Integral involving exponential, power and Bessel function

Question

Is there any formula for calculating the following definite integral, including exponential and Bessel function?

$$ \int_0^{a}x^{-1} e^{x}I_2(bx)dx$$

Thanks in advance

Answer 1

In fact it mainly plays tricks by those formulae in http://www.efunda.com/math/bessel/modifiedbessel.cfm.

For $\int\dfrac{e^xI_2(bx)}{x}dx$ ,

$\int\dfrac{e^xI_2(bx)}{x}dx$

$=\int e^x~d\left(\dfrac{I_1(bx)}{bx}\right)$

$=\dfrac{e^xI_1(bx)}{bx}-\int\dfrac{I_1(bx)}{bx}d(e^x)$

$=\dfrac{e^xI_1(bx)}{bx}-\int\dfrac{e^xI_1(bx)}{bx}dx$

$=\dfrac{e^xI_1(bx)}{bx}+\int e^x\dfrac{d}{dx}(I_1(bx))~dx-\int e^xI_0(bx)~dx$

$=\dfrac{e^xI_1(bx)}{bx}+\int e^x~d(I_1(bx))-\int e^xI_0(bx)~dx$

$=\dfrac{e^xI_1(bx)}{bx}+e^xI_1(bx)-\int I_1(bx)~d(e^x)-\int e^xI_0(bx)~dx$

$=\dfrac{e^xI_1(bx)}{bx}+e^xI_1(bx)-\int e^xI_1(bx)~dx-\int e^xI_0(bx)~dx$

$=\dfrac{e^xI_1(bx)}{bx}+e^xI_1(bx)-\dfrac{1}{b}\int e^x~d(I_0(bx))-\int e^xI_0(bx)~dx$

$=\dfrac{e^xI_1(bx)}{bx}+e^xI_1(bx)-\dfrac{e^xI_0(bx)}{b}+\dfrac{1}{b}\int I_0(bx)~d(e^x)-\int e^xI_0(bx)~dx$

$=\dfrac{e^xI_1(bx)}{bx}+e^xI_1(bx)-\dfrac{e^xI_0(bx)}{b}+\dfrac{1}{b}\int e^xI_0(bx)~dx-\int e^xI_0(bx)~dx$

$=\dfrac{e^xI_1(bx)}{bx}+e^xI_1(bx)-\dfrac{e^xI_0(bx)}{b}+\dfrac{1-b}{b}\int e^xI_0(bx)~dx$

$\therefore$ Wolfram Alpha can get the close-form when $b=1$ .

When $b=-1$ , $\because I_2(-x)=I_2(x)$ , $\therefore$ Wolfram Alpha can get the close-form when $b=-1$ .

When $b\neq\pm1$ , the term of $\int e^xI_0(bx)~dx$ should be left.

In fact when $b\neq\pm1$ , $\int e^xI_0(bx)~dx$ can only be solved by this approach:

$\int e^xI_0(bx)~dx$

$=\int e^x\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{bx}{2}\right)^{2n}}{(n!)^2}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{b^{2n}x^{2n}e^x}{4^n(n!)^2}dx$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^kb^{2n}(2n)!x^ke^x}{4^n(n!)^2k!}+C$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{b^{2n}(2n)!x^{2k}e^x}{4^n(n!)^2(2k)!}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{b^{2n}(2n)!x^{2k-1}e^x}{4^n(n!)^2(2k-1)!}+C$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{b^{2n}(2n)!x^{2k}e^x}{4^n(n!)^2(2k)!}-\sum\limits_{k=1}^\infty\sum\limits_{n=k}^\infty\dfrac{b^{2n}(2n)!x^{2k-1}e^x}{4^n(n!)^2(2k-1)!}+C$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{b^{2n+2k}(2n+2k)!x^{2k}e^x}{4^{n+k}((n+k)!)^2(2k)!}-\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{b^{2n+2k}(2n+2k)!x^{2k+1}e^x}{4^{n+k}((n+k)!)^2(2k+1)!}+C$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{b^{2n+2k}\left(\dfrac{1}{2}\right)_{n+k}x^{2k}e^x}{4^k(1)_{n+k}k!\left(\dfrac{1}{2}\right)_k}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{b^{2n+2k}\left(\dfrac{1}{2}\right)_{n+k}x^{2k+1}e^x}{4^k(1)_{n+k}k!\left(\dfrac{3}{2}\right)_k}+C$ (according to http://mathworld.wolfram.com/PochhammerSymbol.html)