Express $\int_0^{\infty } \frac{e^{-x^2}}{(a+x)^2+b^2} \ dx$ in terms of special functions

Question

My question is: How to express the integral $$\int_0^{\infty } \frac{e^{-x^2}}{(a+x)^2+b^2} \ dx$$ in terms of special functions, where a and b are positive real numbers? Notice that if a vanishes, the integral can be expressed by Error function (using a variation of parametric differentiation). Some tests in Mathematica supports that the integral can be expressed by Erf, Erfi, Si and Ci functions and elementary ones, for instance: $$\int_0^{\infty } \frac{e^{-x^2}}{x^2+2 x+2} \, dx=\frac{1}{8} e^{-2 i} \left(4 e^{2 i} (\text{Si}(2) \cos (2)-\text{Ci}(2) \sin (2))+e^{4 i} \pi (1-2 \text{erf}(1+i))+2 i \pi \text{erfi}(1+i)+\pi \right)$$ But I haven't got the full pattern yet. How would you solve this problem? Thank you for helping me.

Answer 1

Using (for sure) a CAS, if $$I_{a,b}=\int_0^{\infty } \frac{e^{-x^2}}{(a+x)^2+b^2} \, dx$$ then

$$-4 \,i \,b\, e^{(a+i b)^2}\, I_{a,b}=e^{4 i a b} \left(\text{Chi}\left((a-i b)^2\right)+\text{Shi}\left((a-i b)^2\right)-\pi \text{erfi}(a-i b)\right)-$$ $$\text{Chi}\left((a+i b)^2\right)-\text{Shi}\left((a+i b)^2\right)+\pi \text{erfi}(a+i b)$$ and $$I_{0,b}=-\frac{\pi e^{b^2} \text{erf}(b)}{2 b}$$ $$I_{a,0}=-a e^{-a^2} \left(\text{Chi}\left(a^2\right)+\text{Shi}\left(a^2\right)-\pi \text{erfi}(a)\right)+\frac{1}{a}-\sqrt{\pi }$$

Edit

You could have simpler expression writing $(a+x)^2+b^2=(x+r)(x+s)$ and then $$\frac 1{(a+x)^2+b^2}=\frac 1{(x+r)(x+s)}=\frac 1{r-s} \left(\frac 1{x+s} -\frac 1{x+r} \right)$$ and face integrals $$J=\int_0^{\infty } \frac{e^{-x^2}}{x+t} \, dx=\frac{1}{2} e^{-t^2} \left(\pi \, \text{erfi}(t)-\text{Ei}\left(t^2\right)\right)$$ provided $\Im(t)\neq 0\lor \Re(t)>0$.

Answer 2

At first it was a comment, but it was too long. Just some thoughts.

Well, we're looking at the following integral:

$$\mathcal{I}_{\space\text{n}}\left(\alpha\right):=\int_0^\infty\frac{\exp\left(-x^2\right)}{\left(x+\alpha\right)^2+\text{n}}\space\text{d}x\tag1$$

Now, we can use the 'evaluating integrals over the positive real axis' property of the Laplace transform:

$$\mathcal{I}_{\space\text{n}}\left(\alpha\right)=\int_0^\infty\mathcal{L}_x\left[\exp\left(-x^2\right)\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[\frac{1}{\left(x+\alpha\right)^2+\text{n}}\right]_{\left(\text{s}\right)}\space\text{ds}\tag2$$

Using:

$$\sum_{\text{k}=0}^\infty\frac{\left(-x^2\right)^\text{k}}{\text{k}!}=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot x^{2\text{k}}\tag3$$

We can write (using the table of selected Laplace transforms):

$$\mathcal{I}_{\space\text{n}}\left(\alpha\right)=\int_0^\infty\mathcal{L}_x\left[\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot x^{2\text{k}}\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[\frac{1}{\left(x+\alpha\right)^2+\text{n}}\right]_{\left(\text{s}\right)}\space\text{ds}=$$ $$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\int_0^\infty\mathcal{L}_x\left[x^{2\text{k}}\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[\frac{1}{\left(x+\alpha\right)^2+\text{n}}\right]_{\left(\text{s}\right)}\space\text{ds}=$$ $$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\int_0^\infty\frac{\Gamma\left(1+2\text{k}\right)}{\text{s}^{1+2\text{k}}}\cdot\frac{\exp\left(-\alpha\text{s}\right)\cdot\sin\left(\text{s}\sqrt{\text{n}}\right)}{\sqrt{\text{n}}}\space\text{ds}=$$ $$\frac{1}{\sqrt{\text{n}}}\cdot\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}\cdot\Gamma\left(1+2\text{k}\right)}{\text{k}!}\cdot\int_0^\infty\frac{\exp\left(-\alpha\text{s}\right)\cdot\sin\left(\text{s}\sqrt{\text{n}}\right)}{\text{s}^{1+2\text{k}}}\space\text{ds}=$$ $$\frac{1}{\sqrt{\text{n}}}\cdot\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}\cdot\Gamma\left(1+2\text{k}\right)}{\text{k}!}\cdot\mathcal{L}_\text{s}\left[\frac{\sin\left(\text{s}\sqrt{\text{n}}\right)}{\text{s}^{1+2\text{k}}}\right]_{\left(\alpha\right)}\tag4$$