# Evaluation of another definite integral

I have a definite integral that I am trying to solve. Any hint or reference is urgently sought.

,

where $r$ is any positive integer while $\psi$ and $\nu$ are positive real numbers.

• 1. Do you have a reason to believe it has a closed form? 2. Are less-than general results (for example for specific $\psi$ or $r$) of any use? 3. You can type LateX in your question. May 14, 2014 at 13:45
• I am not sure it has a closed form . less than general results is very welcome. and I will edit my question in a minute in latex. May 14, 2014 at 13:57

By substituting $x = \nu y$, the integral becomes

$$\frac{1}{\nu^r} \int_0^{\infty} \frac{x^r e^x \,dx} {\left[1+\psi(e^x-1)\right]^{1+1/\psi}}$$ so the problem is reduced to finding

$$I(a,n)= \int_0^{\infty} \frac{x^n e^x \,dx} {\left[1+\frac{1}{a}(e^x-1)\right]^{1+a}}$$ where I relabeled some letters.

The case $n = 0$ is solved as follows: substituting $u = 1+ \frac{1}{a}(e^x - 1)$ gives $$I(a,0) =\int_1^{\infty} \frac{a}{u^{a+1}}\,du=1$$

After some more testing, I conjecture that for all $n$ and $a$, $$I(a,n)= n!\ a^{a-n} \, {}_pF_q \left(\underbrace{a,\ldots,a}_{n+1};\underbrace{1+a,\ldots,1+a}_{n};1-a \right)$$ While some may not consider this to be a closed form, its evaluation (for given $n$ and $a$) is about an order of magnitude faster than numerical integration.

For your convenience: if the above conjecture is true (it probably is) the result of the original integral becomes:

$$I(r,\nu,\psi)= \frac{r!}{\nu^r} \psi^{r-1/\psi} \, {}_pF_q \left(\underbrace{\frac{1}{\psi},\ldots,\frac{1}{\psi}}_{r+1};\underbrace{1+\frac{1}{\psi},\ldots,1+\frac{1}{\psi}}_{r};1-\frac{1}{\psi} \right)$$ (I corrected a small copying error: I first had 2 in the exponent instead of $r$)

Mathematica input form:

i[r_, [Nu]_ , [Psi]_] := r!/[Nu]^r [Psi]^(r - 1/[Psi])HypergeometricPFQ[Array[1/`[Psi] &, r + 1], Array[1 + 1/[Psi] &, r], 1 - 1/[Psi]]

• @111187 After substituting u=1+1/a(exp(x)-1) what happens with x^n for n not equal to zero. May 17, 2014 at 4:39
• @SA-255525 Well, then you get something nasty that can only be evaluated in terms of hypergeometric functions. That is why I only used this substitution for $n = 0$ and left the rest to Mathematica. May 20, 2014 at 12:48
• @111187 You didn't explain what happened with x^n after substitution. May 20, 2014 at 13:18
• @111187 I would highly appreciate it if you give me Mathematica input form May 20, 2014 at 16:09
• @SA-255525 The $x^n$ becomes $\ln^n(1 + a(u-1))$, which is not a nice enough form to do anything else with. May 20, 2014 at 19:20

This definite integral is very complicated. I doubt that a simple closed form exists. Theoritically, the integral can be expressed as a finite sum of $r$ derivatives of an hypergeometric function (in attachment). But writting explicitly the derivatives would be arduous, all the more so the derivatives must be considered with respect to a parameter insead of the variable of the hypergeometric function.