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I am running MMA to find the solution, but it is taking a long time. Before I wait for long time, Do we have solution for this integral?

$$\int_{-\infty }^{\infty } \frac{ (v-{\alpha_1}) (v-{\alpha_2})}{(v-{\beta_1}) (v-{\beta_2}) (v-{\beta_3})} e^{-v^2/u^2}\, dv$$

where all the roots are complex number.

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  • $\begingroup$ Do any of the roots lie on the real line? Also, you can get rid of the $/ u^2$ by rescaling $v$ and all the roots. $\endgroup$ – avs Sep 14 '16 at 0:54
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I am not surprised that a CAS has problems.

Thinking about the antiderivative and using partial fraction decomposition we have $$\frac{ (v-{\alpha_1}) (v-{\alpha_2})}{(v-{\beta_1}) (v-{\beta_2}) (v-{\beta_3})}=\sum_{i=1}^3\frac{A_i}{v-{\beta_i}} $$ and so we face the problem of $$\int \frac{e^{-\frac{v^2}{u^2}}}{v-\beta} \, dv$$ for which I suspect that no closed form exists.

However, using a CAS, what it produced is $$\int_{-\infty}^\infty \frac{e^{-\frac{v^2}{u^2}}}{v-\beta} \, dv=e^{-\frac{\beta^2}{u^2}} \left(-\pi \, \text{erfi}\left(\beta \sqrt{\frac{1}{u^2}}\right)+\log \left(-\frac{1}{\beta}\right)+\log (\beta)\right)$$ provided $\Re\left(u^2\right)>0\land \Im(\beta)\neq 0$.

So, it seems that you could obtain your result.

Edit

I was too lazy to compute the $A_i$'s which look quite nice $$A_1=\frac{(\alpha_1-\beta_1) (\alpha_2-\beta_1)}{(\beta_1-\beta_2) (\beta_1-\beta_3)} \qquad A_2=\frac{(\alpha_1-\beta_2) (\beta_2-\alpha_2)}{(\beta_1-\beta_2) (\beta_2-\beta_3)} \qquad A_3=\frac{(\alpha_1-\beta_3) (\beta_3-\alpha_2)}{(\beta_1-\beta_3) (\beta_3-\beta_2)} $$

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