I want to integrate this:

$$\int_0^{\infty} dt \exp{\left ( a \, t^b\right)} \, I_v {\left ( a \, t^b\right)} $$

where $I_v(.)$ is the modified bessel function of arbitrary order $v$.

Can someone help me with this please?

  • 1
    $\begingroup$ I'm pretty sure this integral converges only for $a<0$ and probably for some non obvious conditions on $b$ and $\ni$. Any other info you'ld care to share about this problem? $\endgroup$ Commented Jun 28, 2013 at 20:52
  • $\begingroup$ Where did this problem come from? $\endgroup$ Commented Jun 28, 2013 at 21:39
  • $\begingroup$ You can have a closed form solution in terms of the $H$-function. See this problem. $\endgroup$ Commented Jun 28, 2013 at 21:42
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    $\begingroup$ @nbubis: the value of a>0.5, and b can be any positive constant...I know how to get a closed-form expression for such integral without the b, but I try substitution and (let x=at^b, , then dx=abt^(b-1) dt ==> dt=dx/(ab*t^(b-1)),...which made things more complicated. $\endgroup$
    – kazekage
    Commented Jun 29, 2013 at 5:14
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    $\begingroup$ @MhenniBenghorbal: this actually came after long derivation of a wireless channel model involving modified bessel function. How can I get I closed-form expression from this H-function, can you please explain more, this is the first time I see such a function. $\endgroup$
    – kazekage
    Commented Jun 29, 2013 at 5:18

1 Answer 1


$$\int_0^\infty e^{at^b}I_v(at^b)~dt$$

$$=\int_0^\infty e^{at}I_v(at)~d\left(t^\frac{1}{b}\right)$$

$$=\dfrac{1}{b}\int_0^\infty t^{\frac{1}{b}-1}e^{at}I_v(at)~dt$$

$$=\dfrac{(-1)^v}{b}\int_0^\infty t^{\frac{1}{b}-1}e^{at}I_v(-at)~dt$$


$$=\dfrac{(-1)^v}{(-a)^\frac{1}{b}~b}\int_0^\infty t^{\frac{1}{b}-1}e^{-t}I_v(t)~dt$$

$$=\dfrac{(-1)^v~\Gamma\left(\dfrac{1}{b}+v\right)\Gamma\left(\dfrac{1}{2}-\dfrac{1}{b}\right)}{\sqrt\pi(-a)^\frac{1}{b}~2^\frac{1}{b}~b~\Gamma\left(v-\dfrac{1}{b}+1\right)}$$ (according to http://people.math.sfu.ca/~cbm/aands/page_486.htm)


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