37
$\begingroup$

Is it possible to find a closed form (possibly using known special functions) for this integral? $$\int_0^\infty\left(5\,x^5+x\right)\operatorname{erfc}\left(x^5+x\right)\,dx$$ where $\operatorname{erfc}$ is the complementary error function $$\operatorname{erfc} x=\frac{2}{\sqrt{\pi}}\int_x^{\infty}e^{-z^2}dz.$$

$\endgroup$
  • $\begingroup$ The Maple command $$evalf(Int((5*x^5+x)*erfc(x^5+x), x = 0 .. infinity), 50); $$ outputs $0.21195392584162642127934513122200307521311778875987 $. $\endgroup$ – user64494 Sep 28 '13 at 19:13
  • $\begingroup$ Integrating by parts a few times, the integral under consideration can be expressed through $erfc(x,n)$. $\endgroup$ – user64494 Sep 28 '13 at 19:18
  • 1
    $\begingroup$ $\operatorname{erfc}(n,\,x)$ for every $n\in\mathbb{N}$ can be expressed using just $\operatorname{erfc}(x)$ and elementary functions. $\endgroup$ – Vladimir Reshetnikov Sep 29 '13 at 0:38
24
$\begingroup$

With some help from Mathematica I got this result: $$\int_0^\infty\left(5\,x^5+x\right)\operatorname{erfc}\left(x^5+x\right)dx=J_{\frac25}\left(\frac8{25\sqrt5}\right)\left(\frac8{375}\sqrt{\frac25\left(5-\sqrt5\right)}\,\pi\,J_{\frac45}\left(\frac8{25\sqrt5}\right)-\frac8{375}\sqrt{\frac25\left(5+\sqrt5\right)}\,\pi\,J_{\frac65}\left(\frac8{25\sqrt5}\right)\right)+J_{-\frac25}\left(\frac8{25\sqrt5}\right)\left(\frac1{75}\sqrt{2\left(5+\sqrt5\right)}\,\pi\,J_{-\frac15}\left(\frac8{25\sqrt5}\right)+\frac1{25}\left(\sqrt5-5\right)\sqrt{\frac1{10}\left(5+\sqrt5\right)}\,\pi\,J_{\frac15}\left(\frac8{25\sqrt5}\right)-\frac8{375}\sqrt{\frac25\left(5+\sqrt5\right)}\,\pi\,J_{\frac45}\left(\frac8{25\sqrt5}\right)-\frac{4\left(\sqrt5-5\right)\sqrt{2\left(5+\sqrt5\right)}\,\pi\,J_{\frac65}\left(\frac8{25\sqrt5}\right)}{1875}\right)+J_{-\frac15}\left(\frac8{25\sqrt5}\right)\left(\frac1{25}\sqrt{2\left(5-\sqrt5\right)}\,\pi\,J_{\frac25}\left(\frac8{25\sqrt5}\right)+\frac8{375}\sqrt{\frac25\left(5+\sqrt5\right)}\,\pi\,J_{\frac35}\left(\frac8{25\sqrt5}\right)-\frac8{375}\sqrt{\frac25\left(5-\sqrt5\right)}\,\pi\,J_{\frac75}\left(\frac8{25\sqrt5}\right)\right)+J_{\frac15}\left(\frac8{25\sqrt5}\right)\left(-\frac1{75}\sqrt{2\left(5+\sqrt5\right)}\,\pi\,J_{\frac25}\left(\frac8{25\sqrt5}\right)+\frac{4\left(\sqrt5-5\right)\sqrt{2\left(5+\sqrt5\right)}\,\pi\,J_{\frac35}\left(\frac8{25\sqrt5}\right)}{1875}+\frac8{375}\sqrt{\frac25\left(5+\sqrt5\right)}\,\pi\,J_{\frac75}\left(\frac8{25\sqrt5}\right)\right),$$ where $J_\nu(x)$ is the Bessel function of the first kind.


A solution outline:

Note that $(5\,x^5+x)=x\frac{d}{dx}(x^5+x)$. Change the integration variable $y=x^5+x$, then the integral takes the form $$\int_0^\infty\mathcal{BR}(y)\cdot\operatorname{erfc}y\,dy,$$ where $y\mapsto\mathcal{BR}(y)$ is the inverse (properly selected to satisfy $\mathcal{BR}(y)>0$ for $y>0$) of the polynomial function $x \mapsto x^5+x$. This is a well-known non-elementary function called Bring radical, it can be used to express solutions of quintic equations in an explicit form.

An important fact, it has a representation via a generalized hypergeometric function (I used it to answer another question awhile ago). If we plug the hypergeometric representation into the integral and feed it to Mathematica, it produces the result in terms of Bessel functions shown above. I leave this step without a rigorous proof and rely on Mathematica here. The result agrees with a numerical integration to a very high precision. I would be very glad if anybody could write down an explicit derivation of the formula.

$\endgroup$
  • 3
    $\begingroup$ (+1) ... but how do you get that result with Mathematica while I am getting nothing?? $\endgroup$ – Santosh Linkha Sep 30 '13 at 12:23
  • $\begingroup$ @SantoshLinkha It depends: do you use the serious Mathematica or Alpha? Also it depends on your computer power. $\endgroup$ – Von Neumann Nov 21 '17 at 11:50
8
$\begingroup$

Hint:

$\int_0^\infty(5x^5+x)\text{erfc}(x^5+x)~dx$

$=\int_0^\infty\text{erfc}(x^5+x)~d\left(\dfrac{5x^6}{6}+\dfrac{x^2}{2}\right)$

$=\left[\left(\dfrac{5x^6}{6}+\dfrac{x^2}{2}\right)\text{erfc}(x^5+x)\right]_0^\infty-\int_0^\infty\left(\dfrac{5x^6}{6}+\dfrac{x^2}{2}\right)d(\text{erfc}(x^5+x))$

$=\int_0^\infty\left(\dfrac{5x^6}{6}+\dfrac{x^2}{2}\right)(5x^4+1)\dfrac{2e^{-x^2(x^8+1)}}{\sqrt\pi}dx$

$=\dfrac{2}{\sqrt\pi}\int_0^\infty e^{-(\sinh^\frac{1}{2}x)(\sinh^2x+1)}\left(\dfrac{5\sinh^\frac{3}{2}x}{6}+\dfrac{\sinh^\frac{1}{2}x}{2}\right)(5\sinh x+1)~d(\sinh^\frac{1}{4}x)$

$=\dfrac{1}{2\sqrt\pi}\int_0^\infty e^{-\sinh^\frac{1}{2}x\cosh^2x}~\left(\dfrac{5\sinh^\frac{3}{2}x}{6}+\dfrac{\sinh^\frac{1}{2}x}{2}\right)(5\sinh x+1)\sinh^{-\frac{3}{4}}x\cosh x~dx$

That's why it may relate to Bessel function.

$\endgroup$

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.