# How can I solve special Fermi-Dirac integral at Physics?

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1. Feb 22, 2016

### msenay

1. The problem statement, all variables and given/known data
I need to solve this integral,

$$\int _{-\infty }^{\infty }x\left( \dfrac {1} {1-e^{-x}}+\dfrac {1} {1+qe^{-x}}\right) dx$$

My advisor said its solution will be zero. But i havent improved it yet. There is important case. This integral is divergent at x=0. So, i should separate two parts to teh integral. First part should be from -infinty to zero and second one should be from zero to infinty.

2. Relevant equations
I solved this integral $$\int _{-\infty }^{\infty }x^{2}\left( \dfrac {1} {1-e^{-x}}+\dfrac {1} {1+qe^{-x}}\right) dx=\sum _{n=1}^{\infty }\dfrac {1} {n^{3}}-\sum _{n=1}^{\infty }\dfrac {\left( -q^{-1}\right) ^{n}} {n^{3}}$$ This integral is even function (0 < q < 1). So, i can change the integral from zero to infinty instead of from -infinty to infinty.

3. The attempt at a solution
How can i solve this integral? Do you have any idea ?

2. Feb 22, 2016

### Samy_A

Since $\displaystyle \lim_{x\rightarrow +\infty} (\dfrac {1} {1-e^{-x}}+\dfrac {1} {1+qe^{-x}})=2$, I don't see how the integrals can converge.

The series $\displaystyle \sum _{n=1}^{\infty }\dfrac {\left( -q^{-1}\right) ^{n}} {n^{3}}$ doesn't converge for $0<q<1$.

Question: which function is even?

Last edited: Feb 22, 2016
3. Feb 22, 2016

### msenay

thansk for comment dear Samy_A. This kind of integral is called deformed Fermi-Dirac integral. if we can think limit q→1, these integral go to standard Fermi-Dirac integral. Also, you can find these kind of integral in many statistical physics books like Kerson Huang and R.K.Patriha. So, we can think that second integral is even. This is not so important. My aim is to calculate first integral. How i can solve it? Do you have any idea ?

4. Feb 22, 2016

### Samy_A

Ok, sorry, I thought these were usual integrals and series. I'll leave you in more competent hands.

5. Feb 22, 2016

### blue_leaf77

I agree with Samy_A, try to check on your own the value of the integrand as $x$ approaches infinity.
I don't see how that integral will become the standard Fermi-Dirac integral even when $q\rightarrow 1$ for this integral reads
$$F_s(\mu) = \int_0^\infty \frac{x^s}{1+e^{x-\mu}}dx$$
Even when you set $s=1$ and $\mu=0$ and also for some unknown reason deliberately change the sign in front of the exponential term to minus, it does not match the integral you have there. I suspect that you misread the sign of the variable $x$ in the exponential, though.

6. Feb 22, 2016

### msenay

you should do some thing that these deforme Fermi-Dirac integral reduce to standard Fermi-Dirac integral. I didnt mention here. You can examine this article: http://www.sciencedirect.com/science/article/pii/S0378437114003124?np=y. This is only one example. I know how these deforme integral can reduce to standard Fermi-Dirac integral limit q→1. But, this is not my question. I asked to you, how can calculate these integral?

7. Feb 22, 2016

### blue_leaf77

$\displaystyle \lim_{x\rightarrow +\infty} (\dfrac {1} {1-e^{-x}}+\dfrac {1} {1+qe^{-x}})=2$
Have you been able to prove the above equation?
Could you possibly be talking about the equations (23) and (24)? If yes, can you identify the difference between these equations with the one you posted above apart from the power of the polynomial $x$?

8. Feb 22, 2016

### msenay

(23) and (24) equations have same form with mine, but not equal. I tried to solve power of polynomial x. But, there is some problem. We think that my integral has two part like from -infinity to zero and from zero to infinity.
Equations (23) and (24) have same form mine, but not equal. I tried to solve the power of the polynomial x. But, there is some problem. I can explain these. we think that we have two parts integral like from -infinty to zero and from zero to infinty. This is first part: http://texify.com/$%5Cint%20_%7B-%5Cinfty%20%7D%5E%7B0%7D%5Cdfrac%20%7Bxdx%7D%20%7B1-e%5E%7B-x%7D%7D%2B%5Cint%20_%7B-%5Cinfty%20%7D%5E%7B0%7D%5Cdfrac%20%7Bxdx%7D%20%7B1%2Bqe%5E%7B-x%7D%7D$ [Broken]. This is second part. http://texify.com/$%5Cint%20_%7B0%7D%5E%7B%5Cinfty%20%7D%5Cdfrac%20%7Bxdx%7D%20%7B1-e%5E%7B-x%7D%7D%2B%5Cint%20_%7B0%7D%5E%7B%5Cinfty%20%7D%5Cdfrac%20%7Bxdx%7D%20%7B1%2Bqe%5E%7B-x%7D%7D$ [Broken] .
The first part has two integral. First one can solve the power of the polynominal x. But second one can not solve because there is q parameter which is interval 0<q<1.
The second part has also two integral. Here, to define polynomial these integral, exp(x) should be small than 1. But, we don't have this condition here. So, we can not define these integral the power of the polynomial x.

Last edited by a moderator: May 7, 2017
9. Feb 22, 2016

### blue_leaf77

The real problem with the integral is that the integrand blows up to infinity for $x\rightarrow \infty$. Use a function plotter software to plot the integrand if you are not able to prove this mathematically.

10. Feb 22, 2016

### msenay

Yes, i know. Also, x=0 is problem. When i look limit at x=0, there is an divergent. I dont know how i can solve this integral. Thanks for best comment.

11. Feb 22, 2016

### blue_leaf77

The integrand at $x=0$ actually has a limit, i.e. it does not diverge there.
Well, what needs to be solved if we know that the integral diverges. It's pointless to find a closed form of the integral.