# How to rewrite this integral $I = \int e^{ - \left( {ax + \frac{b}{x}} \right)} dx$ as non-elementary function?

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)?

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 $$2$$:

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

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

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

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

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

• 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 – Tuong Nguyen Minh Sep 18 at 11:14