오늘의 계산 55 - An integral involving Fresnel integrals

2013/05/24 01:50

Problem #. Show that[각주:1]

$$\label{p1:wts} \int_{0}^{\infty} x \left\{ (2S(x) - 1)^2 + (2C(x) - 1)^2 \right\}^{2} \, dx = \frac{16 \log 2 - 8}{\pi^2},$$

where $S(x)$ and $C(x)$ denote the Fresnel integrals defined by

\begin{align*} S(x) = \int_{0}^{x} \sin \left( \tfrac{1}{2} \pi t^2 \right) \, dt \quad \text{and} \quad C(x) = \int_{0}^{x} \cos \left( \tfrac{1}{2} \pi t^2 \right) \, dt. \end{align*}

Proof. We divide the proof into several steps.

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 $$\label{p1:repr_1} I = \frac{4}{\pi^{3}} \int_{0}^{\infty} \left| A(v) \right|^{4} \, dv$$ where $A(v)$ denotes the function defined by $$A(v) = \int_{v}^{\infty} \frac{e^{iu}}{\sqrt{u}} \, du.$$

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 $\Re u > 0$, $$\label{p1:iden_1} \frac{1}{\sqrt{u}} = \frac{1}{\Gamma\left(\frac{1}{2}\right)} \frac{\Gamma\left(\frac{1}{2}\right)}{u^{1/2}} = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-ux}}{\sqrt{x}} \, dx = \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} e^{-ux^{2}} \, dx$$ 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 \eqref{p1:iden_1} 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 \nonumber \\ &= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-v s}}{s} \int_{0}^{\infty} \frac{e^{-u} \sin (su)}{\sqrt{u}} \, du\, ds \nonumber \\ &= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} \int_{0}^{\infty} \frac{\sin (su)}{s} \, e^{-v s} \, ds\, du \nonumber \\ &= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} \arctan \left( \frac{u}{v} \right) \, du \nonumber \\ &= \frac{4\sqrt{v}}{\sqrt{\pi}} \int_{0}^{\infty} e^{-vx^{2}} \arctan (x^2) \, dx \qquad (u = vx^2) \label{p1:iden_2} \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 $s$.

3. Evaluation of $I$

Plugging \eqref{p1:iden_2} to \eqref{p1:repr_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 \nonumber \\ &= \frac{64}{\pi^{4}} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\arctan (x^2) \arctan (x^2)}{(x^2 + y^2)^2} \, dx dy \nonumber \\ &= \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 \nonumber \\ &= \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) \label{p1:repr_2} \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 \eqref{p1:repr_2}, 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 it follows that \eqref{p1:wts} reduces to \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 $$\psi_{0}(z+1) = -\gamma + \int_{0}^{1} \frac{1 - x^{z}}{1 - x} \, dx.$$