Evaluar integral con exponente

Question

Cómo evaluar la integralidad:

$$\int^\infty_0e^{-\tfrac{At^2}{t+1}}~dt , \quad A>0$$

Answer 1

$\int_0^\infty e^{-\frac{At^2}{t+1}}~dt$

$=\int_1^\infty e^{-\frac{A(t-1)^2}{t}}~d(t-1)$

$=\int_1^\infty e^{-\frac{A(t^2-2t+1)}{t}}~dt$

$=e^{2A}\int_1^\infty e^{-A\left(t+\frac{1}{t}\right)}~dt$

Considere $\int_0^\infty e^{-A\left(t+\frac{1}{t}\right)}~dt=2K_1(2A)$ ,

$$ \begin{align} \int_0^1e^{-A\left(t+\frac{1}{t}\right)}~dt+\int_1^\infty e^{-A\left(t+\frac{1}{t}\right)}~dt&=2K_1(2A)\\ \int_\infty^1e^{-A\left(\frac{1}{t}+t\right)}~d\left(\dfrac{1}{t}\right)+\int_1^\infty e^{-A\left(t+\frac{1}{t}\right)}~dt&=2K_1(2A)\\ \int_1^\infty\dfrac{1}{t^2}e^{-A\left(t+\frac{1}{t}\right)}~dt+\int_1^\infty e^{-A\left(t+\frac{1}{t}\right)}~dt&=2K_1(2A)\\ 2\int_1^\infty e^{-A\left(t+\frac{1}{t}\right)}~dt+\int_1^\infty\left(\dfrac{1}{t^2}-1\right)e^{-A\left(t+\frac{1}{t}\right)}~dt&=2K_1(2A)\\ 2\int_1^\infty e^{-A\left(t+\frac{1}{t}\right)}~dt+\left[\dfrac{e^{-A\left(t+\frac{1}{t}\right)}}{A}\right]_1^\infty&=2K_1(2A)\\ 2\int_1^\infty e^{-A\left(t+\frac{1}{t}\right)}~dt-\dfrac{e^{-2A}}{A}&=2K_1(2A)\\ \int_1^\infty e^{-A\left(t+\frac{1}{t}\right)}~dt&=K_1(2A)+\dfrac{e^{-2A}}{2A}\\ e^{2A}\int_1^\infty e^{-A\left(t+\frac{1}{t}\right)}~dt&=e^{2A}K_1(2A)+\dfrac{1}{2A}\\ \end{align} $$