# How do I evaluate this integral using Beta functions?

How do I evaluate this integral using Beta functions?

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

Edit 1:The answer given for this question is : $$e^{a^2}* \frac{\sqrt \pi}{2}$$

How do I get that using Beta functions? I have been breaking my head over it for a long time.

Edit 2: There's a possibility that the answer given is wrong. Even then is it possible to solve it using Beta functions?

• Please can you check the integrand? Is it $e^{2ax-x}=e^{(2a-1)x}$? – Olivier Oloa May 8 '18 at 18:49
• Yes. That's right. – Hari May 8 '18 at 18:50
• Then we must have $2a-1<0$ to ensure convergence. – Olivier Oloa May 8 '18 at 18:51
• Are you sure you typed the integral correctly? because $$\frac{e^{a(2a-1)}}{1-2a} \neq e^{a^2}\frac{\sqrt\pi}{2}$$ – Dando18 May 8 '18 at 19:22
• The answer you suggest fits with $\int_a^{+\infty}e^{2ax-x^2}\,dx$. Perhaps there is a misprint? – mickep May 8 '18 at 19:30

Hint. One has $$\int e^{cx}dx=\frac{e^{cx}}{c},\qquad c\ne0.$$
Integrate using $\int e^{ax}dx = \frac{1}{a}e^{ax}+c$.
\begin{align*} \int e^{(2a-1)x}dx &= \left[ \frac{e^{(2a-1)x}}{2a-1} \right]^\infty_{a} \\ &= \frac{e^{a(2a-1)}}{1-2a}, \quad \textrm{for } \Re[a]<1/2 . \\ \end{align*}
Where we require $a$ be less than $1/2$ so that the integrand converges at $\infty$ and the denominator is not $0$.