3
$\begingroup$

I want to solve any of the two integrals for the complex number $a$ \begin{aligned} I_1 & = \int\limits_{0}^{\infty} xe^{a/x^2 - x^2}\text{Erfi}(x/\sqrt{2}) dx\\ I_2 & = \int\limits_{0}^{\infty} \frac{1}{x}e^{a/x^2 - x^2}\text{Erfi}(x/\sqrt{2}) dx, \end{aligned} where \begin{aligned} a & \in \mathbb{C}\\ \text{Re}[a] & < 0. \end{aligned}

Solving one integral, I can get the other through the relation $$I_2 = \frac{d I_1}{da}.$$

My approach

I can not get an answer out of Mathematica. Below are some of the approaches I tried.

a) Integration by parts
I can simplify the integrand enormously by taking the derivative of Erfi, which can be done by integration of parts. I know that the definite integral $\int_{0}^{\infty}e^{a/z^2-z^2}z^n dz$ is related to the Bessel functions, but I need the indefinite integral when doing partial integration, which I can not find.

b) Using a known integral
I know that \begin{aligned} \int\limits_{0}^{\infty}e^{1/x^2 -b^2x^2}\text{Erfc}(\frac{1}{x})dx & = \frac{1}{b\sqrt{\pi}}\left[\sin(2b\cdot\text{Ci}(2b))-\cos(2b\cdot\text{si}(2b))\right] \\ \int\limits_{0}^{\infty}e^{1/x^2 -b^2x^2}\text{Erfc}(\frac{1}{x})xdx & = \frac{\pi}{2b}\left[H_1(2b) - Y_1(2b)-\frac{1}{b}\right], \quad |\text{arg}\,b| < \frac{\pi}{4} \\ \int\limits_{0}^{\infty}e^{1/x^2 -b^2x^2}\text{Erfc}(\frac{1}{x})\frac{dx}{x} & = \frac{\pi}{2}\left[H_0(2b) - Y_0(2b)\right], \quad |\text{arg}\,b| < \frac{\pi}{4} \\ \end{aligned} where Ci and si are the Cosine and Sine integrals, and $H$ is the Struve function. What I try to do is to make the variable substitution $z\to\frac{1}{z}$, then to turn Erfi into Erf (and Erfc) with the same argument as in the above equation $z \to \frac{iz}{\sqrt{2}}$. I can not figure out the limits however when doing this last substitution.

c) Rewriting the integrand
I can rewrite the integrals as \begin{aligned} I_1 & = \int\limits_{0}^{\infty} e^{a(1/x - x)}\text{Erfi}(\sqrt{\sqrt{a}x/2}) dx\\ I_2 & = \int\limits_{0}^{\infty} \frac{1}{x}e^{a(1/x - x)}\text{Erfi}(\sqrt{\sqrt{a}x/2}) dx. \end{aligned}

$\endgroup$
1
$\begingroup$

Let us take $a>0$ and $b>0$ and denote : \begin{eqnarray} I_2(a,b)&:=&\int\limits_0^\infty \frac{1}{x} \exp( -\frac{a}{x^2}-x^2 ) erfi(b x) dx\\ I_1(a,b)&:=&\int\limits_0^\infty x \exp( -\frac{a}{x^2}-x^2 ) erfi(b x) dx \end{eqnarray}

Then clearly we have: \begin{eqnarray} \partial_b I_2(a,b) &=& \frac{2}{\sqrt{\pi}} \int\limits_0^\infty \exp( -\frac{a}{x^2}-(1-b^2)x^2 ) dx = \\ &=& \left. \frac{ \left(e^{-2 \sqrt{a} \sqrt{1-b^2}} \left(1-\text{erf}\left(\frac{\sqrt{a}}{x}-\sqrt{1-b^2} x\right)\right)+e^{2 \sqrt{a} \sqrt{1-b^2}} \left(\text{erf}\left(\frac{\sqrt{a}}{x}+\sqrt{1-b^2} x\right)-1\right)\right)}{2 \sqrt{1-b^2}} \right|_{0}^\infty=\\ &=& \frac{ e^{-2 \sqrt{a-a b^2}}}{ \sqrt{1-b^2}} \end{eqnarray} Now we integrate: \begin{eqnarray} I_2(a,b) &=& \int\limits_0^b \frac{ e^{-2 \sqrt{a} \sqrt{1- \xi^2}}}{ \sqrt{1-\xi^2}} d\xi\\ &=& \int\limits_0^{\arcsin(b)} \exp(-2 \sqrt{a} \cos(\theta)) d\xi \\ &=& J_0(2 \imath \sqrt{a}) \arcsin(b) + 2 \sum\limits_{n=1}^M \frac{i^n}{n} J_n(2 \imath \sqrt{a}) \cdot \sin(n \arcsin(b)) \end{eqnarray} where in the last line we took some integer $M >> 1$ and we used the generating function for Bessel functions https://en.wikipedia.org/wiki/Bessel_function . The series converge quite rapidly and taking $M=10$ suffices completely as the code below demonstrates:

 In[1412]:= {a, b} = 
 RandomReal[{0, 1}, 2, WorkingPrecision -> 50]; M = 10;
NIntegrate[1/x Exp[-a/x^2 - x^2] Erfi[b x], {x, 0, Infinity}]
 NIntegrate[Exp[-2 Sqrt[a] Sqrt[1 - xi^2]]/Sqrt[1 - xi^2], {xi, 0, b}]
 NIntegrate[Exp[-2 Sqrt[a] Cos[th]], {th, 0, ArcSin[b]}]
BesselJ[0, 2 I Sqrt[a]] ArcSin[b] + 
 2 Sum[  I^n/n BesselJ[n, 2 I Sqrt[a]] Sin[n ArcSin[b]], {n, 1, M}]

Out[1413]= 0.0832502

Out[1414]= 0.0832502

Out[1415]= 0.0832502

Out[1416]= 0.0832501731615361275633199191686069096124674656008

Now the other integral is obtained by differentiation. We have: \begin{eqnarray} I_1(a,b)&=&-\left.\frac{d}{d c} I_2(a \cdot c, \frac{b}{\sqrt{c}}) \right|_{c=1}\\ &=&\frac{b J_0\left(2 i \sqrt{a}\right)}{2 \sqrt{1-b^2}}+i \sqrt{a} J_1\left(2 i \sqrt{a}\right) \sin ^{-1}(b) + \\ && \sum\limits_{n=1}^M \frac{\imath^n}{n} \left( i \sqrt{a} \left(J_{n-1}\left(2 i \sqrt{a}\right)-J_{n+1}\left(2 i \sqrt{a}\right)\right) \sin \left(n \sin ^{-1}(b)\right)-\frac{b n J_n\left(2 i \sqrt{a}\right) \cos \left(n \sin ^{-1}(b)\right)}{\sqrt{1-b^2}}\right) \end{eqnarray}

In[1473]:= {a, b} = 
 RandomReal[{0, 1}, 2, WorkingPrecision -> 50]; M = 10;
NIntegrate[x Exp[-a/x^2 - x^2] Erfi[b x], {x, 0, Infinity}]
b/2 1/Sqrt[1 - b^2] - 
 Sum[I^n NIntegrate[BesselJ[n, 2 I Sqrt[xi]], {xi, 0, a}] If[n == 0, 
    ArcSin[b], (Exp[I n ArcSin[b]] - 1)/(I n)] , {n, -M, M}]
b/2 1/Sqrt[1 - b^2] - 
 NIntegrate[(BesselJ[0, 2 I Sqrt[xi]] ArcSin[b] + 
    Sum[ BesselJ[n, 
       2 I Sqrt[xi]] ( -((
        I I^-n E^(-I n ArcSin[b]) (-1 + E^(I n ArcSin[b])) ((-1)^n + 
           E^(I n (\[Pi] + ArcSin[b]))))/n)), {n, 1, M}]), {xi, 0, a}]
c =.; (-D[BesselJ[0, 2 I Sqrt[a c]] ArcSin[b/Sqrt[c]], c] - 
   2 Sum[  I^n/n D[BesselJ[n, 2 I Sqrt[a c]] Sin[n ArcSin[b/Sqrt[c]]],
        c], {n, 1, M}] /. c :> 1)
(b BesselJ[0, 2 I Sqrt[a]])/(2 Sqrt[1 - b^2]) + 
 I Sqrt[a] ArcSin[b] BesselJ[1, 2 I Sqrt[a]] - 
 Sum[  I^n/
    n (-((b n BesselJ[n, 2 I Sqrt[a]] Cos[n ArcSin[b]])/ Sqrt[
      1 - b^2]) + 
     I Sqrt[a] (BesselJ[-1 + n, 2 I Sqrt[a]] - 
        BesselJ[1 + n, 2 I Sqrt[a]]) Sin[n ArcSin[b]]), {n, 1, M}]



Out[1474]= 0.193704

Out[1475]= 0.193704 - 8.40823*10^-20 I

Out[1476]= 0.193704 + 0. I

Out[1477]= 0.1937040115438800579341627452243157963570649708871

Out[1478]= 0.1937040115438800579341627452243157963570649708871

Update: Finally let us consider a third integral: \begin{equation} I_3(a,b) := \int\limits_0^\infty \exp(-\frac{a}{x^2}-x^2) erfi(b x) dx \end{equation} Then by differentiating the quantity in question with respect to $b$ then by using Integral involving a power function and $\exp(-a/x^2-b x^2)$. and finally by integrating the result over $b$ we obtained the following result: \begin{eqnarray} I_3(a,b) = \frac{2 \sqrt{a}}{\sqrt{\pi}} \int\limits_0^{\arcsin(b)} K_1(2 \sqrt{a} \cos(\phi) ) d\phi \end{eqnarray} Unfortunately I know too little about Bessel function now to be able to simplify that any further.

$\endgroup$

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.