I found an integral calculated from what I understand with “differentation under the integration sign” method.
$$
\int_{-\infty}^{\infty}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x= {\frac{\sqrt\pi}{b}}\mathrm{erf}\left(\frac{ab\left(c-d\right)}{\sqrt{a^{2}+b^{2}}}\right)
$$
for $b>0$.
The author of the original post explains how he obtained the solution:
“got it by differentiating the integrand w.r.t. a, then integrated over x=-inf..inf, then substituted a=sqrt(b*z)/sqrt(1-z) and integrated over z and then - most important - checked the result numerically.”
So I tried to follow that procedure and I get:
$$I\left(a\right)=\int_{-\infty}^{\infty}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x$$
$$\frac{\mathrm{d}I\left(a\right)}{\mathrm{d}a}=\frac{\mathrm{d}}{\mathrm{d}a}\int_{-\infty}^{\infty}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x$$
$$\frac{\mathrm{d}I\left(a\right)}{\mathrm{d}a}=\int_{-\infty}^{\infty}\frac{\partial}{\partial a}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x$$
$$\frac{\mathrm{d}I\left(a\right)}{\mathrm{d}a}=\int_{-\infty}^{\infty}\frac{2\exp\left(-b^{2}(-c+x)^{2}-a^{2}(-d+x)^{2}\right)(-d+x)}{\sqrt{\pi}}\,\mathrm{d}x$$
$$\frac{\mathrm{d}I\left(a\right)}{\mathrm{d}a}=\frac{2b^{2}(c-d)}{\left(a^{2}+b^{2}\right)^{3/2}}\exp\left(-\frac{a^{2}b^{2}(c-d)^{2}}{a^{2}+b^{2}}\right)$$
Now I substitute $z=\frac{a^2}{a^2+b^2}$ and after some manipulations I get the right side of the last equation:
$$
2\sqrt{b}(c-d)(1-z)^{3/2}\exp\left(-zb^{2}(c-d)^{2}\right)
$$
I’m not sure what my next step should be so I would appreciate any suggestions.
Answer
Let's try a close variant : differentiate relatively to $d$ instead of $a$ at the start (to avoid the additional factor $(x-d)$ in the integral and complications in the final integration) :
$$I\left(d\right)=\int_{-\infty}^{\infty}e^{-b^{2}\left(x-c\right)^{2}}\ \mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x$$
$$\frac{\mathrm{d}I\left(d\right)}{\mathrm{d}d}=\frac {-2a}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-b^{2}(x-c)^{2}-a^{2}(x-d)^{2}}\,\mathrm{d}x$$
$$\frac{\mathrm{d}I\left(d\right)}{\mathrm{d}d}=\frac {-2a}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-b^{2}(x-c)^{2}-a^{2}(x-d)^{2}}\,\mathrm{d}x$$
$$\frac{\mathrm{d}I\left(d\right)}{\mathrm{d}d}=\frac {-2a}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-\left(a^2+b^2\right)\left(x-\frac{b^2c+a^2d}{a^2+b^2}\right)^2+\frac{\left(b^2c+a^2d\right)^2}{a^2+b^2}-\left(b^2c^2+a^2d^2\right)}\,\mathrm{d}x$$
$$\frac{\mathrm{d}I\left(d\right)}{\mathrm{d}d}=\frac {-2a}{\sqrt{\pi}}e^{\frac{\left(b^2c+a^2d\right)^2}{a^2+b^2}-\left(b^2c^2+a^2d^2\right)}\int_{-\infty}^{\infty}e^{-\left(a^2+b^2\right)y^2}\,\mathrm{d}y$$
$$\frac{\mathrm{d}I\left(d\right)}{\mathrm{d}d}=\frac {-2a}{\sqrt{\pi}}e^{\frac{-a^2b^2\left(c-d\right)^2}{a^2+b^2}}\frac{\sqrt{\pi}}{\sqrt{a^2+b^2}}=-2a\frac{e^{\frac{-a^2b^2\left(c-d\right)^2}{a^2+b^2}}}{\sqrt{a^2+b^2}}$$
At this point we have to integrate again relatively to $d$ to get (up to a function $C$ independent of $d$) :
$$I\left(d\right)=\frac {\sqrt{\pi}}b\operatorname{erf}\left(\frac{ab(c-d)}{\sqrt{a^2+b^2}}\right)+C(a,b,c)$$
(Alpha integration check : note that the denominator is $b$ and not $\sqrt{b}$ nor my earlier $ab$ as I checked numerically!)
After that you'll just have to prove that $C(a,b,c)\equiv 0$
Note that the integration relatively to $d$ seems more straightforward than relatively to $a$ in your case (I'm not saying it can't be done your way!).
Hoping it clarified things a little even if it didn't answer your question,
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