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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

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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
up vote 3 down vote accepted

In fact it mainly plays tricks by those formulae in

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


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



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

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

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

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

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

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

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

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

$\therefore$ 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:

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

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


$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^kb^{2n}(2n)!x^ke^x}{4^n(n!)^2k!}+C$ (can be obtained from

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Excellent stuff, Harry! – Ahaan S. Rungta Oct 21 '13 at 15:55

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