So I want to solve the following integral:
$$\varphi(t) = \int_{0}^{\infty} \frac{ x }{\sigma^{2}} e^{ \frac{-(x^{2} + v^{2})}{2 \sigma^{2}} } I_{0} \left( \frac{xv}{\sigma^{2}} \right) e^{jxt} dx$$
where $I_{0}(\cdot)$ is the modified Bessel function and $j$ is the imaginary unity.
To start, we bring out the exponential and expand the modified Bessel function:
$$\varphi(t) = \frac{1}{\sigma^{2}} e^{\frac{-v^{2}}{2 \sigma^{2}}} \int_{0}^{\infty} x e^{ -\frac{x^2}{2\sigma^2}} I_{0} \left( \frac{xv}{\sigma^2} \right) e^{jxt} dx $$
$$\varphi(t) = \frac{1}{\sigma^{2}} e^{\frac{-v^{2}}{2 \sigma^{2}}} \int_{0}^{\infty} x e^{ -\frac{x^2}{2\sigma^2}} \sum_{k=0}^{\infty} \frac{ \left( \frac{xv}{2 \sigma^2}\right)^{2k}}{k! \Gamma(k+1)} e^{jxt} dx $$
$$\varphi(t) = \frac{1}{\sigma^{2}} e^{\frac{-v^{2}}{2 \sigma^{2}}} \sum_{k=0}^{\infty} \frac{ \left( \frac{v}{2 \sigma^2}\right)^{2k}}{k! \Gamma(k+1)} \int_{0}^{\infty} x^{2k+1} e^{ -\frac{x^2}{2\sigma^2}} e^{jxt} dx $$
$$\varphi(t) = \frac{1}{\sigma^{2}} e^{\frac{-v^{2}}{2 \sigma^{2}}} \sum_{k=0}^{\infty} \frac{ \left( \frac{v}{2 \sigma^2}\right)^{2k}}{k! \Gamma(k+1)} \varphi_{1}(\infty) $$
So we focus on $\varphi_{1}(\infty)$. We let $u = jx$ such that $j^{-1}du = dx$. Noticing that $j^{-n} = 1$ when $n$ is even, we have:
Thus:
$$\varphi_{1}(\infty) = \int_{0}^{j \infty} u^{2k+1} e^{\frac{1}{2 \sigma^2} u^2 - tu} du $$
If we complete the square, we get:
$$\varphi_{1}(\infty) = e^{-\frac{1}{2} t^{2} \sigma^2}\int_{0}^{j \infty} u^{2k+1} e^{\frac{1}{2\sigma^2} (u-\sigma^2 t)^2} du $$
Now we can expand the exponential function:
$$e^{\frac{1}{2\sigma^2} (u-\sigma^2 t)^2} = \sum_{n=0}^{\infty} \frac{(u-\sigma^2 t)^{2n}}{n!} $$
We can use the binomial theorem to expand $(u-\sigma^2 t)^{2n}$ but I think then we end up with more sums than we really need.
So this is where I am stuck. I probably went wrong where I let $u=jx$ but I don't know what else I should do...
Any help?
By the way, the answer can be found here:
https://en.wikipedia.org/wiki/Rice_distribution#cite_note-2
And is:
$$ I(t) = e^{\frac{-v^{2}}{2 \sigma^{2}}} \left[ \Psi_{2}\left( 1; 1; \frac{1}{2}; \frac{v^{2}}{2\sigma^{2}}, -\frac{1}{2} \sigma^{2} t^{2} \right) + j \sqrt{2} \sigma t \Psi_{2} \left( \frac{3}{2}; 1; \frac{3}{2}; \frac{v^{2}}{2 \sigma^{2}}; -\frac{1}{2} \sigma^{2}t^{2} \right) \right]$$
where:
$$ \Psi_{2} \left( \alpha; \beta; \gamma; x; y \right) = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{ (\alpha)_{m+n} }{(\beta)_{m} (\gamma)_{n}} \frac{x^{m}y^{n}}{m!n!} $$
and
$$ (\alpha)_{n} = \alpha (\alpha + 1) (\alpha+2) \ldots (\alpha + n-1) $$
is the rising factorial.
Although, a simpler answer would be better!
According to Mathematica:
$$\int_{0}^{\infty} x^{2 k+1} e^{-\frac{x^2}{2 s^2}+i t x} dx = 2^k s^{2 k+2} \left(\Gamma (k+1) \, _1F_1\left(k+1;\frac{1}{2};-\frac{1}{2} s^2 t^2\right)+i \sqrt{2} s t \Gamma \left(k+\frac{3}{2}\right) \, _1F_1\left(k+\frac{3}{2};\frac{3}{2};-\frac{1}{2} s^2 t^2\right)\right) $$
So I know I am on the right track...just...how do I get these Hypergeometric functions out of there? And how do I get the imaginary unit out of the exponent?
Okay so a bit of progress. We let $u = \frac{x^{2}}{2s^{2}}$. This gives us:
$$ 2^{k}s^{2k+2} \left( \int_{0}^{\infty} u^k e^{-u}e^{j \sqrt{2 s^{2}u}t} du \right) = 2^{k} s^{2 k+2} \left( \left(\Gamma (k+1) \, _1F_1\left(k+1;\frac{1}{2};-\frac{1}{2} s^2 t^2\right)+i \sqrt{2} s t \Gamma \left(k+\frac{3}{2}\right) \, _1F_1\left(k+\frac{3}{2};\frac{3}{2};-\frac{1}{2} s^2 t^2\right)\right) \right)$$
Now we split up the complex exponential: $e^{j \sqrt{2 s^{2} u} t} = cos(\sqrt{2 s^{2} u} t) + jsin(\sqrt{2 s^{2} u} t)$. This give us:
$$ \int_{0}^{\infty} u^k e^{-u} e^{ \sqrt{2 s^{2}u}t} du = \int_{0}^{\infty} u^k e^{-u}cos( \sqrt{2 s^{2}u}t) du + 1j\int_{0}^{\infty} u^k e^{-u}sin(\sqrt{2 s^{2}u}t) du $$
But now I am having trouble reducing both sides to the Hypergeometric function. Mathematica does it for me, but how do I do it myself?
Thanks!