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Is it possible to rewrite or evaluate this integral $I = \int\limits_1^p e^{ - \left( {ax + \frac{b}{x}} \right)} dx$ where $a,b,p > 0$ as some known non-elementary function (For example $\operatorname{Ei}(x)$, $\operatorname{Li}(x)$, etc)?

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Similar to Closed form of $\int_{x = 0}^{C} \exp\left(-\frac{x}{A}-\frac{B}{x}\right)\,dx$,

Approach $1$:

$\int_1^pe^{-ax-\frac{b}{x}}~dx$

$=\int_p^\infty e^{-ax-\frac{b}{x}}~dx-\int_1^\infty e^{-ax-\frac{b}{x}}~dx$

$=\int_1^\infty e^{-apx-\frac{b}{px}}~d(px)-\int_1^\infty e^{-ax-\frac{b}{x}}~dx$

$=p\int_1^\infty e^{-apx-\frac{b}{px}}~dx-\int_1^\infty e^{-ax-\frac{b}{x}}~dx$

$=pK_{-1}\left(ap,\dfrac{b}{p}\right)-K_{-1}(a,b)$ (according to https://core.ac.uk/download/pdf/81935301.pdf)

Approach $2$:

$\int_1^pe^{-ax-\frac{b}{x}}~dx$

$=\int_\frac{\sqrt{a}}{\sqrt{b}}^\frac{p\sqrt{a}}{\sqrt{b}}e^{-a\frac{\sqrt{b}u}{\sqrt{a}}-\frac{b}{\frac{\sqrt{b}u}{\sqrt{a}}}}~d\left(\sqrt{\dfrac{b}{a}}u\right)$

$=\sqrt{\dfrac{b}{a}}\int_\sqrt{\frac{a}{b}}^{p\sqrt{\frac{a}{b}}}e^{-\sqrt{ab}\left(u+\frac{1}{u}\right)}~du$

$=\sqrt{\dfrac{b}{a}}\int_{\ln\sqrt{\frac{a}{b}}}^{\ln p\sqrt{\frac{a}{b}}}e^{-\sqrt{ab}\left(e^v+\frac{1}{e^v}\right)}~d(e^v)$

$=\sqrt{\dfrac{b}{a}}\int_{\ln\sqrt{a}-\ln\sqrt{b}}^{\ln p+\ln\sqrt{a}-\ln\sqrt{b}}e^{v-2\sqrt{ab}\cosh v}~dv$

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  • $\begingroup$ I am sorry but can the term $\int {{e^{v - 2\sqrt {ab} \cosh v}}dv}$ be written as some non - elementary function or in series form. I afraid that with the given bound the integral could not be written as non-elementary function since the post that you have pointed to exploit the fact that the Bessel Function come from an integral with bound approach to infinity $\endgroup$ – Tuong Nguyen Minh Sep 18 at 11:14

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