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I am seeking to evaluate

$\int_0^{\infty} f(x)/x^2 \, dx$

with

$f(x)=1-\sqrt{\pi/6} \left(\cos (x) C\left(\sqrt{\frac{6 x}{\pi }} \right)+S\left(\sqrt{\frac{6 x}{\pi }} \right) \sin (x)\right)/\sqrt{x}$.

$C(x)$ and $S(x)$ are the Fresnel integrals. Numerical integration suggests that the integral equals $2 \pi/(3 \sqrt{3})$, which would also be desirable within the (physical) context it arose. How can this be proved?

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  • $\begingroup$ Could you give the details of your numerical integration? It maybe give some idea. $\endgroup$
    – vesszabo
    Nov 2, 2012 at 17:02
  • $\begingroup$ @vesszabo I have used Mathematica's NIntegrate. In detail, NIntegrate[(1 - Sqrt[Pi/6/x]*(Cos[x]*FresnelC[Sqrt[6*x/Pi]] + Sin[x]*FresnelS[Sqrt[6*x/Pi]]))/x^2, {x, 0, Infinity}, WorkingPrecision -> 40, AccuracyGoal -> 10, MaxRecursion -> 100, Method -> {GlobalAdaptive, MaxErrorIncreases -> 10000, Method -> "GaussKronrodRule"}] (not optimized) yields the integrand with an accuracy of approximately 1e-10. The significant digits (11) so obtained match $2 Pi/(3 \sqrt{3})$. Due to the oscillatory nature of the integrand it is demanding to obtain more accurate approximations of the integral. $\endgroup$
    – dan
    Nov 2, 2012 at 21:10
  • $\begingroup$ Thanks. Maple gives the equality for 20 digits. Your question is interesting. $\endgroup$
    – vesszabo
    Nov 2, 2012 at 21:35
  • $\begingroup$ @vesszabo Excellent. Thank you for checking. It seems that your hardware is more apt (likely), you are more patient (possible), or your integration strategy is more suitable (interesting). What integration rule did you use? $\endgroup$
    – dan
    Nov 3, 2012 at 11:55
  • $\begingroup$ I typed symbolically ('clickable calculus') and used context menu to calculate approximately. $\endgroup$
    – vesszabo
    Nov 3, 2012 at 14:17

1 Answer 1

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First of all, $$ C(x)=\int_{0}^{x}\cos\left(\frac{1}{2}\pi t^{2}\right)dt=\sqrt{\frac{2}{\pi}}\int_{0}^{\sqrt{\pi/2} \,\left (x\right )}\cos(z^{2})\, dz, $$ and $$ S(x)=\int_{0}^{x}\sin\left(\frac{1}{2}\pi t^{2}\right)dt=\sqrt{\frac{2}{\pi}}\int_{0}^{\sqrt{\pi/2} \,\left (x\right )}\sin(z^{2})\, dz. $$ So we have $$ C\left(\sqrt{\frac{6x}{\pi}}\right)=\sqrt{\frac{2}{\pi}}\int_{0}^{\sqrt{3x}}\cos(z^{2})\, dz, $$ and $$ S\left(\sqrt{\frac{6x}{\pi}}\right)=\sqrt{\frac{2}{\pi}}\int_{0}^{\sqrt{3x}}\sin(z^{2})\, dz. $$ Using these we can write $$ \begin{eqnarray*} f(x) & = & 1-\sqrt{\pi/6}\frac{\cos(x)C\left(\sqrt{\frac{6x}{\pi}}\right)+\sin(x)S\left(\sqrt{\frac{6x}{\pi}}\right)}{\sqrt{x}} \\ & = & \frac{\int_{0}^{\sqrt{3x}}(1-\cos(x-z^{2}))\, dz}{\sqrt{3x}} \end{eqnarray*} $$ and $$ \int_{0}^{\infty}\frac{f(x)}{x^{2}}\, dx=\frac{2}{\sqrt{3}}\int_{0}^{\infty}\int_{0}^{\sqrt{3x}}\frac{\sin^{2}((x-z^{2})/2)}{x^{5/2}}\, dz\, dx. $$ Let's introduce the new variable $z=t\sqrt{x}$. Then $dz=\sqrt{x}dt$ and $$ \begin{eqnarray*} \int_{0}^{\infty}\int_{0}^{\sqrt{3x}}\frac{\sin^{2}((x-z^{2})/2)}{x^{5/2}}\, dz\, dx & = & \int_{0}^{\infty}\int_{0}^{\sqrt{3}}\frac{\sin^{2}(x(1-t^{2})/2)}{x^{2}}\, dt\, dx\\ & = & \frac{1}{2}\int_{0}^{\infty}\int_{0}^{\sqrt{3}}\frac{\sin^{2}(x(1-t^{2}))}{x^{2}}\, dt\, dx\\ & = & \frac{1}{2}\int_{0}^{\infty}\int_{0}^{1}\frac{\sin^{2}(x(1-t^{2}))}{x^{2}}\, dt\, dx\\ & & +\frac{1}{2}\int_{0}^{\infty}\int_{1}^{\sqrt{3}}\frac{\sin^{2}(x(t^{2}-1))}{x^{2}}\, dt\, dx\\ & = & \frac{1}{2}\int_{0}^{1}\int_{0}^{\infty}\frac{\sin^{2}(x(1-t^{2}))}{x^{2}}\, dx\, dt\\ & & +\frac{1}{2}\int_{1}^{\sqrt{3}}\int_{0}^{\infty}\frac{\sin^{2}(x(t^{2}-1))}{x^{2}}\, dx\, dt\\ & = & \int_{0}^{1}\frac{\pi}{4}(1-t^{2})\, dt+\int_{1}^{\sqrt{3}}\frac{\pi}{4}(t^{2}-1)\, dt\\ & = & \frac{\pi}{3}, \end{eqnarray*} $$ where the formula $$ \int_0^{\infty}\frac{\sin^2(Ax)}{x^2}\,dx=\frac{1}{2}A\pi,\quad(A>0) $$ was applied. Your conjecture was excellent.

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  • $\begingroup$ Excellent and elegant.There is a minor typesetting error: $x^{3/2}$ should read $x^{5/2}$ in the denominator of the left-hand-side expression of the core block. I also wonder based on what rational you are permitted to pull out the factor 1/2 from the argument of the $\sin^2$ (second equality, main block). Clearly, one could retain the 1/2 until eventually evaluating the integral over x (and find that the equality indeed holds). Thus, do you pull out the 1/2 in anticipation of what follows below or is there some other reasoning behind this step? Many thanks! $\endgroup$
    – dan
    Nov 3, 2012 at 19:04
  • $\begingroup$ Thanks, the typesetting error is corrected. I introduced $x:=2u$, $dx=2du$ and $\int_0^{\infty}\frac{\sin^2(ax/2)}{x^2}\,dx=\int_0^{\infty}\frac{\sin^2(a(2u)/2)}{(2u)^2}\,2du=\frac{1}{2}\int_0^{\infty}\frac{\sin^2(au)}{u^2}\,du$ and wrote again $x$ instead of $u$. $\endgroup$
    – vesszabo
    Nov 3, 2012 at 20:36
  • $\begingroup$ Obviously ... now. Thanks. $\endgroup$
    – dan
    Nov 3, 2012 at 20:54

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