# Seeking help with an error function Integral

I am trying to compute the following Integral $$I = \int_{0}^\infty x \exp \left(-2 x \right) \operatorname{erf}\left(\frac{x}{t^{H}\sqrt[4]{2}}-\frac{t^H}{2^{3/4}}\right) \, dx$$ where $\operatorname{erf} \left(z\right)$ is the error function given by $$\operatorname{erf}(z) = \frac{2}{\sqrt{\pi}}\int_0^z \mathrm{e}^{-t^2} \,dt .$$

Edit To provide some context to the problem I have a probability density function, which I am trying to verify that it is well behaved and integrates to 1. I am also interested in computing the expectation and the integral above is the part I am struggling to integrate. In order to reduce the noise, I only posted part of the formula I was struggling with. The below integral should work out to be 1, but I have not been able to establish that, as of yet. \begin{eqnarray} I &=& \int_0^{\infty}\mathrm{e}^{-2 x} \left(\Psi\left(x, t\right) + \frac{2^{3/4} \mathrm{e}^{x-\frac{t^{-2 H} \left(t^{4 H}+2 x^2\right)}{2 \sqrt{2}}}}{t^{H}\sqrt{\pi }}\right){\kern 1pt} \,dx \\ &=& \int_0^{\infty} \mathrm{e}^{-2 x}\Psi\left(x, t\right){\kern 1pt} \,dx + \frac{2^{3/4}}{t^{H}\sqrt{\pi }}\int_0^{\infty}\mathrm{e}^{-x-\frac{t^{-2 H} \left(t^{4 H}+2 x^2\right)}{2 \sqrt{2}}}{\kern 1pt} \,dx\\ &=& \int_0^{\infty} \mathrm{e}^{-2 x}\Psi\left(x, t\right){\kern 1pt} \,dx + \operatorname{erfc}\left(\frac{t^H}{2^{3/4}}\right)\\ \end{eqnarray} where $$\Psi\left(x, t\right) = \operatorname{erfc} \left(\frac{x_0}{t^{H}\sqrt[4]{2}}-\frac{t^H}{2^{3/4}}\right),$$

Formulation 2 $$\lim_{x\to\infty} \frac{1}{2}\left(\Xi\left(x, t\right) - \mathrm{e}^{2 x} \Psi\left(x, t\right)+1\right) = 1$$ where $$$$\Xi\left(x_0, t\right) = \operatorname{erf} \left(\frac{x_0}{t^{H}\sqrt[4]{2}}+\frac{t^H}{2^{3/4}} \right),$$$$ Any help would be much appreciated.

• @Michael Hardy, Thank you very much for the edit. I found it educational. May 6 '16 at 5:19
• As $x \to -\infty$ the first factor inside the integral tends to $+\infty$ and the second to $-2$. This smells wrong to me. Isn't some bad sign somewhere? May 6 '16 at 12:44
• @leonbloy actually I did find an error in my derivation a few steps, before I got to this integral. I have revised version. May 6 '16 at 12:46
• Did you try to change the order of integration? Actually, what did you try?
– Did
May 6 '16 at 12:56
• @Did, I am not sure how to solve it in this case by order of integration. I do have another formulation, of my problem I will add, maybe that is easier. May 6 '16 at 13:10

If $a=\frac1{t^H\sqrt[4]2}$, $b=\frac{t^H}{2^{3/4}}$, then we need \begin{align}\int_0^{\infty}xe^{-2x}\cdot\frac2{\sqrt{\pi}}\int_0^{ax-b}e^{-y^2}dy&=\left.\frac2{\sqrt{\pi}}\left(-\frac x2-\frac14\right)e^{-2x}\int_0^{ax-b}e^{-y^2}dy\right|_0^{\infty}\\ &+\frac a{\sqrt{\pi}}\int_0^{\infty}\left(x+\frac12\right)e^{-2x}e^{-(ax-b)^2}dx\\ &=\frac14\frac2{\sqrt{\pi}}\int_0^{-b}e^{-y^2}dy\\ &+\frac a{\sqrt{\pi}}e^{-\frac{2b}a+\frac1{a^2}}\int_0^{\infty}\left(x-\frac ba+\frac1{a^2}+\frac ba-\frac1{a^2}+\frac12\right)e^{-a^2\left(x-\frac ba+\frac1{a^2}\right)^2}dx\\ &=\frac14\text{erf}(-b)+\frac a{\sqrt{\pi}}e^{-\frac{2b}a+\frac1{a^2}}\int_{-\frac ba+\frac1{a^2}}^{\infty}\left(u+\frac ba-\frac1{a^2}\right)e^{-a^2u^2}du\\ &=\frac14\text{erf}(-b)+\frac a{\sqrt{\pi}}e^{-\frac{2b}a+\frac1{a^2}}\left\{\left.-\frac1{2a^2}e^{-a^2u^2}\right|_{-\frac ba+\frac1{a^2}}^{\infty}+\frac1a\left(\frac ba-\frac1{a^2}\right)\int_{-b+\frac1a}^{\infty}e^{-v^2}dv\right\}\\ &=\frac14\text{erf}(-b)+e^{-\frac{2b}a+\frac1{a^2}}\left\{\frac1{2a\sqrt{\pi}}e^{-\left(b-\frac1a\right)^2}+\frac12\left(\frac ba-\frac1{a^2}\right)\text{erfc}\left(-b+\frac1a\right)\right\}\end{align} Is that the kind of stuff you're looking for?
EDIT: I should have gone a little farther since $ab=\frac12$. Then $\frac1a=2b$, so we can write this result as $$\int_0^{\infty}xe^{-2x}\cdot\frac2{\sqrt{\pi}}\int_0^{ax-b}e^{-y^2}dy= \frac14\text{erf}(-b)+\frac1{2a\sqrt{\pi}}e^{-b^2}-b^2\text{erfc}(b)$$ Also by popular demand, \begin{align}\int_0^{\infty}e^{-2x}\text{erfc}(ax-b)dx+\text{erfc}(b)&= \text{erfc}(b)+\frac2{\sqrt{\pi}}\int_0^{\infty}e^{-2x}\int_{ax-b}^{\infty}e^{-y^2}dy\\ &=\text{erfc}(b)+\frac2{\sqrt{\pi}}\left\{\left.-\frac12e^{-2x}\int_{ax-b}^{\infty}e^{-y^2}dy\right|_0^{\infty}-\frac a2\int_0^{\infty}e^{-2x}e^{-(ax-b)^2}dx\right\}\\ &=\text{erfc}(b)+\frac12\text{erfc}(-b)-\frac a{\sqrt{\pi}}e^{-\frac{2b}a+\frac1{a^2}}\int_0^{\infty}e^{-a^2\left(x-\frac ba+\frac1{a^2}\right)^2}dx\\ &=\text{erfc}(b)+\frac12\text{erfc}(-b)-\frac a{\sqrt{\pi}}e^{-\frac{2b}a+\frac1{a^2}}\int_{-\frac ba+\frac1{a^2}}^{\infty}e^{-a^2u^2}du\\ &=\text{erfc}(b)+\frac12\text{erfc}(-b)-\frac1{\sqrt{\pi}}e^{-\frac{2b}a+\frac1{a^2}}\int_{-b+\frac1{a}}^{\infty}e^{-v^2}vu\\ &=\text{erfc}(b)+\frac12\text{erfc}(-b)-\frac12e^{-\frac{2b}a+\frac1{a^2}}\text{erfc}\left(-b+\frac1a\right)\\ &=\text{erfc}(b)+\frac12\text{erfc}(-b)-\frac12\text{erfc}\left(b\right)\\ &=\frac12\text{erfc}(b)+\frac12\text{erfc}(-b)\\ &=\frac1{\sqrt{\pi}}\int_b^{\infty}e^{-y^2}dy+\frac1{\sqrt{\pi}}\int_{-b}^{\infty}e^{-y^2}dy\\ &=\frac1{\sqrt{\pi}}\int_b^{\infty}e^{-v^2}dv-\frac1{\sqrt{\pi}}\int_b^{-\infty}e^{-v^2}dv\\ &=\frac1{\sqrt{\pi}}\int_b^{\infty}e^{-v^2}dv+\frac1{\sqrt{\pi}}\int_{-\infty}^be^{-v^2}dv\\ &=\frac1{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-v^2}dv\\ &=\frac1{\sqrt{\pi}}(2)\left(\frac12\right)\Gamma\left(\frac12\right)=1\end{align} Verification complete, but I noticed that my error rate was kind of high in the above. Please check and let me know of any further edits required.
• Are you referring to the step $$-\frac12e^{-\frac{2b}a+\frac1{a^2}}\text{erfc}\left(-b+\frac1a\right)=-\frac12\text{erfc}\left(b\right)$$ That step is actually a no-op given that $\frac1a=2b$: $$e^{-\frac{2b}a+\frac1{a^2}}=e^{-4b^2+4b^2}=e^0=1$$ and $$\text{erfc}\left(-b+\frac1a\right)=\text{erfc}(-b+2b)=\text{erfc}(b)$$ May 7 '16 at 2:41