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I want to calculate this integral:

$$I = -k\int_0^\infty \text{erfc}\left({\frac{\eta + z + 2akt}{2\sqrt{at}}}\right)\exp\left[k(\eta + z + akt)\right]\ d\eta$$

In his Handbook of Linear Partial Differential Equations on pg. 181, Polyanin seems to get the following answer (I have to work backwards a little bit):

$$I = -\text{erfc}\left(\frac{z}{2\sqrt{at}}\right) + \exp\left[k(z + akt)\right]\ \text{erfc}\left({\frac{z + 2akt}{2\sqrt{at}}}\right) $$

So the $\eta$ s just disappeared, I'm not sure about the $k$ out in front, and a new $\text{erfc}$ function appeared. I saw this question and this question which make it seem impossible. What did he do?

HiddenBabel
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1 Answers1

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Making notations simpler $$I=\int \text{erfc}(a x+b) \, e^{c x+d}\, dx=\int e^{c x+d}\,dx-\int \text{erf}(a x+b)\, e^{c x+d}\,dx$$ $$\int e^{c x+d}\,dx=\frac 1c\, e^{c x+d}$$ One integration by parts $$\int \text{erf}(a x+b)\, e^{c x+d}\,dx=\frac 1c\, \text{erf}(a x+b)\, e^{c x+d}-\frac 1c\,e^{\frac{c (c-4 a b)}{4 a^2}+d}\,\text{erf}\left(\frac{2 a b-c}{2 a}+a x\right)$$ Recombining, using the bounds (assuming $a>0$ and $c<0$), the definite integral is $$J=\frac {e^d}c\left(e^{\frac{c (c-4 a b)}{4 a^2}} \text{erfc}\left(\frac{2 a b-c}{2 a}\right)+\text{erf}(b)-1\right) $$

Now, make $$a=\frac{1}{2 \sqrt{a t}} \qquad b=\frac{z+2 a k t}{2 \sqrt{a t}}\qquad c=k \qquad d=k(z+kat)$$

$$- k \,J=e^{k (a k t+z)}\, \text{erfc}\left(\frac{z+2 a k t}{2 \sqrt{a t}}\right)-\text{erfc}\left(\frac{z}{2 \sqrt{a t}}\right)$$ which is your formula.

Claude Leibovici
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