For $t = 0$, let $$f_t(x) = e^{-tx}(1+\frac{x}{t})^{t^2} ; (x>-t)$$ and $0$ otherwise.
a) For $s > -1$, proof that $$ \Gamma(s+1) = s^{s+0.5}e^{-s}\int^{\infty}_{-\sqrt{s}} f_\sqrt{s}(x) dx$$
b) Show that for all real number x $$\lim_{t\to \infty}f_t(x) = e^{\frac{-x^2}{2}}$$
c) Show that ${t\to f_t(x)}$ is monotone decreasing when $x > 0$ ans monotone increasing when $x<0$
I can actually do c for now (figure that I can just check the movement of $lnf_t(x)$) but having no clue in doing a nor b. Any help is appreciated. Thank you.
For question part (a), if we take the following integral $$\int_{-\sqrt{s}}^\infty f_\sqrt{s}(x)dx=\int_{-\sqrt{s}}^\infty e^{-x\sqrt{s}}\left(1+\frac{x}{\sqrt{s}}\right)^sdx$$ and apply the change of variables $y=s+x\sqrt{s}$, the integral becomes $$\int_0^\infty e^{-y}e^s\left(\frac{y}{s}\right)^s\frac{dy}{\sqrt{s}}=\frac{e^s}{s^{s+0.5}}\int_0^\infty y^se^{-y}dy=\frac{e^s}{s^{s+0.5}}\Gamma(s+1)$$ Thus $$\begin{align}\int_{-\sqrt{s}}^\infty f_\sqrt{s}(x)dx&=\frac{e^s}{s^{s+0.5}}\Gamma(s+1)\\\Rightarrow \Gamma(s+1)&=s^{s+0.5}e^{-s}\int_{-\sqrt{s}}^\infty f_\sqrt{s}(x)dx\end{align}$$
As regards part (b), we have $$\lim_{t\to \infty}f_t(x) =\lim_{t\to \infty}e^{-xt}\left(1+\frac{x}{t}\right)^{t^2}=\lim_{t\to \infty}e^{-xt+t^2\log(1+\frac{x}{t})}$$ We can apply a Taylor series expansion to the log function, as $|\frac{x}{t}|<1$, resulting in $$\large \lim_{t\to \infty}f_t(x) =\lim_{t\to \infty}e^{-xt+t^2\left(\frac{x}{t}-\frac{x}{2t^2}+O(\frac{x^3}{t^3})\right)}=\lim_{t\to \infty}e^{\left(-\frac{x^2}{2}+\frac{1}{t}O(\frac{x^3}{t^3})\right)}=e^{-\frac{x^2}{2}}$$