We have the p.d.f. of first passage time, $T_a$, to a fixed level $a$ as:$$f_{T_a}(t)=\frac a{\sqrt{2π}}\cdot t^\frac{-3}{2}\cdot e^\frac{-a^2}{2t},\;t>0.$$ We need to find the p.d.f. of $T_{a+b}$. For this purpose, we may first find the Laplace transform of $T_{a+b}$ and then taking the inverse Laplace transform we can get the p.d.f. of $T_{a+b}. $ Since $\mathscr{L}\{f_{T_{a+b}}(t)\}=\mathscr{L}\{f_{T_a}(t)\}\cdot\mathscr{L}\{f_{T_b}(t)\}$, where $T_a$ and $(T_{a+b}-T_a)$ are two independent random variables and $T_{a+b}=T_a+(T_{a+b}-T_a).$
In a book on Stochastic processes, I found out that $\mathscr L\{f_{T_a}(t)\}=\bar h_a(s)=e^{-a\sqrt{2s}}$. But I want to know how to find it actually. I am not able to do that yet based on my knowledge regarding Laplace transform. Mainly the $t$ in the denominator in the exponent of $e$ in $f_{T_a}$ is causing problems for me.
So please help me to find it. Any kind of suggestion will be useful.
$$F(s)=\frac a{\sqrt{2π}}\int_0^\infty t^\frac{-3}{2}e^\frac{-a^2}{2t}e^{-st}dt=(x=\sqrt t)\,\, \frac a{\sqrt{2π}}\int_0^\infty e^\frac{-a^2}{2x^2}e^{-sx^2}\frac{dx}{x^2}$$ $$=(t=\sqrt s x)\,\,\frac {2a\sqrt s}{\sqrt{2π}}\int_0^\infty e^\frac{-sa^2}{2t^2}e^{-t^2}\frac{dt}{t^2}=\frac{2c}{\sqrt\pi}\int_0^\infty e^{-t^2-\frac{c^2}{t^2}}\frac{dt}{t^2} \,\,\, (c=\frac{a\sqrt s}{\sqrt2})$$ $$=\frac{2\,c\,e^{-2c}}{\sqrt\pi}\int_0^\infty e^{-(t-\frac{c}{t})^2}\frac{dt}{t^2} $$ Making change $x=\frac{c}{t}$ $$F(s)=\frac{2\,e^{-2c}}{\sqrt\pi}\int_0^\infty e^{-(x-\frac{c}{x})^2}dx$$ $$2 F(s)=\frac{2\,e^{-2c}}{\sqrt\pi}\Bigl( \int_0^\infty e^{-(t-\frac{c}{t})^2}\frac{c\,dt}{t^2}+\int_0^\infty e^{-(t-\frac{c}{t})^2}dt\Bigr)=\frac{2\,e^{-2c}}{\sqrt\pi}\int_0^\infty e^{-(t-\frac{c}{t})^2}d\Bigl(t-\frac{c}{t}\Bigr)$$ $$2 F(s)=\frac{2\,e^{-2c}}{\sqrt\pi}\int_{-\infty}^\infty e^{-x^2}dx=2\,e^{-2c}$$ $$F(s)=e^{-a\sqrt{2s}}$$