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I can evaluate the following integral

$$ \int_{0}^{\infty}x^{2}\exp\left(-bx^{2}\right)\mathrm{erf}\left(ax\right)\,\mathrm{d}x=\frac{a}{2\sqrt{\pi}b\left(a^{2}+b\right)}+\frac{1}{2\sqrt{\pi}b^{3/2}}\arctan\left(\frac{a}{\sqrt{b}}\right) $$

using differentiation under the integral sign with $a$ as the parameter. I'm curious about the conditions for that procedure, though.

A certain set of necessary conditions can be found here: Will moving differentiation from inside, to outside an integral, change the result?

I have some troubles with the 3rd one:

  • for some integrable function $g$, for all $x\in E$, and for all $a\in U$, $$ \left| \frac{\partial f}{\partial a}(a, x) \right| \le g(x). $$

I want to find $g\left(x\right)$ (independent of $a$):

$$ \left|\frac{\partial}{\partial a} x^{2}\exp\left(-bx^{2}\right)\mathrm{erf}\left(ax\right)\right|=\left|\frac{2}{\sqrt{\pi}} x^{3}\exp\left(-a^{2}x^{2}-bx^{2}\right)\right|\leq \frac{2}{\sqrt{\pi}} \left|x\right|\,x^{2}\exp\left(-bx^{2}\right)=g\left(x\right). $$

Is $g\left(x\right)$ valid here? I'm unsure of $\left|x\right|$ element.


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Since you are considering integration interval $(0,+\infty)$ you can say that $|x|=x$. –  Norbert Jul 5 '12 at 12:42
Thank you, you're obviously right. But if my integration interval was $(-\infty,+\infty)$ – how should I manage that? Define $g(x)$ for $(-\infty,0)$ and $[0,+\infty)$ separately? –  petru Jul 5 '12 at 13:16
Yes you can. But I don't understand why I so much afraid of $|x|$ –  Norbert Jul 5 '12 at 13:22
I'm not affraid, I just needed to be sure :-) Thanks! –  petru Jul 5 '12 at 13:26
Since $x \mapsto \exp x$ decays faster than any (negative) power of $x$ at infinity, your $g$ is integrable, and therefore does the job. –  Siminore Jul 5 '12 at 13:29

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