Integral involving exponential and power

Question

I have bumped into the following integral, which Mathematica is apparently not able to solve. I have tried a couple of change of variables and a series expansion for $x$ close to $1$, but without much success. I am interested in a closed-form solution if existent: it doesn't look as harmful as it might turn out to be, though...

$$ I(c,\xi)=\int_1^\infty\mathrm{d}x\ \mathrm{e}^{-cx}x^{\frac{\xi -1}{\xi }} \left(x^{1/\xi }-1\right)^{\xi }\ , $$ for $c,\xi>0$ (and $\xi$ not an integer in general). Thanks for your help folks.

Answer 1

Hint:

For integer $\xi$,

$$\left(x^{1/\xi}-1\right)^\xi=\sum_{k=0}^\xi\binom \xi k(-1)^kx^{k/\xi}$$

so that

$$I(c,\xi)=\sum_{k=0}^\xi\binom \xi k(-1)^kc^{-(k-1)/\xi-1}\Gamma\left(\frac{k-1}\xi+2,c\right).$$

You can't generalize to fractional $\xi$ because of the summation, but interpolation on $\xi$ might yield good approximations.

Answer 2

The case $\xi=1/2$ can be solved in terms of modified Bessel- and Struvefunctions. For a proof rewrite $x=\cosh(t)$ so that

$$ I(c,\frac{1}2)=\int_0^{\infty}dte^{-c \cosh(t)}\frac{\sinh^2(t)}{\cosh(t)}=\int_0^{\infty}dte^{-c \cosh(t)}\cosh(t)-\int_0^{\infty}dte^{-c \cosh(t)}\frac{1}{\cosh(t)} $$

According to DMLF 10.32.9 we can rewrite this as

$$ I(c,\frac{1}2)=K_1(c)+\int dc K_0(c) $$

the last integral equals (10.43.1 of DMLF) a combination of modified Bessel functions $K_{\nu}(z)$ and modified Struve functions $L_{\mu }(z)$. We obtain

$$ I(c,\frac{1}2)=K_1(c)+\frac{1}{2}\pi c (L_{-1 }(c)K_{0}(c)+L_{0 }(c)K_{1}(c))+C $$

Because $I(\infty,\xi)=0$ and $K_1(\infty) =0$ as well as $\lim_{c\rightarrow\infty} c (L_{-1 }(c)K_{0}(c)+L_{0 }(c)K_{1}(c))=1$ we have $C=-\frac{\pi}{2}$ and finally

$$ I(c,\frac{1}2)=K_1(c)+\frac{1}{2}\pi c (L_{-1 }(c)K_{0}(c)+L_{0 }(c)K_{1}(c))-\frac{\pi}{2} $$

I suspect that this method can't be generalized to other non-integer values of $\xi$ but maybe i'm wrong..