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A difficult integral with exp and erf squared

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petru
#1
Apr29-12, 04:43 AM
P: 6
Hello,

I have big difficulties solving the following integral:
[tex]
\int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x
[/tex]

I tried integration by parts, and also tried to apply the technique called “differentiation under the integration sign” but with no results.

I’m not very good at calculus so my question is if anyone could give me any hint of how to approach this integral. I would be ultimately thankful.

If it could help at all, I know that
[tex]
\int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x=\frac{a}{b^{2}\sqrt{a^{2}+b^{2}}}\exp\left (-\frac{a^{2}b^{2}\left(c-d\right)^{2}}{a^{2}+b^{2}}\right)+\frac{\sqrt{\pi}c}{b}\mathrm{erf}\lef t(\frac{ab\left(c-d\right)}{\sqrt{a^{2}+b^{2}}}\right),
[/tex]

and
[tex]
\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),
[/tex]

both for b>0.
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dextercioby
#2
Apr29-12, 05:03 AM
Sci Advisor
HW Helper
P: 11,894
If Mathematica and Gradshteyn-Rytzhik can't help you with the answer, it means it can't be done. You'd gave to let a,b,c,d have some specific numerical values and then use approximation techniques.
petru
#3
Apr29-12, 07:00 AM
P: 6
dextercioby Thanks for your reply! I spent a lot of time trying to find closed form of that integral, so even if it can’t be done, I would like to learn smoething out of it and thus I have another question.

The last integral in my first post:

[tex]\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),[/tex]
for b>0,

was calculated using “differentation under the integration sign” method.

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:
[tex]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[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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)[/tex]

Now I substitute [tex]z=\frac{a^2}{a^2+b^2}[/tex] and after some manipulations I get the right side of the last equation:
[tex]
2\sqrt{b}(c-d)(1-z)^{3/2}\ \exp\left(-zb^{2}(c-d)^{2}\right)
[/tex]

I would appreciate any suggestions of how I should proceed.


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