$\int_0^\infty e^{2ax-x^2}\,dx$

Question

$$\int_0^\infty e^{2ax-x^2}\,dx$$ The answer is $e^{a^2}\sqrt\pi$, but why?

Answer 1

$$\int_0^\infty e^{2ax-x^2}\,dx\\=e^{a^2}\int_0^\infty e^{-a^2+2ax-x^2}\,dx\\=e^{a^2}\int_0^\infty e^{-(x-a)^2}\,dx\\=e^{a^2}\int_{-a}^\infty e^{-u^2}du$$according to definition of $Q$-function we have:$$Q(x)=\dfrac{1}{\sqrt{2\pi}}\int_{x}^{\infty}e^{-\dfrac{u^2}{2}}du=\dfrac{1}{\sqrt{\pi}}\int_{x\over\sqrt 2}^{\infty}e^{-u^2}du$$So we can write here:$$\int_0^\infty e^{2ax-x^2}\,dx=e^{a^2}\sqrt\pi Q(-a\sqrt 2)$$also $$\int_{-\infty}^\infty e^{2ax-x^2}\,dx=e^{a^2}\sqrt\pi$$using a similar way.