrate $$(1)$$ I get:

$$dv = -\frac{1}{HT} \hspace{5mm} \hspace{5mm}; u = \frac{1}{e^{E/T}+1}$$ so $$v= -\frac{1}{H} \ln(T) \hspace{5mm}; \hspace{5mm} \frac{du}{dT}= \frac{e^{E/T} E}{(e^{E/T}+1)^2 T^2}$$

and using the formula for integration by parts $$\int u dv = uv - \int v du$$ , I obtain:

$$f_s= -\frac{1}{H} \ln(T) \frac{1}{e^{E/T}+1} + \frac{1}{H} \int \ln(T)\frac{e^{E/T}E}{(e^{E/T}+1)^2 T^2}$$

This seems to have made the integration more difficult instead of simplifying it.

If $$T$$ is a variable and $$E/T$$ is constant (say $$k$$), it means $$E$$ is a variable as well. Therefore $$f_{\alpha}$$ and $$f_s$$ are functions of two variables: $$(T,E)$$. Your derivative $$\left( \frac{\partial f_s}{\partial T} \right)_{E/T}$$ is thus actually the directional derivative of $$f_s$$ along the direction given by the vector $$(T,kT)$$ (for a certain $$k$$) in the $$TE$$ plane.
All that to say that, in this direction, $$E/T$$ is constant and therefore you can replace it by $$k$$ in your expression for $$f_{\alpha}$$ and only consider the $$T$$ in the fraction in front of $$f_{\alpha}$$ when you integrate $$f_s$$.