I'm trying to evaluate the following integral via Feynman's method (where erf is the error function):
$ \int_{0}^{\infty} \frac{b}{2} \exp(-bx)\ erf(\sqrt{xc}) dx$
I'm considering the function I(a) defined as:
$I(a) = \int_{0}^{\infty} \frac{b}{2} \exp(-bx)\ erf(a\sqrt{xc}) dx$
Deriving this function:
$\frac{dI(a)}{da} = \frac{b}{2} \int_{0}^{\infty}\exp(-bx)\Big(\frac{2}{\sqrt{\pi}}\exp(a^2xc)\sqrt{xc} \Big) dx = \frac{b\sqrt{c}}{2}\frac{2}{\sqrt{\pi}} \int_{0}^{\infty}\sqrt{x} \exp(-x(b+a^2c)) dx =$
$=\frac{b\sqrt{c}}{2}\frac{2}{\sqrt{\pi}} \frac{1}{(b+a^2c)^{\frac{3}{2}}} \Gamma\big(\frac{3}{2}\big) = \frac{b\sqrt{c}}{2}\frac{1}{(b+a^2c)^{\frac{3}{2}}}$
We conclude that:
$I(a) = \frac{b\sqrt{c}}{2} \int \frac{1}{(b+a^2c)^{\frac{3}{2}}} da = \frac{b\sqrt{c}}{2} \frac{a}{b\sqrt{b+ca^2}}+ Const$
We have that $Const = 0$, because $0+Const = I(0)$. In this way we have:
$I(1) = \int_{0}^{\infty} \frac{b}{2} \exp(-bx)\ erf(\sqrt{xc}) dx = \frac{1}{2\sqrt{1+\frac{b}{c}}}$
The solution in my book gives: $\frac{1}{2}\Big(1-\frac{1}{\sqrt{1+\frac{b}{c}}}\Big)$
Can someone help me understand what I'm doing wrong?
Thanks!
Using erfc instead erf: $ \int_{0}^{\infty} \frac{b}{2} \exp(-bx)\ erfc(\sqrt{xc}) dx$
We have:
$ I(a) = \int_{0}^{\infty} \frac{b}{2}\exp(-bx)\frac{2}{\sqrt{\pi}}\int_{a\sqrt{cx}}^{\infty} \exp(-t^2) dt dx $
$\therefore \frac{dI(a)}{da} = -\frac{b\sqrt{c}}{2}\frac{2}{\sqrt{\pi}}\int_{0}^{\infty} \exp(-x(b+a^2c))\sqrt{x}dx = -\frac{b\sqrt{c}}{2}\frac{2}{\sqrt{\pi}}\Gamma\big(\frac{3}{2}\big)\frac{1}{(b+a^2c)^{\frac{3}{2}}}= -\frac{b\sqrt{c}}{2}\frac{1}{(b+a^2c)^{\frac{3}{2}}}$
Therefore ($\sqrt{\frac{c}{b}}a = tgu)$:
$I(a) = -\frac{b\sqrt{c}}{2}\int \frac{da}{(b+a^2c)^{\frac{3}{2}}} = -\frac{1}{2}\int cos(u)du = -\frac{1}{2}\sin\Big(\tan^{-1}(\sqrt{\frac{c}{b}}a)\Big) =-\frac{a\sqrt{c}}{2\sqrt{b+a^2c}}+ Const$
We have that $I(0) = \frac{1}{2} = Const $. In this way, the integral is obtained evaluating $I(1) = \frac{1}{2}\Bigg(1-\frac{1}{\sqrt{1+\frac{b}{c}}}\Bigg)$