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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?

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  • $\begingroup$ Please can you check the integrand? Is it $e^{2ax-x}=e^{(2a-1)x}$? $\endgroup$ – Olivier Oloa May 8 '18 at 18:49
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    $\begingroup$ Yes. That's right. $\endgroup$ – Hari May 8 '18 at 18:50
  • $\begingroup$ Then we must have $2a-1<0$ to ensure convergence. $\endgroup$ – Olivier Oloa May 8 '18 at 18:51
  • $\begingroup$ Are you sure you typed the integral correctly? because $$ \frac{e^{a(2a-1)}}{1-2a} \neq e^{a^2}\frac{\sqrt\pi}{2} $$ $\endgroup$ – Dando18 May 8 '18 at 19:22
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    $\begingroup$ The answer you suggest fits with $\int_a^{+\infty}e^{2ax-x^2}\,dx$. Perhaps there is a misprint? $\endgroup$ – mickep May 8 '18 at 19:30
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Hint. One has $$ \int e^{cx}dx=\frac{e^{cx}}{c},\qquad c\ne0. $$

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  • $\begingroup$ But, how do I bring Beta functions is play, because simple integration won't work. $\endgroup$ – Hari May 8 '18 at 18:56
  • $\begingroup$ @Hari There is no need for the Beta function here... $\endgroup$ – Olivier Oloa May 8 '18 at 18:56
  • $\begingroup$ @Hari why won't simple integration work? $\endgroup$ – Dando18 May 8 '18 at 18:57
  • $\begingroup$ @Dando18 I got what you two are saying, simple integration, with that condition is right. But I am trying to solve this question using Beta functions. $\endgroup$ – Hari May 8 '18 at 19:17
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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 $0$.

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