Convolution of complementary Error functi0ns

calculusintegrationinverse laplace

I'm interested in the solution to the following integral, given that $a$ is a positive real:
$$I=\int_0^1 \text{erfc}\!\left(\frac{a}{\sqrt{t}}\right) \text{erfc}\!\left(\frac{a}{\sqrt{1-t}}\right) dt=(2 b+1) \text{erfc}\left(\sqrt{b}\right)-\frac{2 \sqrt{b} e^{-b}}{\sqrt{\pi }}$$
where $a=\sqrt{\frac{b}{4}}$. The result (that Mariusz pointed out in the comments and i reproduced using mathematica) does not suffice, I require the step by step solution or a reference.

Actually, it seems that this is the final step in a lengthy calculation of the integral in this question: Evaluation of $\int_0^T \int_0^t \tau^{-1/2} (T-\tau)^{-1/2} \exp(\frac{ix^2}{2(T-\tau)})d\tau dt$

Using the convolution theorem for laplace transformations, one gets:
$$I=\mathcal{L}^{-1}\left[s^{-2} e^{-2\sqrt{bs}}\right](1)$$
Sadly, i could not find a way to perform this inverse laplace transformation, so I tried to tackle the integral directly. Using $erfc=1-erf$ and dropping some easy to calculate terms, I needed to calculate:
$$J=\int_0^1 \text{erf}\!\left(\frac{a}{\sqrt{t}}\right) \text{erf}\!\left(\frac{a}{\sqrt{1-t}}\right) dt=\frac{4}{\pi}\int_0^1\int_0^{\frac{a}{\sqrt{t}}}\int_0^{\frac{a}{\sqrt{1-t}}}e^{-x^2-y^2}dxdydt$$

Or, written a bit differently:
$$J=\frac{4}{\pi}\int_0^1\int_{\eta(t)}e^{-v^2}d^2vdt$$
where $\eta(t)=[0,\frac{a}{\sqrt{t}}]\times[0,\frac{a}{\sqrt{1-t}}]$

The function to be integrated does, in this form, exhibit cylindrical symmetry, thus i tried switching into cylindrical coordinates:

$$J=\frac{4}{\pi}\int_0^{+\infty}\int_0^1l(r|\eta(t))dt \exp(-r^2)dr$$
where $l(r|\eta(t))$ is the arc length of the section of a circle of radius $r$ with $\eta(t)$.
Some geometric reasoning gives($A\leq B$):
$$l(r|[0,A]\times [0,B])=\begin{cases} \frac{\pi}{2}r & r\leq A\\
\frac{\pi}{2}r – r\arctan(\sqrt{\frac{r^2}{A^2}-1}) & A\leq r\leq B\\
\frac{\pi}{2}r – r\arctan(\sqrt{\frac{r^2}{A^2}-1}) -r\arctan(\sqrt{\frac{r^2}{B^2}-1}) & B\leq r\leq \sqrt{A^2+B^2}\\
0 & \sqrt{A^2+B^2}\leq r
\end{cases}$$

Also, it is noteworthy to see:
$$l(r|[0,A]\times [0,B])=Re(\frac{\pi}{2}r – r\arctan(\sqrt{\frac{r^2}{A^2}-1}) -r\arctan(\sqrt{\frac{r^2}{B^2}-1})) \Theta(\sqrt{A^2+B^2}-r)$$

The heaviside step function can be taken into account by adjusting the integration range in $t$ depending on $r$:
for $r\leq \sqrt{A^2+B^2}$, one integrates from 0 to 1.
for $\sqrt{A^2+B^2}\leq r$, one integrates over:
$$[0,1/2-\sqrt{1/4-a^2/r^2}]\cup [1/2+\sqrt{1/4-a^2/r^2},1] (*)$$

Luckily, $\arctan\sqrt{\frac{r^2t}{a^2}-1}$ (appearing in $l(r|\eta(t))$) does posses an easy antiderivative(which i will not write out here)! Also, the antiderivative's value at the intermediate points in (*) are purely imaginary, so that taking the real part cancels them.

So finally, if I'm not totally mistaken, we arrive at:

$$J=\frac{4}{\pi}Re\left[\int_0^{+\infty}e^{-r^2}\left(
\frac{\pi}{2}r\left(1-2\sqrt{1/4-a^2/r^2}\right)
-2a^2/r\left(r^2/a^2\arctan\left(\sqrt{\frac{r^2t}{a^2}-1}\right) -\sqrt{\frac{r^2t}{a^2}-1}\right)
\right)dr\right] $$

I think this transformation had no merit for this integral seem even scarier than the one i started with, but maybe someone has an idea how to work with that.

Best Answer

$$\mathcal{L}_s^{-1}\left[s^v \exp \left(-a \sqrt{s}\right)\right](x)=\frac{x^{-v-1} \exp \left(-\frac{a^2}{8 x}\right) D_{2 v+1}\left(\frac{a}{\sqrt{2 x}}\right)}{2^{v+\frac{1}{2}} \sqrt{\pi }}$$ for: $\Re\left(a^2\right)>0$ and $\Re\left(s\right)>0$ where: $D_{2 v+1}\left(\frac{a}{\sqrt{2} \sqrt{x}}\right)$ is Parabolic Cylinder Function

In yours case: $v=-2$, $a=2 \sqrt{b}$ and $x=1$

$$\color{red}{\mathcal{L}_s^{-1}\left[\frac{\exp \left(-2 \sqrt{b s}\right)}{s^2}\right](x)}=\\2 e^{-\frac{b}{2}} \sqrt{\frac{2}{\pi }} D_{-3}\left(\sqrt{2} \sqrt{b}\right)=\\\frac{8 e^{-b} H_{-3}\left(\sqrt{b}\right)}{\sqrt{\pi }}=\color{red}{\\-\frac{2 \sqrt{b} e^{-b}}{\sqrt{\pi }}+(1+2 b)\text{erfc}\left(\sqrt{b}\right)}$$

where: $H_{-3}\left(\sqrt{b}\right)$ is Hermite polynomial

Reference:

Integrals and series.Volume 5, Inverse Laplace transforms Example 10,page 52.

Related Solutions

Calculus – Integral of Product of Exponential Function and Two Complementary Error Functions

Let $c \in {\mathbb R}$ and $d \in {\mathbb R}$. Without loss of generality we can consider a slightly different function: \begin{equation} f(c,d):= \int\limits_{-\infty}^\infty \exp(-t^2) \mbox{erfc}(t-c) \mbox{erfc}(t-d) dt \end{equation} Now we differentiate with respect to $c$. We have: \begin{eqnarray} \partial_c f(c,d) &=& \lim\limits_{\epsilon \rightarrow 0} \frac{\sqrt{2}}{\sqrt{\pi}} \exp(-\frac{c^2}{2}) \left( \sqrt{\pi} + 2 \sqrt{\pi} T(\epsilon,\frac{1}{\sqrt{2}},\frac{2d-c}{\sqrt{2}}) - 2 \sqrt{\pi} T(\epsilon,\frac{1}{\sqrt{2}},\frac{-2d+c}{\sqrt{2}})\right)\\ &=&\sqrt{2} \exp(-\frac{c^2}{2}) \left( 1-erf(\frac{c-2 d}{\sqrt{6}})\right) \end{eqnarray} where $T(h,a,b)$ is the generalized Owen's T function Generalized Owen's T function .

Now, all we need to do is to integrate. Since $f(-\infty,d)=0$ we integrate over $c$ from minus infinity to $c$. We have: \begin{eqnarray} f(c,d) &=& \sqrt{2} \left( \sqrt{\frac{\pi}{2}}(1+erf(\frac{c}{\sqrt{2}}) + 2erf(\frac{d}{\sqrt{2}})) + \sqrt{2 \pi} 2 T(c,\frac{1}{\sqrt{3}},-\frac{2 d}{\sqrt{3}})\right) \\ &=& \frac{1}{3 \sqrt{\pi}} \left( 12 \pi T\left(c,\frac{c-2 d}{\sqrt{3} c}\right)+12 \pi T\left(d,\frac{d-2 c}{\sqrt{3} d}\right)-6 \arctan\left(\frac{c-2 d}{\sqrt{3} c}\right)-6 \arctan\left(\frac{d-2 c}{\sqrt{3} d}\right)+3 \pi \text{erf}\left(\frac{c}{\sqrt{2}}\right)+3 \pi \text{erf}\left(\frac{d}{\sqrt{2}}\right)+4 \pi\right) \end{eqnarray} where $T(h,a)$ is Owen's T function https://en.wikipedia.org/wiki/Owen%27s_T_function .

In[1060]:= {c, d} = RandomReal[{-2, 2}, 2, WorkingPrecision -> 50];
NIntegrate[Exp[-t^2] Erfc[t - c] Erfc[t - d], {t, -Infinity, Infinity}]
(4 \[Pi] - 6 ArcTan[(c - 2 d)/(Sqrt[3] c)] - 
 6 ArcTan[(-2 c + d)/(Sqrt[3] d)] + 3 \[Pi] Erf[c/Sqrt[2]] + 
 3 \[Pi] Erf[d/Sqrt[2]] + 12 \[Pi] OwenT[c, (c - 2 d)/(Sqrt[3] c)] + 
 12 \[Pi] OwenT[d, (-2 c + d)/(Sqrt[3] d)])/(3 Sqrt[\[Pi]])

Out[1061]= 0.483318

Out[1062]= 0.48331807775609703646923225386370751256977344715

Update: Now let us take four real numbers $a_1 \in {\mathbb R}$, $a_2 \in {\mathbb R}$, $c\in {\mathbb R}$ and $d \in {\mathbb R}$ and consider a more general integral: \begin{equation} f^{(a_1,a_2)}(c,d):= \int\limits_{-\infty}^\infty \exp(-t^2) \mbox{erfc}(a_1 t-c) \mbox{erfc}(a_2 t-d) dt \end{equation} Then by doing the same calculations as above we easily arrive at the following formula: \begin{eqnarray} &&f^{(a_1,a_2)}(c,d)= \frac{1}{\sqrt{\pi}} \left(\right.\\ && 4 \pi T\left(\frac{\sqrt{2} c}{\sqrt{a_1^2+1}},\frac{a_1 a_2 c-\left(a_1^2+1\right) d}{c \sqrt{a_1^2+a_2^2+1}}\right)+4 \pi T\left(\frac{\sqrt{2} \sqrt{a_1^2+1} d}{\sqrt{a_1^2 a_2^2+a_1^2+a_2^2+1}},\frac{a_1 a_2 d-\left(a_2^2+1\right) c}{d \sqrt{a_1^2+a_2^2+1}}\right) +\\ &&-2 \arctan\left(\frac{a_1 a_2 c-\left(a_1^2+1\right) d}{c \sqrt{a_1^2+a_2^2+1}}\right)-2 \arctan\left(\frac{a_1 a_2 d-\left(a_2^2+1\right) c}{d \sqrt{a_1^2+a_2^2+1}}\right)+\\ &&\pi \text{erf}\left(\frac{\sqrt{a_1^2+1} d}{\sqrt{a_1^2 a_2^2+a_1^2+a_2^2+1}}\right)+\pi \text{erf}\left(\frac{c}{\sqrt{a_1^2+1}}\right)+\\ &&\pi +2 \arctan\left(\frac{a_1 a_2}{\sqrt{a_1^2+a_2^2+1}}\right)\\ &&\left.\right) \end{eqnarray}

For[count = 1, count <= 200, count++,
  {a1, a2, c, d} = RandomReal[{-5, 5}, 4, WorkingPrecision -> 50];
  I1 = NIntegrate[
    Exp[-t^2] Erfc[a1 t - c] Erfc[a2 t - d], {t, -Infinity, Infinity}];
  I2 = 1/Sqrt[\[Pi]] (\[Pi] + 
      2 ArcTan[(a1  a2)/Sqrt[(1 + a1^2 + a2^2)]] - 
      2 ArcTan[ (a1 a2 c - (1 + a1^2) d)/(
        Sqrt[(1 + a1^2 + a2^2)] c)] - 
      2 ArcTan[ (a1 a2 d - (1 + a2^2) c)/(
        Sqrt[(1 + a1^2 + a2^2)] d)] + \[Pi] Erf[c/Sqrt[
        1 + a1^2]] + \[Pi] Erf[(Sqrt[(1 + a1^2)] d)/ Sqrt[
        1 + a1^2 + a2^2 + a1^2 a2^2]] + 
      4 \[Pi] OwenT[(Sqrt[2] c)/Sqrt[
        1 + a1^2],  (a1 a2 c - (1 + a1^2) d)/(
        Sqrt[(1 + a1^2 + a2^2)] c)] + 
      4 \[Pi] OwenT[(Sqrt[2] Sqrt[(1 + a1^2)] d)/ Sqrt[
        1 + a1^2 + a2^2 + a1^2 a2^2],  (a1 a2 d - (1 + a2^2) c)/(
        Sqrt[(1 + a1^2 + a2^2)] d)]);
  If[Abs[I2/I1 - 1] > 10^(-3), 
   Print["results do not match", {a1, a2, c, d, {I1, I2}}]; Break[]];
  If[Mod[count, 10] == 0, PrintTemporary[count]];
  ];

[Math] integrals with error function

Sometimes just using the integral form of error function can save you.

$\int_0^t\dfrac{1}{x}e^{-\frac{a^2}{x}} \text{erf}\biggl(\dfrac{b}{\sqrt{x}}\biggr)~dx$

$=\dfrac{2}{\sqrt{\pi}}\int_0^t\dfrac{1}{x}e^{-\frac{a^2}{x}}\int_0^\frac{b}{\sqrt{x}}e^{-u^2}~du~dx$

$=\dfrac{2}{\sqrt{\pi}}\int_0^t\dfrac{1}{x}e^{-\frac{a^2}{x}}\int_0^1e^{-\left(\frac{bu}{\sqrt{x}}\right)^2}~d\biggl(\dfrac{bu}{\sqrt{x}}\biggr)~dx$

$=\dfrac{2}{\sqrt{\pi}}\int_0^t\dfrac{1}{x}e^{-\frac{a^2}{x}}\int_0^1\dfrac{b}{\sqrt{x}}e^{-\frac{b^2u^2}{x}}~du~dx$

$=\dfrac{2b}{\sqrt{\pi}}\int_0^t\int_0^1\dfrac{e^{-\frac{b^2u^2+a^2}{x}}}{x\sqrt{x}}du~dx$

$=\dfrac{2b}{\sqrt{\pi}}\int_0^1\int_0^t\dfrac{e^{-\frac{b^2u^2+a^2}{x}}}{x\sqrt{x}}dx~du$

$=\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_0^t-e^{-\frac{b^2u^2+a^2}{x}}~d\biggl(\dfrac{1}{\sqrt{x}}\biggr)~du$

$=\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_\infty^\frac{1}{\sqrt{t}}-e^{-(b^2u^2+a^2)x^2}~dx~du$

$=\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_\frac{1}{\sqrt{t}}^\infty e^{-(b^2u^2+a^2)x^2}~dx~du$

$=\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_0^\infty e^{-(b^2u^2+a^2)x^2}~dx~du-\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_0^\frac{1}{\sqrt{t}}e^{-(b^2u^2+a^2)x^2}~dx~du$

$=2b\int_0^1\dfrac{1}{\sqrt{b^2u^2+a^2}}du-\dfrac{4b}{\sqrt{\pi}}\int_0^1\biggl[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(b^2u^2+a^2)^nx^{2n+1}}{n!(2n+1)}\biggr]_0^\frac{1}{\sqrt{t}}~du$

$=2b\int_0^1\dfrac{1}{\sqrt{b^2u^2+a^2}}du-\dfrac{4b}{\sqrt{\pi}}\int_0^1\sum\limits_{n=0}^\infty\dfrac{(-1)^n(b^2u^2+a^2)^n}{t^{n+\frac{1}{2}}n!(2n+1)}du$

$=2b\int_0^1\dfrac{1}{\sqrt{b^2u^2+a^2}}du-\dfrac{4b}{\sqrt{\pi}}\int_0^1\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^na^{2n-2k}b^{2k}u^{2k}}{t^{n+\frac{1}{2}}n!(2n+1)}du$

$=2b\biggl[\dfrac{\ln\left(b^2u+b\sqrt{b^2u^2+a^2}\right)}{b}\biggr]_0^1-\dfrac{4b}{\sqrt{\pi}}\biggl[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^na^{2n-2k}b^{2k}u^{2k+1}}{t^{n+\frac{1}{2}}k!(n-k)!(2n+1)(2k+1)}\biggr]_0^1$

$=2\ln\left(b^2+b\sqrt{a^2+b^2}\right)-2\ln(|a|b)-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n4a^{2n-2k}b^{2k+1}}{\sqrt{\pi}t^{n+\frac{1}{2}}k!(n-k)!(2n+1)(2k+1)}$

$\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\text{erfc}\left(\sqrt{\dfrac{a^2+x^2}{t}}\right)~dx$

$=\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\int_\sqrt{\frac{a^2+x^2}{t}}^\infty e^{-u^2}~du~dx$

$=\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\int_0^\infty e^{-u^2}~du~dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\int_0^\sqrt{\frac{a^2+x^2}{t}}e^{-u^2}~du~dx$

$=\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\biggl[\sum\limits_{n=0}^\infty\dfrac{(-1)^nu^{2n+1}}{n!(2n+1)}\biggr]_0^\sqrt{\frac{a^2+x^2}{t}}~dx$

$=\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(a^2+x^2)^{n+\frac{1}{2}}}{t^{n+\frac{1}{2}}n!(2n+1)}dx$

$=\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\sum\limits_{n=0}^\infty\dfrac{(-1)^n(a^2+x^2)^n}{t^{n+\frac{1}{2}}n!(2n+1)}dx$

$=\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^na^{2n-2k}x^{2k}}{t^{n+\frac{1}{2}}n!(2n+1)}dx$

$=\left[\ln\left(x+\sqrt{a^2+x^2}\right)\right]_0^b-\dfrac{2}{\sqrt{\pi}}\biggl[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^na^{2n-2k}x^{2k+1}}{t^{n+\frac{1}{2}}k!(n-k)!(2n+1)(2k+1)}\biggr]_0^b$

$=\ln\left(b+\sqrt{a^2+b^2}\right)-\ln|a|-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2a^{2n-2k}b^{2k+1}}{\sqrt{\pi}t^{n+\frac{1}{2}}k!(n-k)!(2n+1)(2k+1)}$

Best Answer

Related Solutions

Calculus – Integral of Product of Exponential Function and Two Complementary Error Functions

[Math] integrals with error function

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