오늘의 계산 55 - An integral involving Fresnel integrals

Problem 1. Show that[각주:1]

$$\begin{array}{}\text{(1)}& {\int }_0^{\mathrm{\infty }}x{\{(2S(x)-1{)}^2+(2C(x)-1{)}^2\}}^2\mathrm{d}x=\frac{16\log 2-8}{{\pi }^2},\end{array}$$

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

$$\begin{array}{r}S(x)={\int }_0^x\sin \left(\frac{1}{2}\pi t^2\right)\mathrm{d}t\text{and}C(x)={\int }_0^x\cos \left(\frac{1}{2}\pi t^2\right)\mathrm{d}t.\end{array}$$

 

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

$$I=\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}{\{(1-2C(x){)}^2+(1-2S(x){)}^2\}}^2\mathrm{d}v$$

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

$$1-2S(x)=\sqrt{\frac{2}{\pi }}{\int }_v^{\mathrm{\infty }}\frac{\sin u}{\sqrt{u}}\mathrm{d}u\text{and}1-2C(x)=\sqrt{\frac{2}{\pi }}{\int }_v^{\mathrm{\infty }}\frac{\cos u}{\sqrt{u}}\mathrm{d}u,$$

we can write $I$ as

$$\begin{array}{}\text{(2)}& I=\frac{4}{{\pi }^3}{\int }_0^{\mathrm{\infty }}{|A(v)|}^{4}dv\end{array}$$

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

$$A(v)={\int }_v^{\mathrm{\infty }}\frac{e^{iu}}{\sqrt{u}}\mathrm{d}u.$$

2. Simplification of ${|A(v)|}^2$

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

$$\begin{array}{}\text{(3)}& \frac{1}{\sqrt{u}}=\frac{1}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{u^{1/2}}=\frac{1}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{e^{-ux}}{\sqrt{x}}\mathrm{d}x=\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{e}^{-u{x}^2}\mathrm{d}x\end{array}$$

Using this identity,

$$\begin{array}{rl}A(v)& =\frac{2}{\sqrt{\pi }}{\int }_v^{\mathrm{\infty }}{e}^{iu}{\int }_0^{\mathrm{\infty }}{e}^{-u{x}^2}\mathrm{d}x\mathrm{d}u=\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{\int }_v^{\mathrm{\infty }}{e}^{-(x^2-i)u}\mathrm{d}u\mathrm{d}x\\ & =\frac{2e^{iv}}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{e}^{-v{x}^2}{\int }_0^{\mathrm{\infty }}{e}^{-(x^2-i)u}\mathrm{d}u\mathrm{d}x=\frac{2e^{iv}}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{e^{-v{x}^2}}{x^2-i}\mathrm{d}x.\end{array}$$

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

$$\begin{array}{rl}{|A(v)|}^2& =A(v)\overline{A(v)}=\frac{4}{\pi }{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{e^{-v(x^2+y^2)}}{(x^2-i)(y^2+i)}\mathrm{d}x\mathrm{d}y\\ & =\frac{4}{\pi }{\int }_0^{\mathrm{\infty }}{\int }_0^{\frac{\pi }{2}}\frac{r{e}^{-v{r}^2}}{(r^2{\cos }^2\theta -i)(r^2{\sin }^2\theta +i)}\mathrm{d}\theta \mathrm{d}r\\ & =\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}{\int }_0^{\frac{\pi }{2}}\frac{e^{-vs}}{(s{\cos }^2\theta -i)(s{\sin }^2\theta +i)}\mathrm{d}\theta \mathrm{d}s\\ & =\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}\frac{e^{-vs}}{s}{\int }_0^{\frac{\pi }{2}}(\frac{1}{s{\cos }^2\theta -i}+\frac{1}{s{\sin }^2\theta +i})\mathrm{d}\theta \mathrm{d}s\\ & =\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}\frac{e^{-vs}}{s}{\int }_0^{\mathrm{\infty }}(\frac{1}{s-i(t^2+1)}+\frac{1}{s{t}^2+i(t^2+1)})\mathrm{d}t\mathrm{d}s.\end{array}$$

Evaluation of the inner integral is easy, and we obtain

$$\begin{array}{rl}{|A(v)|}^2& =2{\int }_0^{\mathrm{\infty }}\frac{e^{-vs}}{s}\mathrm{\Re }\left(\frac{i}{\sqrt{1+is}}\right)\mathrm{d}s.\end{array}$$

Applying $\text{(3)}$ again, we find that

$$\begin{array}{rl}{|A(v)|}^2& =2{\int }_0^{\mathrm{\infty }}\frac{e^{-vs}}{s}\mathrm{\Re }\left(\frac{i}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{e^{-(1+is)u}}{\sqrt{u}}\mathrm{d}u\right)ds\\ & =\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{e^{-vs}}{s}{\int }_0^{\mathrm{\infty }}\frac{e^{-u}\sin (su)}{\sqrt{u}}\mathrm{d}uds\\ & =\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{e^{-u}}{\sqrt{u}}{\int }_0^{\mathrm{\infty }}\frac{\sin (su)}{s}{e}^{-vs}\mathrm{d}s\mathrm{d}u\\ & =\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{e^{-u}}{\sqrt{u}}\arctan \left(\frac{u}{v}\right)\mathrm{d}u\\ \text{(4)}& & =\frac{4\sqrt{v}}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{e}^{-v{x}^2}\arctan (x^2)\mathrm{d}x(u=v{x}^2)\end{array}$$

Here, we exploited the identity

$${\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}{e}^{-sx}\mathrm{d}x=\arctan \left(\frac{1}{s}\right),$$

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

3. Evaluation of $I$

Plugging $\text{(4)}$ to $\text{(2)}$ and applying the polar coordinate change, $I$ reduces to

$$\begin{array}{rl}I& =\frac{64}{{\pi }^4}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}v{e}^{-v(x^2+y^2)}\arctan (x^2)\arctan (x^2)\mathrm{d}x\mathrm{d}y\mathrm{d}v\\ & =\frac{64}{{\pi }^4}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{\arctan (x^2)\arctan (x^2)}{(x^2+y^2{)}^2}\mathrm{d}x\mathrm{d}y\\ & =\frac{64}{{\pi }^4}{\int }_0^{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}\frac{\arctan (r^2{\cos }^2\theta )\arctan (r^2{\sin }^2\theta )}{r^3}\mathrm{d}r\mathrm{d}\theta \\ \text{(5)}& & =\frac{32}{{\pi }^4}{\int }_0^{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}\frac{\arctan (s{\cos }^2\theta )\arctan (s{\sin }^2\theta )}{s^2}\mathrm{d}s\mathrm{d}\theta .(s=r^2)\end{array}$$

Now let us denote

$$J(u,v)={\int }_0^{\mathrm{\infty }}\frac{\arctan (us)\arctan (vs)}{s^2}\mathrm{d}s.$$

Then a simple calculation shows that

$$\frac{{\partial }^{2}J}{\partial u\partial v}(u,v)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{d}s}{(u^2{s}^2+1)(v^2{s}^2+1)}=\frac{\pi }{2(u+v)}.$$

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

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

Plugging this to $\text{(5)}$, it follows that

$$\begin{array}{rl}I& =-\frac{64}{{\pi }^3}{\int }_0^{\frac{\pi }{2}}{\sin }^2\theta \log \sin \theta \mathrm{d}\theta =-\frac{16}{{\pi }^3}\frac{\partial \beta }{\partial z}(\frac{3}{2},\frac{1}{2})\end{array}$$

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

$$\beta (z,w)=2{\int }_0^{\frac{\pi }{2}}{\sin }^{2z-1}\theta {\cos }^{2w-1}\theta \mathrm{d}\theta =\frac{\mathrm{\Gamma }(z)\mathrm{\Gamma }(w)}{\mathrm{\Gamma }(z+w)}.$$

Therefore it follows that $\text{(1)}$ reduces to

$$\begin{array}{rl}I& =\frac{16}{{\pi }^3}\frac{\mathrm{\Gamma }\left(\frac{3}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(2)}\{{\psi }_0(2)-{\psi }_0\left(\frac{3}{2}\right)\}=\frac{8}{{\pi }^2}{\int }_0^1\frac{\sqrt{x}-x}{1-x}\mathrm{d}x=\frac{8(2\log 2-1)}{{\pi }^2},\end{array}$$

where ${\psi }_0(z)={\displaystyle \frac{{\mathrm{\Gamma }}^{\prime }(z)}{\mathrm{\Gamma }(z)}}$ is the digamma function, satisfying the following identity

$${\psi }_0(z+1)=-\gamma +{\int }_0^1\frac{1-x^z}{1-x}\mathrm{d}x.$$

References

1. Marty Colos, An integral involving Fresnel integrals - Math StackExchange [본문으로]