I need to calculate the following integral: $$\int_0^\infty \left(\left(2\ S(x)-1\right)^2+\left(2\ C(x)-1\right)^2\right)^2 x\ \mathrm dx,$$ where $$S(x)=\int_0^x\sin\frac{\pi z^2}{2}\mathrm dz,$$ $$C(x)=\int_0^x\cos\frac{\pi z^2}{2}\mathrm dz$$ are the Fresnel integrals.

Numerical integration gives an approximate result $0.31311841522422385...$ that is close to $\frac{16\log2-8}{\pi^2}$, so it might be the answer.

  • 1
    $\begingroup$ Just out of curiosity, how did you guess the answer? It seems correct to 30+ digits... $\endgroup$ May 21 '13 at 19:41
  • $\begingroup$ Is this the integral you are considering. $\endgroup$ May 23 '13 at 1:27
  • $\begingroup$ Not quite, here the OP is talking about Fresnel integrals rather than $\sin$ and $\cos$ integrals, they're not quite the same: $\int _{0}^{y}\!{\frac {\sin \left( x \right) }{\sqrt {x}}}{dx}=\sqrt { 2}\sqrt {\pi }{\it FresnelS} \left( {\frac {\sqrt {2}\sqrt {y}}{\sqrt {\pi }}} \right)$, $\int _{0}^{y}\!{\frac {\sin \left( x \right) }{x}}{dx}={\it Si} \left( y \right) $. $\endgroup$ May 23 '13 at 8:37
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    $\begingroup$ @GrahamHesketh There are plenty of inverse symbolic calculators nowadays. For example, you can do it in Mathematica: WolframAlpha["0.31311841522422385", IncludePods -> "PossibleClosedForm", TimeConstraint -> Infinity] or directly at WolframAlpha if you're lucky enough not to hit timeout. $\endgroup$ May 23 '13 at 17:22
  • $\begingroup$ Oh, cool cheers. $\endgroup$ May 23 '13 at 17:29

Step 1. Reduction of the integral

Let $I$ denote the integral in question. With the change of variable $v = \frac{\pi x^2}{2}$, we have

$$ I = \frac{1}{\pi} \int_{0}^{\infty} \left\{ (1 - 2 C(x) )^{2} + (1 - 2S(x))^{2} \right\}^{2} \, dv $$

where $x = \sqrt{2v / \pi}$ is understood as a function of $v$. By noting that

$$ 1-2 S(x) = \sqrt{\frac{2}{\pi}} \int_{v}^{\infty} \frac{\sin u}{\sqrt{u}} \, du \quad \text{and} \quad 1-2 C(x) = \sqrt{\frac{2}{\pi}} \int_{v}^{\infty} \frac{\cos u}{\sqrt{u}} \, du, $$

we can write $I$ as

$$ I = \frac{4}{\pi^{3}} \int_{0}^{\infty} \left| A(v) \right|^{4} \, dv \tag{1} $$

where $A(v)$ denotes the function defined by

$$ A(v) = \int_{v}^{\infty} \frac{e^{iu}}{\sqrt{u}} \, du. $$

Step 2. Simplification of $\left| A(v) \right|^2$.

Now we want to simplify $\left| A(v) \right|^2$. To this end, we note that for $\Re u > 0$,

$$ \frac{1}{\sqrt{u}} = \frac{1}{\Gamma\left(\frac{1}{2}\right)} \frac{\Gamma\left(\frac{1}{2}\right)}{u^{1/2}} = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-ux}}{\sqrt{x}} \, dx = \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} e^{-ux^{2}} \, dx \tag{2} $$

Using this identity,

\begin{align*} A(v) &= \frac{2}{\sqrt{\pi}} \int_{v}^{\infty} e^{iu} \int_{0}^{\infty} e^{-u x^2} \, dx du = \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \int_{v}^{\infty} e^{-(x^2-i)u} \, du dx \\ &= \frac{2 e^{iv}}{\sqrt{\pi}} \int_{0}^{\infty} e^{-v x^2} \int_{0}^{\infty} e^{-(x^2-i)u} \, du dx = \frac{2 e^{iv}}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-v x^2}}{x^2-i} \, dx. \end{align*}

Thus by the polar coordinate change $(x, y) \mapsto (r, \theta)$ followed by the substitutions $r^2 = s$ and $\tan \theta = t$, we obtain

\begin{align*} \left| A(v) \right|^2 &= A(v) \overline{A(v)} = \frac{4}{\pi} \int_{0}^{\infty} \int_{0}^{\infty} \frac{e^{-v (x^2+y^2)}}{(x^2-i)(y^2 + i)} \, dxdy \\ &= \frac{4}{\pi} \int_{0}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{r e^{-v r^2}}{(r^2 \cos^{2}\theta-i)(r^2 \sin^{2}\theta + i)} \, d\theta dr \\ &= \frac{2}{\pi} \int_{0}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{e^{-v s}}{(s \cos^{2}\theta-i)(s \sin^{2}\theta + i)} \, d\theta ds \\ &= \frac{2}{\pi} \int_{0}^{\infty} \frac{e^{-v s}}{s} \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{s \cos^{2}\theta-i} + \frac{1}{s \sin^{2}\theta + i} \right) \, d\theta ds \\ &= \frac{2}{\pi} \int_{0}^{\infty} \frac{e^{-v s}}{s} \int_{0}^{\infty} \left( \frac{1}{s -i(t^2 + 1)} + \frac{1}{s t^2 + i (t^2 + 1)} \right) \, dt ds. \end{align*}

Evaluation of the inner integral is easy, and we obtain

\begin{align*} \left| A(v) \right|^2 &= 2 \int_{0}^{\infty} \frac{e^{-v s}}{s} \Re \left( \frac{i}{\sqrt{1 + is}} \right) \, ds. \end{align*}

Applying $(2)$ again, we find that

\begin{align*} \left| A(v) \right|^2 &= 2 \int_{0}^{\infty} \frac{e^{-v s}}{s} \Re \left( \frac{i}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-(1+is)u}}{\sqrt{u}} \, du \right) \, ds \\ &= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-v s}}{s} \int_{0}^{\infty} \frac{e^{-u} \sin (su)}{\sqrt{u}} \, du\, ds \\ &= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} \int_{0}^{\infty} \frac{\sin (su)}{s} \, e^{-v s} \, ds\, du \\ &= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} \arctan \left( \frac{u}{v} \right) \, du \\ &= \frac{4\sqrt{v}}{\sqrt{\pi}} \int_{0}^{\infty} e^{-vx^{2}} \arctan (x^2) \, dx \qquad (u = vx^2) \tag{3} \end{align*}

Here, we exploited the identity

$$ \int_{0}^{\infty} \frac{\sin x}{x} e^{-sx} \, dx = \arctan \left(\frac{1}{s}\right), $$

which can be proved by differentiating both sides with respect to $s$.

Step 3. Evaluation of $I$.

Plugging $(3)$ to $(1)$ and applying the polar coordinate change, $I$ reduces to

\begin{align*} I &= \frac{64}{\pi^{4}} \int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} v e^{-v(x^{2}+y^{2})} \arctan (x^2) \arctan (x^2) \, dx dy dv \\ &= \frac{64}{\pi^{4}} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\arctan (x^2) \arctan (x^2)}{(x^2 + y^2)^2} \, dx dy \\ &= \frac{64}{\pi^{4}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{\arctan (r^2 \cos^2 \theta) \arctan (r^2 \sin^2 \theta)}{r^3} \, dr d\theta \\ &= \frac{32}{\pi^{4}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{\arctan (s \cos^2 \theta) \arctan (s \sin^2 \theta)}{s^2} \, ds d\theta. \qquad (s = r^2) \tag{4} \end{align*}

Now let us denote

$$ J(u, v) = \int_{0}^{\infty} \frac{\arctan (us) \arctan (vs)}{s^2} \, ds. $$

Then a simple calculation shows that

$$ \frac{\partial^{2} J}{\partial u \partial v} J(u, v) = \int_{0}^{\infty} \frac{ds}{(u^2 s^2 + 1)(v^2 s^2 + 1)} = \frac{\pi}{2(u+v)}. $$

Indeed, both the contour integration method or the partial fraction decomposition method work here. Integrating, we have

$$ J(u, v) = \frac{\pi}{2} \left\{ (u+v) \log(u+v) - u \log u - v \log v \right\}. $$

Plugging this to $(4)$, it follows that

\begin{align*} I &= -\frac{64}{\pi^{3}} \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \log \sin \theta \, d\theta = -\frac{16}{\pi^{3}} \frac{\partial \beta}{\partial z}\left( \frac{3}{2}, \frac{1}{2} \right) \end{align*}

where $\beta(z, w)$ is the beta function, satisfying the following beta function identity

$$ \beta (z, w) = 2 \int_{0}^{\infty} \sin^{2z-1}\theta \cos^{2w-1} \theta \, d\theta = \frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)}. $$

Therefore we have

\begin{align*} I &= \frac{16}{\pi^{3}} \frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(2)} \left\{ \psi_{0} (2) - \psi_{0} \left(\tfrac{3}{2} \right) \right\} = \frac{8}{\pi^2} \int_{0}^{1} \frac{\sqrt{x} - x}{1 - x} \, dx = \frac{8 (2 \log 2 - 1)}{\pi^2}, \end{align*}

where $\psi_0 (z) = \dfrac{\Gamma'(z)}{\Gamma(z)}$ is the digamma function, satisfying the following identity

$$ \psi_{0}(z+1) = -\gamma + \int_{0}^{1} \frac{1 - x^{z}}{1 - x} \, dx. $$

  • 9
    $\begingroup$ This is madness! $\phantom{}$ $\endgroup$
    – Norbert
    May 24 '13 at 11:06
  • 5
    $\begingroup$ it's very nice!+1, $\endgroup$
    – math110
    May 24 '13 at 15:09
  • 5
    $\begingroup$ This is absolutely amazing. $\endgroup$
    – Pedro Tamaroff
    May 25 '13 at 20:01
  • 3
    $\begingroup$ Your answer is amazing, thanks! Sorry, I missed the time when I could manually award the bounty, so the half of it was automatically awarded to (also very good) Graham Hesketh's answer. I started a new bounty worth +200 to reward your answer. The system only allows me to award it tomorrow. $\endgroup$ May 27 '13 at 2:18
  • 4
    $\begingroup$ Thank you everyone! Even I can't believe that a solution like this is possible. And also @MartyColos, thank you for a new bounty! $\endgroup$ May 27 '13 at 2:41

Following on from Ron...

Make the substitutions: $x=\sqrt{y}$, $p=1/2+1/2\,\sqrt {1+4\,r}$, to get:

$\displaystyle \dfrac{64}{{\pi }^{2}}\,\int _{0}^{\infty }\!x \left( \int _{1}^{\infty }\!{\frac {\sin \left( 2\,\pi \,{x}^{2}p \left( p-1 \right) \right) }{p}}{dp} \right) ^{2}{dx}$,

$\displaystyle=\dfrac{32}{{\pi }^{2}}\,\int _{0}^{\infty }\! \left( \int _{0 }^{\infty }\!{\frac {2\,\sin \left( 2\,\pi\, y\, r \right) }{\sqrt {1+4\, r} \left( 1+\sqrt {1+4\,r} \right) }}{dr} \right) ^{2}{dy} $, see Appendix,

$\displaystyle=\dfrac{16}{{\pi }^{2}}\,\int _{-\infty}^{\infty }\! \left( \int _{0 }^{\infty }\!{\frac {2\,\sin \left( 2\,\pi \,y \,r \right) }{\sqrt {1+4\, r} \left( 1+\sqrt {1+4\,r} \right) }}{dr} \right) ^{2}{dy} $,

$\displaystyle=\dfrac{16}{{\pi }^{2}}\,\int _{-\infty}^{\infty }\! \left( \int _{-\infty }^{\infty }\!{\frac {H(r)\,\sin \left( 2\,\pi \,\,y\,r \right) }{\sqrt {1+4\, |r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right) ^{2}{dy} $ : $H \left( r \right) =\cases{1&$0\leq x$\cr -1&$x<0$\cr}$.

Note then that:

$\displaystyle\left(\int _{-\infty }^{\infty }\!{\frac {H(r)\,\sin \left( 2\,\pi \,\,y\,r \right) }{\sqrt {1+4\, |r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right) ^{2}=\displaystyle\left|-\dfrac{i}{2}\int _{-\infty }^{\infty }\!{\frac {H(r)\,e^{ -i2\pi \,y\,r } }{\sqrt {1+4\, |r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr}+\dfrac{i}{2}\int _{-\infty }^{\infty }\!{\frac {H(r)\,e^{ i2\pi\, y\,r } }{\sqrt {1+4\, |r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right| ^{2}$

$=\displaystyle\left|\int _{\infty }^{\infty }\!{\frac {\left(H(r)-H(-r)\right)\,e^{ i2\pi \,y\,r } }{2\sqrt {1+4\, |r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right| ^{2}=\displaystyle\left|\int _{\infty }^{\infty }\!{\frac {\,e^{ i2\pi \,y\,r } }{\sqrt {1+4\, |r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right| ^{2}.$

We can now use Plancherel's theorem which states:

$\displaystyle \int _{-\infty }^{\infty }\! \left| f \left( y \right) \right| ^{2}{dy}=\int _{-\infty }^{\infty }\! \left| F \left( r \right) \right| ^{2}{dr} $ : $ \displaystyle F \left( r \right) =\int _{-\infty }^{\infty }\!f \left( y \right) { e^{-i2\pi r y}}{dy} $,

and thus:

$\displaystyle\dfrac{16}{{\pi }^{2}}\,\int _{-\infty}^{\infty }\! \displaystyle\left|\int _{\infty }^{\infty }\!{\frac {\,e^{ i2\pi \,y\,r } }{\sqrt {1+4\, |r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right| ^{2}{dy}$
$=\displaystyle\dfrac{16}{{\pi }^{2}}\,\int _{-\infty}^{\infty }\! \displaystyle\left|{\frac {1 }{\sqrt {1+4\, |r|} \left( 1+\sqrt {1+4\,|r|} \right) }} \right| ^{2}{dr}$,

$=\dfrac{16}{{\pi }^{2}}\displaystyle\int _{0}^{\infty }\!{\frac {2}{ \left( 1+4\,r \right) \left( 1+ \sqrt {1+4\,r} \right) ^{2}}}{dr}$.

Now if we undo the substitution made earlier by letting $r = p^2-p$:

$\dfrac{16}{{\pi }^{2}}\displaystyle\int _{0}^{\infty }\!{\frac {2}{ \left( 1+4\,r \right) \left( 1+ \sqrt {1+4\,r} \right) ^{2}}}{dr}=\dfrac{16}{{\pi }^{2}}\int _{1}^{\infty }\!{ \frac {1}{2{p}^{2} \left( 2\,p-1 \right) }}{dp}$,

and then make one final substitution $p=\dfrac{1}{q+2}$ we get:

$\displaystyle\dfrac{16}{{\pi }^{2}}\int _{1}^{\infty }\!{ \frac {1}{2{p}^{2} \left( 2\,p-1 \right) }}{dp}=-\dfrac{16}{{\pi }^{2}}\int_{-2}^{-1}\dfrac{1}{2}+\dfrac{1}{q}{dq}=\dfrac{16\ln(2)-8}{\pi^2}$.


Note that the integral: $$\int _{0 }^{\infty }\!{\frac {2\,\sin \left( 2\,\pi \,y \,r \right) }{\sqrt {1+4\, r} \left( 1+\sqrt {1+4\,r} \right) }}{dr} $$

converges by the Chartier-Dirichlet test because: $$f(r)={\frac {2\, }{\sqrt {1+4\, r} \left( 1+\sqrt {1+4\,r} \right) }}$$

is monotonic and continuous on $ \mathbb R^+ $ and $f(r)\rightarrow0$ as $r\rightarrow \infty$, and because: $$\left|\int_{0}^{b}\sin\left(2\pi\,y\,r\right){dr}\right|$$ is bounded as $b\rightarrow\infty$.


This is not a solution yet, but I think it is a way forward.


$$f(x) = \left (\frac12 - S(x)\right) \cos{\left (\frac{\pi}{2} x^2\right)} - \left (\frac12 - C(x)\right) \sin{\left (\frac{\pi}{2} x^2\right)} $$

$$g(x) = \left (\frac12 - C(x)\right) \cos{\left (\frac{\pi}{2} x^2\right)} + \left (\frac12 - S(x)\right) \sin{\left (\frac{\pi}{2} x^2\right)} $$

Then it is a straightforward exercise to show that

$$g(x) + i f(x) = e^{-i \pi x^2/2} \int_x^{\infty} dt \, e^{i \pi t^2/2}$$

and that

$$(2 S(x)-1)^2 + (2 C(x)-1)^2 = 4 [g(x)^2+f(x)^2] = 4 x^2 \int_1^{\infty} du \, \int_1^{\infty} dv \, e^{i \pi x^2 (u^2-v^2)/2}$$

I can convert the double integral to a single integral by changing coordinates to $p=u+v$, $q=u-v$, $p \in [2,\infty)$, $q \in [-(p-2),p-2]$. The Jacobian is $1/2$ and we have

$$(2 S(x)-1)^2 + (2 C(x)-1)^2 = 2 x^2 \int_2^{\infty} dp \, \int_{-(p-2)}^{p-2} dq \, e^{i \pi x^2 p q/2}$$

which after evaluation of the inner integral and some rescaling, we get

$$(2 S(x)-1)^2 + (2 C(x)-1)^2 = \frac{8}{\pi} \int_1^{\infty} dp \, \frac{\sin{[2 \pi x^2 p (p-1)]}}{p}$$

I am not sure how to evaluate this integral analytically, nor am I sure that that would be the best move here. The desired integral is therefore

$$\int_0^{\infty} dx \, x\, [(2 S(x)-1)^2 + (2 C(x)-1)^2]^2 =\\ \frac{64}{\pi^2} \int_0^{\infty} dx \, x\, \int_1^{\infty} dp \, \frac{\sin{[2 \pi x^2 p (p-1)]}}{p} \int_1^{\infty} dq \, \frac{\sin{[2 \pi x^2 q (q-1)]}}{q}$$

From here, I am not quite sure what to do. Naturally, I would like to reverse the order of integration so that the integral over $x$ is interior, but I am not sure how to justify that, given that the integrals are not absolutely convergent. Further, even if I can justify that, I think the resulting integral is some delta functions which do not look promising. I am continuing to look at this, but I figure that maybe someone else can also contribute from here, or just tell me if I am off the mark.


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