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I have a double integral, which I want to solve:

$\mathcal{I}_{au}=\int_0^{t'}e^{(a-c)t''}\int_0^{t''}e^{(-b+c)t'''}\text{exp}\big(-\frac{\lambda^2}{2}\left( S_2(t''')^2+S_1(t')^2+S_1(t'')^2+2S_{12}t't'''-2S_{12}t''t'''-2S_1t't''\right) \big)dt'''dt''$

I am able to solve the inner integral as follows:

$\mathcal{I}_{au}=\int_0^{t'}e^{(a-c)t''} \mathcal{I}_{in} dt''$

with $\mathcal{I}_{in}=\frac{\sqrt{\pi}}{\sqrt{2S_{2}}\lambda}\text{exp}\bigg( \frac{b^2+c^2+2cS_{12}\lambda^2(t''-t')+(S_{12}^2-S_1S_2)(t'-t'')^2\lambda^4-2b(c+S_{12}\lambda^2(t''-t'))}{2S_2\lambda^2}\bigg)\times \bigg( \text{erf}\left( \frac{b-c+(S_{12}\lambda^2(t'-t'')+S_2t'')\lambda^2}{\sqrt{2S_2}\lambda}\right)-\text{erf}\left( \frac{b-c+(S_{12}\lambda^2(t'-t'')+S_2t'')\lambda^2}{\sqrt{2S_2}\lambda}\right)\bigg)$

but afterwards I am not able to integrate this thing in the form of $\int_0^{t'}\text{exp}(ax+bx^2)(\text{erf}(cx+d)-\text{erf}(gx+d))dx$.

With my variables it seems to be equal to zero, at least numerically. The variables I tested are always giving zero, all, which do not form the following exceptions:

(1) $S_{12}=0$ leads to something nonzero. (2) $S_{12}=\sqrt{S_1S_2}$ does also lead to a nonzero result ($S_1=S_2=S_{12}$ do also lead to nonzero.)

Is there an analytical expression for this integral? Why does it get zero in these special cases? How do I solve this? (Partial Integration does not work.)

Thanks for your help!!!

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  • $\begingroup$ For $b=0$, Mathematica can compute the integral analytically. Try FullSimplify[LaplaceTransform[(Erf[c x+d]-Erf[g x + d])UnitBox[x/(2t)],x,a,Assumptions->t>0]] $\endgroup$ Commented Sep 22 at 16:00

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If you goal is to compute the antiderivative$$I=\int\text{exp}(ax+bx^2)\,\text{erf}(cx+d)\,dx$$ you could try $$\text{erf}(cx+d)=\frac{2}{\sqrt{\pi }}\sum_{n=0}^\infty (-1)^n\, \frac {(cx+d)^{2n+1}}{(2 n+1)\, n! }$$

The antiderivatives $$J_n=\int \text{exp}(ax+bx^2)\,\,(cx+d)^{2n+1}\,dx$$ are given by Wolfram Alpha for $n=0,1,2,3$ but any CAS would provide them (they are quite nasty).

To make life a bit easier, let $$x=\frac{y-d}{c} \qquad\qquad \alpha=\frac{a c-2 b d}{c^2}\qquad\qquad \beta=\frac b {c^2}$$

$$J_n=\frac 1c\,\exp\Big(\frac{d (b d-a c)}{c^2}\Big)\,\int \text{exp}(\alpha\,y+\beta\, y^2)\,\,y^{2n+1}\,dy$$ are a bit more pleasant.

Edit

If $b$ is sufficiently small, we could write the integrand as $$\sum_{n=0}^\infty \frac 1{n!} e^{a x}\, x^{2 n}\, \text{erf}(c x+d)\,b^n$$

The required integrals

$$J_n=\int e^{a x}\, x^{2 n}\, \text{erf}(c x+d)\,dx$$ have closed form.

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  • $\begingroup$ Thank you for your answer, this holds true, if cx+d is small and real. Unfortunately, this is not the case in my situation. $\endgroup$ Commented Sep 30 at 6:37
  • $\begingroup$ @NickAndersson1411.As always when you use series expansion. Good luck with the monster ! By the way is $a$ or $b$ small ? $\endgroup$ Commented Sep 30 at 6:45
  • $\begingroup$ both are complex and have a big imaginary part. $\endgroup$ Commented Oct 1 at 7:47

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