IBP Struggles: Solving Integrals of \frac{x^2}{e^x+1} \& \frac{x^3}{e^x+1}

[Eats Dirt]

Homework Statement

Find the Integrals of \frac{x^2}{e^x+1}\\ \frac{x^3}{e^x+1}

Homework Equations

Integration by parts

The Attempt at a Solution

I did IBP twice and it seemed to just get bigger and uglier and now I am stuck. I found the solutions online of the integrals but still do not know how to do them myself.

 

[verty]

We had a similar thread not too long ago, have a look here.

 

[Saitama]

As said in the other thread, you need polylogarithms to express the final result. Instead, the definite integral from 0 to infinity can be easily evaluated and you have the following general result:

$$\int_0^{\infty} \frac{x^{s-1}}{e^x+1}\,dx=\left(1-2^{1-s}\right)\zeta(s)\Gamma(s)$$

 

[Eats Dirt]

Hey, sorry for the atrociously late reply. I am not familiar with the two functions in the general form you posted. I recognize the one as a gamma function but have never seen the other.

 

[Orodruin]

It is the Riemann zeta function: http://en.wikipedia.org/wiki/Riemann_zeta_function
$$
\zeta(s) = \sum_{n=1}^\infty n^{-s}
$$

 

[Eats Dirt]

As said in the other thread, you need polylogarithms to express the final result. Instead, the definite integral from 0 to infinity can be easily evaluated and you have the following general result:

$$\int_0^{\infty} \frac{x^{s-1}}{e^x+1}\,dx=\left(1-2^{1-s}\right)\zeta(s)\Gamma(s)$$

I have another integral where the term in the exponential is divided by a constant T so it takes the form of \frac{x^2}{e^\frac{x}{T}+1}\\ \frac{x^3}{e^\frac{x}{T}+1}

Is there some way to get a new variation of this formula that can take these constants into account?

 

[Orodruin]

Change variables to $y = x/T$.

 

[Eats Dirt]

Change variables to $y = x/T$.

Then we just use the same integration formula from pranav? Is this because the limits of integration involve an infinity so any constant T in the denominator will have no effect?

 

[Orodruin]

It will have an effect, namely multiplying the result by a factor $T^{n+1}$, where n is the exponent of x, since
$$
x^n\,dx = T^{n+1} y^n\,dy
$$