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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 $$\operatorname{erfc} x=\frac{2}{\sqrt{\pi}}\int_x^{\infty}e^{-z^2}dz.$$

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  • $\begingroup$ The Maple command $$evalf(Int((5*x^5+x)*erfc(x^5+x), x = 0 .. infinity), 50); $$ outputs $0.21195392584162642127934513122200307521311778875987 $. $\endgroup$
    – user64494
    Sep 28, 2013 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, 2013 at 19:18
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    $\begingroup$ $\operatorname{erfc}(n,\,x)$ for every $n\in\mathbb{N}$ can be expressed using just $\operatorname{erfc}(x)$ and elementary functions. $\endgroup$ Sep 29, 2013 at 0:38

2 Answers 2

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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.

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    $\begingroup$ (+1) ... but how do you get that result with Mathematica while I am getting nothing?? $\endgroup$ Sep 30, 2013 at 12:23
  • $\begingroup$ @SantoshLinkha It depends: do you use the serious Mathematica or Alpha? Also it depends on your computer power. $\endgroup$
    – Enrico M.
    Nov 21, 2017 at 11:50
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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.

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