Closed-form Solution of an Integral Equation

Question

Recently, I get stuck in an equation, and I would like to obtain an closed-form solution of an equation, which is given as $$ \int_1^\infty\frac{x}{t^\alpha-x}dt=c, $$ where $\alpha>1$ and $c>0$ are constants. Is it possible to obtain the expression of $x$, where $x\in\left(0,1\right)$? If not, can we get the scaling relation between $x$ and parameters $\alpha$ and $c$? Thank you for your help.

Answer 1

Your integral can be written as $\dfrac{x}{a} \Phi\left(x,1,1 - 1/a\right)$ where $\Phi$ is the Lerch Phi function. I doubt that there's a closed-form solution for $x$ in general, or a scaling relation. You could get a series expansion:

$$ \eqalign{x &= \left( a-1 \right) c-{\frac { \left( a-1 \right) ^{3}}{2\,a-1}}{c} ^{2}+{\frac { \left( 2\,{a}^{2}-4\,a+1 \right) \left( a-1 \right) ^{ 4}}{ \left( 2\,a-1 \right) ^{2} \left( 3\,a-1 \right) }}{c}^{3}\cr &-{ \frac { \left( 4\,{a}^{4}-19\,{a}^{3}+25\,{a}^{2}-9\,a+1 \right) \left( a-1 \right) ^{5}}{ \left( 2\,a-1 \right) ^{3} \left( 3\,a-1 \right) \left( 4\,a-1 \right) }}{c}^{4}\cr &+{\frac { \left( 24\,{a}^{7}- 210\,{a}^{6}+624\,{a}^{5}-789\,{a}^{4}+450\,{a}^{3}-128\,{a}^{2}+18\,a -1 \right) \left( a-1 \right) ^{6}}{ \left( 3\,a-1 \right) ^{2} \left( 4\,a-1 \right) \left( 5\,a-1 \right) \left( 2\,a-1 \right) ^ {4}}}{c}^{5}+\ldots} $$