closed-form convolution integral of two pdfs (Hypoexponential and Lévy)

Question

I would need a closed-form expression for the convolution integral of these two probability density functions:

$$p_1(x) = (\lambda_1\lambda_2/(\lambda_1-\lambda_2))\cdot(\exp(-\lambda_2x)-\exp(-\lambda_1x))$$

$$p_2(x) = \frac{C}{\sqrt{2\pi}}{\frac{1}{x^{3/2}}}\exp\left(\frac{-C^2}{2x}\right);$$

for $x\geq0$, while $p_{1,2}(x) = 0$ for $x<0$

where

$p_1(x)$ is the pdf of the hypoexponential distribution with parameters $\lambda_1$ and $\lambda_2$

$p_2(x)$ is the pdf of the Lévy distribution with parameters $c = C$ and $\mu = 0$

The aim is to obtain the pdf $p_3(x)$ of the sum of two independent random variables with pdf $p_1(x)$ and $p_2(x)$, respectively.

$$p_3(x) = \operatorname{heaviside}(x)\int_0^{x}{p_1(t)p_2(x-t)dt}$$

I was not able to get a closed-form solution with Matlab symbolic toolbox, which produces:

$$p_3(x) = \operatorname{heaviside}(x)\int_0^{x}{ -\frac{\sqrt{2}\,\sqrt{C}\,l_{1}\,l_{2}\,{e}^{\frac{C}{2\,\left(t-x\right)}}\,\left({e}^{-l_{1}\,t}-{e}^{-l_{2}\,t}\right)}{2\,\sqrt{\pi }\,\left(l_{1}-l_{2}\right)\,{\left(x-t\right)}^{3/2}}dt }$$

Can someone solve it manually or confirm that it does not have a closed-form?

Answer 1

Mathematica gives the result

$$\int\limits_0^x p_1(t)\,p_2(x-t)\,dt=\frac{(\lambda_1 \lambda_2)}{2 (\lambda_1-\lambda_2)} \left(e^{-\lambda_2 x} \left(e^{i C \sqrt{2 \lambda_2}}\ \text{erfc}\left(\frac{C}{\sqrt{2 x}}+i \sqrt{\lambda_2 x}\right)+e^{-i C \sqrt{2 \lambda_2}}\ \text{erfc}\left(\frac{C}{\sqrt{2 x}}-i \sqrt{\lambda_2 x}\right)\right)-e^{-\lambda_1 x} \left(e^{i C \sqrt{2 \lambda_1}}\ \text{erfc}\left(\frac{C}{\sqrt{2 x}}+i \sqrt{\lambda_1 x}\right)+e^{-i C \sqrt{2 \lambda_1}}\ \text{erfc}\left(\frac{C}{\sqrt{2 x}}-i \sqrt{\lambda_1 x}\right)\right)\right)$$

which is valid for $\lambda_1,\lambda_2,C,x>0\land\lambda_1\ne\lambda_2$.

Answer 2

Based on @StevenClark answer, the closed form of the convolution integral:

$$f(x) = \int_0^x\frac{\sqrt{c} \lambda_1\lambda_2 e^{\frac{c}{2 t-2 x}} \left(e^{-\lambda_2 t}-e^{-\lambda_1 t}\right)}{\sqrt{2 \pi } (\lambda_1-\lambda_2) (x-t)^{3/2}}dt$$

where $$c = C^2$$

was obtained using Mathematica with:

$$\text{Assuming}\left[\lambda_1>0\land \lambda_2>0\land c>0\land x\geq 0,\text{FullSimplify}\left[f(x)\right]\right]$$ as:

$$\frac{\lambda_1 \lambda_2 \left(e^{-\lambda_2 x-i \sqrt{2} \sqrt{c \lambda_2}} \left(e^{2 i \sqrt{2} \sqrt{c \lambda_2}} \text{erfc}\left(\frac{\sqrt{\frac{c}{x}}}{\sqrt{2}}+i \sqrt{\lambda_2 x}\right)+\text{erfc}\left(\frac{\sqrt{\frac{c}{x}}}{\sqrt{2}}-i \sqrt{\lambda_2 x}\right)\right)-e^{-\lambda_1 x-i \sqrt{2} \sqrt{c \lambda_1}} \left(e^{2 i \sqrt{2} \sqrt{c \lambda_1}} \text{erfc}\left(\frac{\sqrt{\frac{c}{x}}}{\sqrt{2}}+i \sqrt{\lambda_1 x}\right)+\text{erfc}\left(\frac{\sqrt{\frac{c}{x}}}{\sqrt{2}}-i \sqrt{\lambda_1 x}\right)\right)\right)}{2 (\lambda_1-\lambda_1)}$$