Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
For $b=1$ Alpha returns a closed form for the indefinite integral ($x\mapsto a$) : $$\frac 12-e^x\;I_0(x)+\frac{1+x}x\;e^x\;I_1(x)$$ for $b=-1$ this becomes $$\frac 12-e^{-x}\;I_0(x)+\frac{1-x}x\;e^{-x}\;I_1(x)$$ Such kind of integrals are handled in Luke Y.L. book of 1962 'Integrals of Bessel functions) –  Raymond Manzoni May 13 '13 at 13:24
    
Thank you Raymond for your kind and swift reply, yet, I am looking for the integral using a general beta parameter. –  dioxen May 13 '13 at 13:27
    
In the book I saw only the cases $b=\pm 1$ from a quick look (with the parameter $b$ transposed to the exponential) so no sure that a general solution in closed form exists... –  Raymond Manzoni May 13 '13 at 13:31

1 Answer 1

up vote 3 down vote accepted

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)

share|improve this answer
    
Excellent stuff, Harry! –  Ahaan S. Rungta Oct 21 '13 at 15:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.