I am trying to solve this integral: $$ \int_{0}^\infty \frac{\exp\left[-\frac{1}{\beta(x_s^2-1)}(u^2+y_p-y_s)\right]\exp\left[-\frac{1}{\beta}(u-\sqrt{y_s})^2\right]}{\sqrt{\frac{\pi}{\beta(x_s^2-1)}(u^2+y_p-y_s)}}\,u\,du, $$ where $\beta,\,x_s,\,y_p$ and $y_s$ are constants and there is no relation between them. I change from $u$ to $t=\frac{u-\sqrt{y_s}}{\sqrt{\beta}}$ and I obtain these two integrals which are quite similar: $$ \int_{-\sqrt{\frac{y_s}{\beta}}}^\infty \frac{\exp\left[-\frac{1}{\beta(x_s^2-1)}(\beta(t+\sqrt{y_s})^2+y_p-y_s)-t^2\right]}{\sqrt{\frac{\pi}{\beta(x_s^2-1)}(\beta(t+\sqrt{y_s})^2+y_p-y_s)}}\,t \,dt $$ and $$ \int_{-\sqrt{\frac{y_s}{\beta}}}^\infty \frac{\exp\left[-\frac{1}{\beta(x_s^2-1)}(\beta(t+\sqrt{y_s})^2+y_p-y_s)-t^2\right]}{\sqrt{\frac{\pi}{\beta(x_s^2-1)}(\beta(t+\sqrt{y_s})^2+y_p-y_s)}}\, dt $$ but I don't know how to carry out the integration after performing the substitution. How can I compute the integrals? Thank you in advance.
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What makes you suspect that this has a nice closed form? – infinitylord Jun 14 '17 at 12:13
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What exactly is the difference between the last two integrals? They both look identical to me...and they both undoable and ugly. – DonAntonio Jun 14 '17 at 12:22
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The fact that the integral involves polynoms and exponentials; but maybe the integral has no solution in terms of mathematical functions. – Miguel Angel Jimenez Herrera Jun 14 '17 at 12:22
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The first integral has a t multiplying the integrand in the second @DonAntonio – Miguel Angel Jimenez Herrera Jun 14 '17 at 12:24
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1Wow, was that hidden or what! Thanks – DonAntonio Jun 14 '17 at 12:27