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Hard Integral

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Homework Statement



Find the Integrals of [tex] \frac{x^2}{e^x+1}\\ \frac{x^3}{e^x+1} [/tex]


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.
 

Answers and Replies

  • #2
verty
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We had a similar thread not too long ago, have a look here.
 
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  • #3
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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:

$$\int_0^{\infty} \frac{x^{s-1}}{e^x+1}\,dx=\left(1-2^{1-s}\right)\zeta(s)\Gamma(s)$$
 
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  • #4
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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.
 
  • #6
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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 [tex] \frac{x^2}{e^\frac{x}{T}+1}\\ \frac{x^3}{e^\frac{x}{T}+1} [/tex]

Is there some way to get a new variation of this formula that can take these constants into account?
 
  • #7
Orodruin
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Change variables to ##y = x/T##.
 
  • #8
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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?
 
  • #9
Orodruin
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It will have an effect, namely multiplying the result by a factor ##T^{n+1}##, where n is the exponent of x, since
$$
x^n\,dx = T^{n+1} y^n\,dy
$$
 

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