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I am trying to evaluate the following integral

$$I = \int_1^{\infty } x \mathop{erfc}(a + b \log (x)) \, dx$$

where $a$, $b$ are some positive constants.

Using the substitution $t = \log (x)$, the integral transforms to $$I = \int_0^{\infty } \mathrm{e}^{2t} \mathop{erfc}(a + b t) \, dt.$$ I am not quite sure how to proceed from here. Any help will be greatly appreciated.

Thank you.

Edit let further $y= a + b t$, then the integral changes to $$I = \frac{1}{b}\int_a^{\infty } \mathrm{e}^{\frac{2}{b}\left(y - a\right)} \mathop{erfc}(y) \, dy.$$

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  • $\begingroup$ A shift $a+bt=y$ followed by an integration by parts will solve ur problem $\endgroup$ – tired Jul 7 '16 at 8:08
  • $\begingroup$ @tired, Thank you I will try that. $\endgroup$ – Comic Book Guy Jul 7 '16 at 8:12
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Hint. Integrating by parts, one easily gets $$ \int_0^{\infty } e^{2t} \mathop{erfc}(a + b t)\,dt=\left. \frac{ e^{2t} }2\mathop{erfc}(a + b t)\right|_0^\infty -\frac12\int_0^{\infty} e^{2t} \left( -\frac{2 b e^{-(a+b t)^2}}{\sqrt{\pi}}\right) dt $$ giving

$$ \begin{align} \int_0^{\infty } e^{2t} \mathop{erfc}(a + b t) \, dt&=-\frac{\mathop{erfc}(a)}2+\frac b{\sqrt{\pi }}\int_0^{\infty } e^{2t}e^{-(a+b t)^2} dt \\\\&=-\frac{\mathop{erfc}(a)}2+\frac12e^{\large \frac{1-2 a b}{b^2}} \mathop{erfc}\left(a-\frac{1}{b}\right). \end{align} $$

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