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In an answer to Integrating a product of exponentials and error functions the integral $I(\gamma)$ below is evaluated using a differentiation technique:

$I(\gamma)= \int_0^\infty x^2\exp(-\delta x^2)\operatorname{erf}(\gamma x)\ \mathrm{d} x$ $I^\prime(\gamma) = \frac{2}{\sqrt{\pi}} \int_0^\infty x^3 \exp(-x^2 \left( \delta + \gamma^2 \right) ) \mathrm{d} x = \frac{1}{(\gamma^2 + \delta)^2} \frac{1}{\sqrt{\pi}}$

where the differentiation is wrt $\gamma$.

Thus:

$$ I(\gamma) = \int_0^\gamma \frac{1}{\sqrt{\pi}} \frac{1}{(\gamma^2+\delta)^2} \mathrm{d} \gamma = \frac{\gamma }{2 \sqrt{\pi } \delta \left(\gamma ^2+\delta \right)}+\frac{1}{2 \sqrt{\pi } \delta ^{3/2}} \arctan\left(\frac{\gamma }{\sqrt{\delta }}\right) $$

why is the final integral integrated between the limits 0 and $\gamma$?

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apg
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    Essentially see the comments by Sasha on the second answer in http://math.stackexchange.com/questions/78615/integrating-a-product-of-exponentials-and-error-functions?rq=1 – apg Feb 15 '14 at 15:52

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For a differentiable function $f(x)$, we know that $\displaystyle \int_0^x f'(t)\mathrm{d}t = f(x) - f(0)$, so that $f(x) = \displaystyle \int_0^xf'(t)\mathrm{d}t + f(0)$.

So in this case, $I(\gamma) = \displaystyle\color{#aa2020}{I(0)} + \int_0^\gamma I'(t)\mathrm{d}t$, and as $\operatorname{erf}(0) = 0$, we also have $\color{#aa2020}{I(0)} = 0$. So that's why.

This is a fairly common occurrence when differentiating under the integral to do this sort of trick.

davidlowryduda
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