I need you help with evaluating this integral: $$I=\int_0^\infty F(z)\,F\left(z\,\sqrt2\right)\frac{e^{-z^2}}{z^2}dz,\tag1$$ where $F(x)$ represents Dawson's integral: $$F(x)=e^{-x^2}\int_0^x e^{y^2}dy.\tag2$$


$$ \begin{align} \int_{0}^{\infty} F(x) F(x \sqrt{2}) \frac{e^{-x^{2}}}{x^{2}} dx &= \int_{0}^{\infty} \int_{0}^{\sqrt{2}} \int_{0}^{1} x e^{-x^{2}} e^{x^{2} y^{2}} x e^{-2x^{2}} e^{x^{2}z^{2}} \frac{e^{-x^{2}}}{x^{2}} \ dy \ dz \ dx \\ &= \int_{0}^{\sqrt{2}} \int_{0}^{1} \int_{0}^{\infty} e^{-(4-y^{2}-z^{2})x^{2}} \ dx \ dy \ dz \\ &= \frac{\sqrt{\pi}}{2} \int_{0}^{\sqrt{2}} \int_{0}^{1} \frac{1}{\sqrt{4-y^{2}-z^{2}}} \ dy \ dz \end{align}$$

Let $y = \sqrt{4-z^{2}} \sin \theta$.

$$\begin{align} &= \frac{\sqrt{\pi}}{2} \int_{0}^{\sqrt{2}} \int_{0}^{\arcsin ( \frac{1}{\sqrt{4-z^{2}}})} \ d \theta \ d z \\ &= \frac{\sqrt{\pi}}{2} \int_{0}^{\sqrt{2}} \arcsin \left( \frac{1}{\sqrt{4-z^{2}}} \right) \ dz \end{align}$$

Now integrate by parts.

$$ = \frac{\sqrt{\pi}}{2} \left( \frac{ \sqrt{2} \pi}{4} - \int_{0}^{\sqrt{2}} \frac{z^{2}}{\sqrt{3-z^{2}} (4-z^{2})} \ dz \right)$$

Let $ \displaystyle z = \frac{1}{u}$.

$$ = \frac{\sqrt{\pi}}{2} \left( \frac{\sqrt{2} \pi}{4} - \int_{1 / \sqrt{2}}^{\infty} \frac{1}{\sqrt{3u^{2}-1} (4u^{2}-1)} \frac{du}{u}\right) $$

Let $ \displaystyle w^{2} = 3u^{2}-1$.

$$ \begin{align} &= \frac{\sqrt{\pi}}{2} \left( \frac{\sqrt{2} \pi}{4} - 3 \int_{1 /\sqrt{2}}^{\infty} \frac{1}{ (4w^{2}+1)(w^{2}+1)} \ dw\right) \\ &=\frac{\sqrt{\pi}}{2} \left( \frac{\sqrt{2} \pi}{2} - 4 \int_{1/ \sqrt{2}}^{\infty} \frac{1}{4w^{2}+1} + \int_{1/ \sqrt{2}}^{\infty} \frac{1}{w^{2}+1} \ dw\right) \\ &= \frac{\sqrt{\pi}}{2} \left[ \frac{\sqrt{2} \pi}{4} - \pi +2 \arctan \left( \sqrt{2} \right) +\frac{\pi}{2} - \arctan \left( \frac{1}{\sqrt{2}} \right) \right] \\ &= \frac{\sqrt{\pi}}{2} \left( \frac{\sqrt{2} \pi}{4} - \pi + 3 \arctan{\sqrt{2}} \right) \end{align}$$


Using the same approach, one can derive the generalization

$$ \int_{0}^{\infty} F(ax) F(bx) \frac{e^{-p^{2}x^{2}}}{x^{2}} \ dx $$

$$ = \frac{\sqrt{\pi}}{2} \left[b \arcsin \left( \frac{a}{\sqrt{a^{2}+p^{2}}} \right) - \sqrt{a^{2}+b^{2}+p^{2}} \arctan \left(\frac{ab}{p \sqrt{a^{2}+b^{2}+p^{2}}} \right) + a \arctan \left( \frac{b}{p} \right)\right]$$

where $a, b,$ and $c$ are all positive parameters.



  • 6
    $\begingroup$ (-1): As with many of your other answers, there is no indication whatsoever of how you found this result, or any suggestion of why it's true. Please share with the result of us exactly how you're finding these integrals (I'm honestly curious, and would like to know the techniques you're using). In its current state, this answer seems to be far more appropriate as a comment. $\endgroup$ – user61527 Mar 16 '14 at 4:56
  • $\begingroup$ It is very unlikely that Cleo will ever share the source of her results. (See math.stackexchange.com/questions/562694/…, meta.math.stackexchange.com/q/11759, meta.math.stackexchange.com/q/11723.) $\endgroup$ – user21820 Nov 28 '16 at 9:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.