solve the integral equation 2

Question

I want to solve the integral .Solve is difficult. I want to use statistical methods to solve them. $$\int_{0}^{+\infty}x \exp\{ ax-b x^2\}d x=\int _{0}^{+\infty} x\exp\{-b(x^2-\frac{a}{b}x)\}dx=\\ exp\{\frac{a^2}{4b^2}\}\int_{0}^{+\infty}x\exp\{-b(x-\frac{a}{b})^2\}dx=\sqrt{\frac{\pi}{4}}\exp{\frac{a^2}{4b^2}}\int_{0}^{+\infty}\frac{\sqrt{2}}{\pi(1\2)}x \exp\frac{-b}{2(1\2)}(x-\frac{a}{b})^2dx=\frac{\pi}{4}\exp\{\frac{a^2}{4b^2}\}.E(HN) $$ Wherever a and b are fixed and HN half normal distribution. What is the final answer

Answer 1

As you obviously understood is that the first thing could be to complete the square. So, $$ax-bx^2=-(\sqrt b x-\frac a{2\sqrt b})^2+\frac{a^2}{4b}$$ So, now change variable $$\sqrt b x-\frac a{2\sqrt b}=y\implies x=\frac{a+2 \sqrt{b} y}{2 b}\implies dx=\frac{dy}{\sqrt{b}}$$ Replacing and expanding, $$I=\int x e^{ ax-b x^2}d x=\frac{a e^{\frac{a^2}{4 b}}}{2 b^{3/2}}\int e^{-y^2}dy +\frac{ e^{\frac{a^2}{4 b}}}{b}\int y\, e^{-y^2}dy$$ The first integral is classical since $$\int e^{-z^2}dz=\frac{1}{2} \sqrt{\pi } \,\text{erf}(z)$$ where appears the error function.

The second integral is simple (noticing that $y$ is close to the derivative of $y^2$ - otherwise change again variable $y^2=t$) $$\int z\,e^{-z^2}dz=-\frac{e^{-z^2}}{2}$$ Now, change the lower bound ($y$ will vary between $-\frac a{2\sqrt b}$ and $\infty$) and get, after rearrangements, $$J=\int_0^\infty x e^{ ax-b x^2}d x=\frac{ a \sqrt{\pi b}\, e^{\frac{a^2}{4 b}} \left(\text{erf}\left(\frac{a}{2 \sqrt{b}}\right)+1\right)+2 b}{4 b^2}$$