1
$\begingroup$

Problem

I would like to compute the integral:

\begin{align} \int_{0}^{+\infty} \text{erf}(ax+b) \exp(-(cx+d)^2) dx \tag{1} \end{align}

I have been looking at this popular table of integral of the error functions, and also found here the following expression:

\begin{align} \int_{-\infty}^{+\infty} e^{-(\alpha x+ \beta)^2} \text{erf}(\gamma x+\delta) dx = \dfrac{\sqrt{\pi}}{\alpha} \text{erf} \left[ \dfrac{\alpha \delta - \beta \gamma}{\sqrt{\alpha^2+ \gamma^2}} \right] \tag{2} \end{align}

as well as:

\begin{align} \int_{0}^{+\infty} e^{-\alpha^2 x^2} \text{erf}(\beta x) dx = \dfrac{\text{arctan}(\beta / \alpha)}{\alpha \sqrt{\pi}} \tag{3} \end{align}

However the latter (3) is only a particular case of (1), which is what I am looking for. Do you know how to prove (2)? This might help me understand how to compute (1)? Or do you know how to compute (1)?

$\endgroup$
  • $\begingroup$ in the second integral : ${{e}^{-{{(\alpha x+\beta )}^{2}}}}$ $\endgroup$ – user547564 Aug 30 '18 at 12:00
1
$\begingroup$

Partial solution

Define: $$I\left( \delta \right)=\int_{-\infty }^{+\infty }{{{e}^{-{{(\alpha x+\beta )}^{2}}}}}\text{erf}(\gamma x+\delta )dx$$ now differentiate with respect to $\delta$ $$\begin{align} & \frac{dI}{d\delta }=\int_{-\infty }^{+\infty }{{{e}^{-{{(\alpha x+\beta )}^{2}}}}}\frac{\partial }{\partial \delta }\left( \text{erf}(\gamma x+\delta ) \right)dx \\ & \quad \quad =\int_{-\infty }^{+\infty }{{{e}^{-{{(\alpha x+\beta )}^{2}}}}\left( \frac{2{{e}^{-{{\left( \gamma x+\delta \right)}^{2}}}}}{\sqrt{\pi }} \right)}dx\ \\ & \quad \quad =\frac{2}{\sqrt{\pi }}\int_{-\infty }^{+\infty }{{{e}^{-{{(\alpha x+\beta )}^{2}}-{{\left( \gamma x+\delta \right)}^{2}}}}}dx=\frac{2}{\sqrt{\pi }}\left( \frac{\sqrt{\pi }{{e}^{-\frac{{{\left( \alpha \delta -\beta \gamma \right)}^{2}}}{{{\alpha }^{2}}+{{\gamma }^{2}}}}}}{\sqrt{{{\alpha }^{2}}+{{\gamma }^{2}}}} \right) \\\end{align}$$ Finally, you can find that: $$I\left( \delta \right)=\frac{\sqrt{\pi }}{\alpha }erf\left( \frac{\alpha \delta -\beta \gamma }{\sqrt{{{\alpha }^{2}}+{{\gamma }^{2}}}} \right)$$

$\endgroup$
  • 1
    $\begingroup$ The proof you presented is known; the problem lies with the lower limit of 0 instead of $-\infty.$ I upvoted anyway because the OP wants a proof of this formula too. $\endgroup$ – skbmoore Aug 30 '18 at 15:01
  • $\begingroup$ you are right, in fact i figured that lately,....may be i Should delete my answer?? $\endgroup$ – user547564 Aug 30 '18 at 15:03
  • 1
    $\begingroup$ Leave the answer -- it may inspire the OP or a reader to derive an answer, though I'm not too hopeful. $\endgroup$ – skbmoore Aug 31 '18 at 1:09

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.