Exponential integral with square root

Question

Can this be solved?

$$Y=\int_{0}^\infty Ae^{-B\sqrt{t}}\mathrm{d}t$$ Where $A$ and $B$ are constants.

Answer 1

Let us do a change of variable: choose $x=-B\sqrt{t}$ then we have $$t=\frac{x^2}{B^2}\\ \mathrm{d}t = \frac{2x}{B^2}\mathrm{d}x$$ Substituting in the integral $$\int_0^\infty Ae^{B\sqrt{t}}\mathrm{d}t = \frac{2A}{B^2}\int_0^\infty xe^{-x}\mathrm{d}x = \frac{2A}{B^2}\int_0^\infty x^{1-1}e^{-x}\mathrm{d}x = \frac{2A}{B^2}\Gamma(1)=\frac{2A}{B^2}$$ where $\Gamma(z)$ is the Euler Gamma function, defined as $$\Gamma(z)= \int_0^\infty t^{z-1}e^{-t}\mathrm{d}t$$

We could solve the last integral without the Euler gamma function by using integration by parts, mainly $$\int_0^\infty \underbrace{x}_{f}\underbrace{e^{-x}}_{g'}\mathrm{d}x = -\underbrace{xe^{-x}|_0^\infty}_{\text{is zero}}+\int_0^\infty e^{-x}\mathrm{d}x = -e^{-x}|_0^\infty = 1$$ Remember that integration by parts says $$\int f(x)g'(x)\mathrm{d}x = f(x)g(x) - \int f'(x)g(x)\mathrm{d}x$$