How to integrate the following integral for all $q$?

Question

How to integrate

$$ \int^{\infty}_{-\infty}\frac{1}{4} \lambda _1 \lambda _2 e^{-\lambda _1 |x|^q-\lambda _2 |x|}\,dx $$

where $0<q<1$?

If it can't be solved for $0<q<1$ then can it be solve for just $q=\frac{1}{2}$?

Answer 1

For $q=\frac 12 :$

$I=\frac 12 \lambda_1 \lambda_2 \int_{0}^{+\infty}e^{-\lambda_1x^{1/2}-\lambda_2x}dx$

Setting $u^2=\lambda_2x$, $2udu=\lambda_2dx$, and :

$I=\lambda_1\int_0^{+\infty}u\times e^{-u^2-\frac{\lambda_1}{\sqrt{\lambda_2}}u}du=-\frac 12\lambda_1\int_0^{+\infty}(-2u-\frac{\lambda_1}{\sqrt{\lambda_2}}+\frac{\lambda_1}{\sqrt{\lambda_2}})\times e^{-u^2-\frac{\lambda_1}{\sqrt{\lambda_2}}u}du=-\frac 12\lambda_1([e^{-u^2-\frac{\lambda_1}{\sqrt{\lambda_2}}u}]_0^{+\infty}+\frac{\lambda_1}{\sqrt{\lambda_2}}\int_0^{+\infty}e^{-u^2-\frac{\lambda_1}{\sqrt{\lambda_2}}u}du)=\frac 12\lambda_1-\frac 12 \frac{\lambda_1^2}{\sqrt{\lambda_2}}\int_0^{+\infty}e^{-(u+\frac{\lambda_1}{2\sqrt{\lambda_2}})^2+\frac{\lambda_1^2}{4\lambda_2}}du=\frac 12\lambda_1(1-\frac{\lambda_1}{\sqrt{\lambda_2}}e^{\frac{\lambda_1^2}{4\lambda_2}}\int_{\frac{\lambda_1}{2\sqrt{\lambda_2}}}^{+\infty}e^{-w^2}dw) $

The last integral can be expressed with the help of the erfc function :

$\int_{\frac{\lambda_1}{2\sqrt{\lambda_2}}}^{+\infty}e^{-w^2}dw=\frac 12 \sqrt{\pi}erfc(\frac{\lambda_1}{2\sqrt{\lambda_2}})$

Therefore,

$$I_{1/2}=\frac 12\lambda_1(1-\frac{\lambda_1}{2\sqrt{\lambda_2}}\sqrt{\pi}e^{\frac{\lambda_1^2}{4\lambda_2}}erfc(\frac{\lambda_1}{2\sqrt{\lambda_2}}))$$