$\int_{0}^{\infty} \text{e}^{c z x} \frac{y}{\sqrt{2 \pi}} \text{e}^{- \frac{y^2}{2x} } x^{- \frac{3}{2}} \, \text{d}x = \text{e}^{y\sqrt{2cz}} $

Question

Let $\nu<0$ and $\xi,c >0$. I want to show the identity $$ \int_{0}^{\infty} \text{e}^{c \nu \eta} \frac{\xi}{\sqrt{2 \pi}} \text{e}^{- \frac{\xi^2}{2 \eta} } \eta^{- \frac{3}{2}} \, \text{d}\eta=\text{e}^{-\xi\sqrt{2c|\nu|}}. $$ My attempt: To evaluate the integral $$ \int_{0}^{\infty} \text{e}^{c \nu \eta} \frac{\xi}{\sqrt{2 \pi}} \text{e}^{- \frac{\xi^2}{2 \eta} } \eta^{- \frac{3}{2}} \, \text{d}\eta = \frac{\xi}{\sqrt{2 \pi}} \int_{0}^{\infty} \text{e}^{\frac{2c\nu \eta^2-\xi^2}{2 \eta}} \eta^{- \frac{3}{2}} \, \text{d}\eta=\frac{\xi}{\sqrt{2 \pi}} \int_{0}^{\infty} \text{e}^{\frac{-2c|\nu| \eta^2-\xi^2}{2 \eta}} \eta^{- \frac{3}{2}} \, \text{d}\eta. $$ we make the substitution $$ u= \frac{\xi-\sqrt{2c|\nu|}\eta}{\sqrt{2\eta}} $$ or equivalently $$ \eta = \frac{u^2+ \xi \sqrt{2 c |\nu|}-\sqrt{2\xi u^2 \sqrt{2c |\nu| } +u^4}}{2c|\nu|}. $$ Because of $$ \text{e}^{-u^2} = \text{e}^{\frac{-\xi^2+2\xi\eta\sqrt{2c|\nu|} -2c|\nu|\eta^2 }{2\eta}}, \qquad \text{e}^{\frac{-2c|\nu|\eta^2-\xi^2}{2\eta}} = \text{e}^{-u^2-\xi\sqrt{2c|\nu|}} $$ and $$ \frac{\text{d}u}{\text{d}\eta}= \frac{- \sqrt{2} \xi - \sqrt{2}\eta \sqrt{2c|\nu|} }{4\eta^{\frac{3}{2}}}, \qquad \eta^{-\frac{3}{2}} \, \text{d}\eta = - \frac{4}{\sqrt{2}\xi + \sqrt{2} \left(\frac{u^2+ \xi \sqrt{2 c |\nu|}-\sqrt{2\xi u^2 \sqrt{2c |\nu| } +u^4}}{2c|\nu|}\right) \sqrt{2c|\nu|}} \, \text{d}u $$ we get $$ \int_{0}^{\infty} \text{e}^{c \nu \eta} \frac{\xi}{\sqrt{2 \pi}} \text{e}^{- \frac{\xi^2}{2 \eta} } \eta^{- \frac{3}{2}} \, \text{d}\eta = \frac{\xi}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \text{e}^{-u^2-\xi\sqrt{2c|\nu|}} \frac{2 \sqrt{2}\sqrt{2c|\nu|}}{u^2+ 2 \xi \sqrt{2 c |\nu|}-\sqrt{2\xi u^2 \sqrt{2c |\nu| } +u^4}} \, \text{d}u. $$ Now we write $$ \frac{2 \sqrt{2}\sqrt{2c|\nu|}}{u^2+ 2 \xi \sqrt{2 c |\nu|}-\sqrt{2\xi u^2 \sqrt{2c |\nu| } +u^4}} =\frac{1}{\xi} + \frac{2 \sqrt{2} \xi \zeta - u^2- 2 \xi \zeta+\sqrt{2\xi u^2 \zeta +u^4} }{u^2 \xi+ 2 \xi^2 \zeta- \xi \sqrt{2\xi u^2 \zeta +u^4}}. $$ To get $$ \int_{0}^{\infty} \text{e}^{c \nu \eta} \frac{\xi}{\sqrt{2 \pi}} \text{e}^{- \frac{\xi^2}{2 \eta} } \eta^{- \frac{3}{2}} \, \text{d}\eta = \text{e}^{-\xi\sqrt{2c|\nu|}} $$ I need that $$ \int_{-\infty}^{\infty} \text{e}^{-u^2} \frac{2 \sqrt{2} \xi \zeta - u^2- 2 \xi \zeta+\sqrt{2\xi u^2 \zeta +u^4} }{u^2 \xi+ 2 \xi^2 \zeta- \xi \sqrt{2\xi u^2 \zeta +u^4}} \, \text{d}u=0. $$ How can I show this?

Answer 1

Hint. If you make the change of variable $t=x^{-1/2}$, $dt=-\dfrac12 x^{-3/2}dx$, in the following equivalent integral you obtain $$ \int_0^{\infty}e^{\large -(b^2x+\frac{a^2}x)}x^{-3/2} dx=2\int_0^{\infty}e^{ -a^2t^2-b^2/t^2}dt=\int_{-\infty}^{\infty}e^{ -a^2t^2-b^2/t^2}dt $$ One may then recall that, for any integrable function $f$, we have

$$ \int_{-\infty}^{+\infty}f\left(t-\frac{s}{t}\right)\mathrm{d}t=\int_{-\infty}^{+\infty} f(t)\: \mathrm{d}t, \quad s>0. \tag1 $$

Applying it to $f(t)=e^{-a^2t^2}$, $a>0$, $b>0$, $s=b/a$, you get

$$ \int_{-\infty}^{+\infty}e^{\large -a^2(t-\frac{b}{a t})^2}\mathrm{d}t=\int_{-\infty}^{+\infty} e^{-a^2t^2} \mathrm{d}t=\frac{\sqrt{\pi}}a, \quad s>0. \tag2 $$

Thus

$$ \int_{-\infty}^{+\infty}e^{-a^2t^2-b^2/t^{2}}\mathrm{d}t=\frac{\sqrt{\pi}}a\:e^{-2ab}.\tag3 $$ This is a path to obtain your initial integral.