# Integrals with the special functions $Ci(x)$ and $erf(x)$

I'm looking for the solutions of the following two integrals:

$$I_1=\int\limits_0^\infty dx\, e^{-x^2}\text{Ci}(ax)$$ and $$I_2=\int\limits_0^\infty dx\, e^{-ax}\text{erf}(x)$$ with $$\text{Ci}(x)=\int\limits_\infty^x\frac{\cos(t)}{t}dt$$ and $$\text{erf}(x)=\frac{2}{\sqrt{\pi}}\int\limits_0^xe^{-t^2}dt$$

Now I'm not 100% sure what is meant by "the solution of the integrals" since these will probably be not-evaluable. But I'm guessing that the question is to reduce the expression to one single special function in stead of the integral of a special function with an elementary function.

Mathematica yields me the answers: $$I_1=-\frac{\sqrt{\pi}}{4}\Gamma\left(0,\frac{a^2}{4}\right)$$ and $$I_2=\exp\left(\frac{a^2}{4}\right)\frac{1-\text{erf}(a/2)}{a}$$

A good first step for evaluating these integrals $$I_1$$ and $$I_2$$ seemed to fill in the integral representations of these special functions and try to switch te integrals over $$x$$ and $$t$$. However this has not (yet) been a success. I also tried to find a differential equation for these integrals, but also this was not so easy to do. Are there any tips/tricks to evaluate these integrals ?

Hint: Differentiate $I_1(a)$ with regard to a. Similarly, define $~I_2(b)~=~\displaystyle\int_0^\infty e^{-ax}~\text{erf}(bx)~dx,~$ and then differentiate it with regard to b.

• Nice, adding a factor in order to eliminate one of the integrations !
– Nick
Mar 23, 2015 at 14:51

For $I_2$, please note that this integral is just the Laplace transform of $\text{Erf}(x)$ w.r.t. to the variable $a$

Furthermore the Laplace transform of an integral $\mathcal{L}(\int f(x) dx)$ is just $\mathcal{L}(f(x))/a$

Using this we find that

$$I_2=\frac{2}{a\sqrt{\pi}}\int_0^{\infty}e^{- a x}e^{-x^2}dx$$

Completing the square in the exponent and changing variables $y=\frac{a}{2}+x$

This turns into

$$I_2=2\frac{ e^{\frac{a^2}{4}}}{a\sqrt{\pi}}\int_{\frac{a}{2}}^{\infty}e^{-y^2}dy$$

Now using straightforwardly the definition of the Errorfunction we obtain

$$I_2=\frac{e^{\frac{a^2}{4}}}{a}\left(1-\text{Erf}(a/2)\right)$$

As promised!

I will have a look on $I_1$ later

• thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
– Nick
Mar 23, 2015 at 13:27