Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods?

Aside from some trigonometric substitutions and identities, we will need the following identity, which can be shown using integration by parts twice: $$ \int_0^{\infty}\cos(\alpha t)e^{-\lambda t}\,\mathrm{d}t=\frac{\lambda}{\alpha^2+\lambda^2}\tag{1} $$ We will also use the standard arctangent integral: $$ \int_0^\infty\frac{\mathrm{d}t}{a^2+t^2}=\frac\pi{2a}\tag{2} $$ Now $$ \begin{align} &\left(\int_0^\infty\color{#C00000}{\sin}(x^2) e^{-\lambda x^2}\,\mathrm{d}x\right)^2\\ &=\int_0^\infty\int_0^\infty \color{#C00000}{\sin}(x^2)\color{#C00000}{\sin}(y^2) e^{-\lambda(x^2+y^2)}\,\mathrm{d}y\,\mathrm{d}x\tag{3.1}\\ &=\frac12\int_0^\infty\int_0^\infty \left(\cos(x^2-y^2) \color{#FF0000}{-}\cos(x^2+y^2)\right) e^{-\lambda(x^2+y^2)}\,\mathrm{d}y\,\mathrm{d}x\tag{3.2}\\ &=\frac12\int_0^{\pi/2}\int_0^\infty \left(\cos(r^2\cos(2\phi)) \color{#FF0000}{-}\cos(r^2)\right)e^{-\lambda r^2} \,r\,\mathrm{d}r\,\mathrm{d}\phi\tag{3.3}\\ &=\frac14\int_0^{\pi/2}\int_0^\infty \left(\cos(s\cos(2\phi)) \color{#FF0000}{-}\cos(s)\right) e^{-\lambda s} \,\mathrm{d}s\,\mathrm{d}\phi\tag{3.4}\\ &=\frac14\int_0^{\pi/2}\left(\frac{\lambda}{\cos^2(2\phi)+\lambda^2} \color{#FF0000}{-}\frac{\lambda}{1+\lambda^2}\right)\,\mathrm{d}\phi\tag{3.5}\\ &=\frac12\int_0^{\pi/4} \frac{\lambda}{\cos^2(2\phi)+\lambda^2}\,\mathrm{d}\phi \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.6}\\ &=\frac14\int_0^{\pi/4} \frac{\lambda\,\mathrm{d}\tan(2\phi)} {1+\lambda^2+\lambda^2\tan^2(2\phi)} \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.7}\\ &=\frac14\int_0^\infty\frac{\mathrm{d}t}{1+\lambda^2+t^2} \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.8}\\ &=\frac{\pi/8}{\sqrt{1+\lambda^2}} \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.9} \end{align} $$

$(3.1)$ change the square of the integral into a double integral

$(3.2)$ use $2\color{#C00000}{\sin}(A)\color{#C00000}{\sin}(B)=\cos(A-B)\color{#FF0000}{-}\cos(A+B)$

$(3.3)$ convert to polar coordinates

$(3.4)$ substitute $s=r^2$

$(3.5)$ apply $(1)$

$(3.6)$ pull out the constant and apply symmetry to reduce the domain of integration

$(3.7)$ multiply numerator and denominator by $\sec^2(2\phi)$

$(3.8)$ substitute $t=\lambda\tan(2\phi)$

$(3.9)$ apply $(2)$

Finally, take the square root of both sides of $(3)$ and let $\lambda\to0^+$ to get $$ \int_0^\infty\color{#C00000}{\sin}(x^2)\,\mathrm{d}x=\sqrt{\frac\pi8}\tag{4} $$

Addendum

I just noticed that the same proof works for $$ \int_0^\infty\cos(x^2)\,\mathrm{d}x=\sqrt{\frac\pi8}\tag{5} $$ if each red $\color{#C00000}{\sin}$ is changed to $\cos$ and each red $\color{#FF0000}{-}$ sign is changed to $+$.

First, use the substitution $t = x^2$ to obtain

$$ I := \int_{0}^{\infty}\sin\left(x^2\right)\;dx=\int_{0}^{\infty}\frac{\sin t}{2\sqrt{t}}\;dt. $$

This integral converges conditionally, thus we make an integration by parts to obtain an absolutely convergent integral as follows:

$$I = \left[\frac{1-\cos t}{2\sqrt{t}}\right]_{0}^{\infty}-\int_{0}^{\infty}\left(\frac{d}{dt}\frac{1}{2\sqrt{t}}\right)(1 - \cos t)\;dt =\int_{0}^{\infty}\frac{1 - \cos t}{4t^{3/2}}\;dt.$$

Now, from gamma integral,

$$ \begin{align*} I &=\frac{1}{4\Gamma(3/2)}\int_{0}^{\infty}\left(\frac{\Gamma(3/2)}{t^{3/2}}\right)(1 - \cos t)\;dt \\ &=\frac{1}{4\Gamma(3/2)}\int_{0}^{\infty}\left(\int_{0}^{\infty}u^{1/2}e^{-tu}\;du\right)(1 - \cos t)\;dt\\ &=\frac{1}{4\Gamma(3/2)}\int_{0}^{\infty}\int_{0}^{\infty}u^{1/2}e^{-tu}(1 - \cos t)\;dudt \\ &=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\int_{0}^{\infty}u^{1/2}e^{-tu}(1 - \cos t)\;dudt, \end{align*} $$

where in the last line we have used the fact that $\Gamma(3/2) = \frac{1}{2}\sqrt{\pi}$, which is a direct consequence of the Gaussian integral. (Of course, this integral can be evaluated by a famous real analysis technique.) By Tonelli's theorem we can change the order of integration, and with the substitution $u = v^2$ we obtain

$$\begin{align*}I &=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\int_{0}^{\infty}u^{1/2}e^{-tu}(1 - \cos t)\;dtdu \\ &= \frac{1}{2\sqrt{\pi}} \int_{0}^{\infty}\frac{u^{1/2}}{u(1+u^2)}\;du = \frac{1}{2\sqrt{\pi}} \int_{0}^{\infty} \frac{2}{1+v^4} \; dv. \end{align*}$$

To evaluate the last integral, we use the following decomposition

$$ \frac{2}{1+v^4} = \frac{1+v^{-2}}{(v-v^{-1})^2-2} - \frac{1-v^{-2}}{(v+v^{-1})^2-2}. $$

Thus with the substitution $z = v - v^{-1}$ and $w = v + v^{-1}$, the integral becmes

$$\begin{align*}I &= \frac{1}{2\sqrt{\pi}} \left( \int_{0}^{\infty} \frac{d(v-v^{-1})}{(v-v^{-1})^2+2} - \int_{0}^{\infty} \frac{d(v+v^{-1})}{(v+v^{-1})^2-2} \right) \\ &= \frac{1}{2\sqrt{\pi}} \left( \int_{-\infty}^{\infty} \frac{dz}{z^2+2} - \int_{\infty}^{\infty} \frac{dw}{w^2-2} \right) = \frac{1}{2\sqrt{\pi}} \left( \frac{\pi}{\sqrt{2}} - 0 \right) = \sqrt{\frac{\pi}{8}}. \end{align*}$$

Only a slight modification of this argument immediately yields

$$ \int_{0}^{\infty} \cos\left(x^2\right) \; dx = \sqrt{\frac{\pi}{8}}. $$

Everybody who has not at least quick-browse throught some pages of Integral Kokeboken brochure (although in norwegian), I definitelly recommend to do so. In this gem one can find : (for improving redability I will adopt the methodology used by @robjohn, I will even use the same apriori identities)

$$ \begin{align} &\int_0^\infty\sin{(x^2)}\;\mathrm{d}x=\\ &=\frac{1}{2}\int_0^\infty\sin{u}\;\frac{\mathrm{d}u}{\sqrt{u}}\tag{1.1}\\ &=\frac{1}{2}\int_0^\infty\sin{u}\;\frac{2}{\sqrt{\pi}}\int_{0}^\infty e^{-uv^2}\;\mathrm{d}v\,\mathrm{d}u\tag{1.2} \\ &=\frac{1}{\sqrt{\pi}}\int_0^\infty\int_{0}^\infty e^{-uv^2}\sin{u}\;\mathrm{d}u \,\mathrm{d}v\tag{1.3}\\ &=\frac{1}{\sqrt{\pi}}\int_0^\infty\frac{\mathrm{d}v}{1+v^4}\tag{1.4}\\ &=\frac{1}{2\sqrt{\pi}}\int_0^\infty\frac{1+v^2}{1+v^4}\;\mathrm{d}v\tag{1.5}\\ &=\frac{1}{2\sqrt{\pi}}\int_0^\infty\frac{d(v-v^{-1})}{(v-v^{-1})^2+2}\tag{1.6}\\ &=\frac{1}{2\sqrt{\pi}}\int_{-\infty}^\infty\frac{dw}{w^2+2}=\frac{1}{2\sqrt{2\pi}}\arctan\left(\frac{w}{\sqrt{2}}\right)\bigg{|}_{-\infty}^{\infty}=\sqrt{\frac{\pi}{8}}\tag{1.7}\\ \end{align} $$

Notes :

$(1.1)$ substitution $u=x^2$

$(1.2)$ Gaussian integral

$(1.3)$ changing the order of integration

$(1.4)$ the second apriori formula, we will get the same integral as @Sangchul Lee got.

$(1.5)$ same method, but slightly modified, wrtite the integral as $I=(I+I)/2$ then in the second make substitution $v\rightarrow 1/v$ to get this form

$(1.6)$ dividing both numeratpr and denominator by $v^2$, completing the differential

$(1.7)$ substitution $w=v-v^{-1}$, doesn't need any commentary :)

Similarly for $\cos(x^2)$.

Let $$F(t)=\int_0^{\infty} \sin(tx^2)\ dx$$ Taking the Laplace Transform of $F(t)$: $$\mathscr{L}\{F(t)\}=\int_0^{\infty}F(t)e^{-st}\ dt$$ By being slightly less than rigorous this can be rewritten as $$\int_0^{\infty}\mathscr{L}\{\sin (tx^2)\}\ dx=\int_0^{\infty}\frac{x^2}{s^2+x^4}\ dx$$ This can be put into partial fractions as $$\frac{x^2}{s^2+x^4}\to\frac{Ax+B}{(s+ix^2)}+\frac{Cx+D}{(s-ix^2)}$$ $$(Ax+B)(s-ix^2)+(Cx+D)(s+ix^2)=x^2$$ $$A=C=0,\ B=\frac{i}{2},\ D=-\frac{i}{2}$$ $$\begin{equation}\begin{aligned} \int_0^{\infty}\frac{x^2}{s^2+x^4}\ dx&=\int_0^{\infty}\frac{i}{2(s+ix^2)}\ dx\ -\int_0^{\infty}\frac{i}{2(s-ix^2)}\ dx \\ &=\frac{1}{2}\int_0^{\infty}\frac{1}{x^2-is}\ dx\ +\frac{1}{2}\int_0^{\infty}\frac{1}{x^2+is}\ dx \\ &=\lim_{c\to\infty}\frac{1}{2}\bigg[\frac{1}{\sqrt{is}}\tan ^{-1} \bigg(\frac{x}{\sqrt{is}}\bigg)-\frac{1}{\sqrt{is}}\tanh ^{-1}\bigg(\frac{x}{\sqrt{is}}\bigg)\bigg]_0^c \\ &=\frac{\pi}{4}\bigg(\frac{1}{\sqrt{is}}+\frac{1}{\sqrt{-is}}\bigg)=\frac{\pi}{4}\bigg(\frac{\sqrt{2}}{2\sqrt{s}}(1-i)+\frac{\sqrt{2}}{2\sqrt{s}}(1+i)\bigg)=\frac{\pi\sqrt{2}}{4\sqrt{s}} \end{aligned}\end{equation}$$ Now taking the inverse Laplace Transform: $$\mathscr{L}^{-1}\bigg\{\frac{\pi\sqrt{2}}{4\sqrt{s}}\bigg\}=\frac{\pi\sqrt{2}}{4}\mathscr{L}^{-1}\bigg\{\frac{1}{\sqrt{s}}\bigg\}=\frac{\pi\sqrt{2}}{4\sqrt{\pi t}}$$ By noting that $$F(1)=\int_0^{\infty}\sin (x^2)\ dx$$ $$\int_0^{\infty}\sin (x^2)\ dx=\frac{\pi\sqrt{2}}{4\sqrt{\pi\times 1}}=\sqrt{\frac{\pi}{8}}$$

Here is paper that addresses this exact question by H. Flanders, "On Fresnel integrals", American Mathematical Monthly, vol. 89, no. 4, 1982, pp. 264-266.

Here is another paper on the topic that does not require a subscription.

Integrate $\displaystyle{J \equiv \int_{0}^{\infty}{\rm e}^{{\rm i}x^{2}}\,{\rm d}x}$. $\quad\Im J = ?$.

\begin{eqnarray*} J^{2} & = & \int_{0}^{\infty}{\rm e}^{{\rm i}x^{2}}\,{\rm d}x \int_{0}^{\infty}{\rm e}^{{\rm i}y^{2}}\,{\rm d}y = {\pi \over 2}\int_{0}^{\infty}{\rm e}^{{\rm i}\,\rho^{2}}\rho\,{\rm d}\rho = {\pi \over 4}\int_{0}^{\infty}{\rm e}^{{\rm i}\,\rho}\,{\rm d}\rho \\[3mm] & = & {\pi \over 4}\int_{-\infty}^{\infty} \left(% {\rm i}\int_{-\infty}^{\infty}{{\rm d}k \over 2\pi}\, {{\rm e}^{-{\rm i}k\rho} \over k + {\rm i}0^{+}} \right) {\rm e}^{{\rm i}\,\rho}\,{\rm d}\rho = {\rm i}\,{\pi \over 4}\int_{-\infty}^{\infty}{{\rm d}k \over k + {\rm i}0^{+}} \int_{-\infty}^{\infty}{{\rm d}\rho \over 2\pi}\,{\rm e}^{{\rm i}\left(1 - k\right)\rho} \\[3mm] & = & {\rm i}\,{\pi \over 4}\int_{-\infty}^{\infty}{{\rm d}k \over k + {\rm i}0^{+}} \delta\left(1 - k\right) = {\rm i}\,{\pi \over 4}{1 \over 1 + {\rm i}0^{+}} = {\rm i}\,{\pi \over 4}\left\lbrack 1 - {\rm i}\pi\,\delta\left(1\right)\right\rbrack = {\rm i}\,{\pi \over 4} = {\rm e}^{{\rm i}\pi/2}\,{\pi \over 4} \\[1cm]&&\mbox{} \end{eqnarray*}

$$ J = \sqrt{\,{\rm e}^{{\rm i}\pi/2}\,{\pi \over 4}\,} = {\rm e}^{{\rm i}\pi/4}\,\sqrt{\,\pi \over 4\,} $$

$$ \int_{0}^{\infty}\sin\left(x^{2}\right)\,{\rm d}x = \Im J = \overbrace{\quad\sin\left(\pi \over 4\right)\quad}^{=\ 1/\sqrt{\,2\,}} \sqrt{\,\pi \over 4\,} = \sqrt{\,\pi \over 8\,} $$

Short Answer

Let set, $$I=\int_0^\infty \cos(x^2) dx \quad\text{and}\quad J=\int_0^\infty \sin(x^2) dx$$

Summary: We will prove that $$ \color{blue}{0= (I+J)(I-J)=I^2 -J^2 =\lim_{t \to 0}I_t^2-J^2_t}$$ Where, $$I_t = \int_0^\infty e^{-tx^2}\cos(x^2) dx~~~~\text{and}~~~ J_t = \int_0^\infty e^{-tx^2}\sin(x^2) dx$$ $t\mapsto I_t$ and $t\mapsto J_t$ are clearly continuous due to the presence of the powerful integrand factor $e^{-tx^2}$.

However, By Fubini we have,

\begin{split} I_t^2-J^2_t&=& \left(\int_0^\infty e^{-tx^2}\cos(x^2) dx\right) \left(\int_0^\infty e^{-ty^2}\cos(y^2) dy\right) - \left(\int_0^\infty e^{-tx^2}\sin(x^2) dx\right) \left(\int_0^\infty e^{-ty^2}\sin(y^2) dy\right) \\ &=& \int_0^\infty \int_0^\infty e^{-t(x^2+y^2)}\cos(x^2+y^2)dxdy\\ &=&\int_0^{\frac\pi2}\int_0^\infty re^{-tr^2}\cos r^2 drd\theta\\&=&\frac\pi4 Re\left( \int_0^\infty \left[\frac{1}{i-t}e^{(i-t)r^2}\right]' dr\right)\\ &=&\color{blue}{\frac\pi4\frac{t}{1+t^2}\to 0~~as ~~~t\to 0} \end{split}

<To end the proof:** Let show that $I\ge 0$ and $J\ge 0$. Performing an integration by part we obtain $$J = \frac{1}{\sqrt{2}} \int^\infty_0\frac{\sin(2x)}{x^{1/2}}\,dx=\frac{1}{\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx\color{red}{\ge0}$$ Given that $\color{red}{\sin 2x= 2\sin x\cos x =(\sin^2x)'}$. Similarly we have, $$I = \frac{1}{\sqrt{2}}\int^\infty_0\frac{\cos(2x)}{\sqrt{x}}\,dx=\frac{1}{2\sqrt{2}}\underbrace{\left[\frac{\sin 2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{4\sqrt{2}} \int^\infty_0\frac{\sin 2 x}{x^{3/2}}\,dx\\= \frac{1}{4\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{3}{8\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx\color{red}{\ge0}$$

Conclusion: $I^2-J^2 =0$.

However using similar technique in above prove one can easily arrives at the following $$I_tJ_t = \frac\pi8\frac{1}{t^2+1}$$ from which one get explicit value of $$I^2=J^2= IJ = \lim_{t\to 0}I_tJ_t =\frac\pi8$$

since, $$\\ 2 I_tJ_t = I_tJ_t+I_tJ_t = \\= \left(\int_0^\infty e^{-tx^2}\sin(x^2) dx\right) \left(\int_0^\infty e^{-ty^2}\cos(y^2) dy\right) + \left(\int_0^\infty e^{-tx^2}\cos(x^2) dx\right) \left(\int_0^\infty e^{-ty^2}\sin(y^2) dy\right) \\= \int_0^\infty \int_0^\infty e^{-t(x^2+y^2)}\sin(x^2+y^2)dxdy $$

for more details see here:Proof only by transformation that : $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $

I'm not sure if this is a "real method" (that's because I know very little about it) but it works.

Let $$I=\int_0^\infty \sin(x^n)\,\mathrm{d}x$$ Therefore $$I=\int_0^\infty \sin(x^n)\,\mathrm{d}x=-\mathrm{Im}\left[\int_0^\infty e^{-ix^n}\,\mathrm{d}x\right]$$ Make the substitution $\,t=ix^n \ \Rightarrow \ x=i^{-1/n}\, t^{1/n} \ \Rightarrow \ \mathrm{d}x=\dfrac{i^{-1/n}}{n}\, t^{1/n-1}\,\mathrm{d}t$. Whence, $$\begin{aligned} I&=-\mathrm{Im}\left[i^{-1/n}\,\frac{1}{n}\int_0^\infty t^{1/n-1} e^{-t}\,\mathrm{d}t\right]\\ &=-\frac{1}{n}\ \mathrm{Im}\left[i^{-1/n}\,\Gamma\left(\frac{1}{n}\right)\right]\\ &=-\frac{1}{n}\,\Gamma\left(\frac{1}{n}\right)\, \mathrm{Im}\left[e^{-\tfrac{i\pi}{2n}}\right] \quad \text{(since $i=e^{-i\pi/2}$)}\\ &=\Gamma\left(1+\frac{1}{n}\right)\sin\left(\frac{\pi}{2n}\right) \quad \left(\text{since} \ \mathrm{Im}\left[e^{-\tfrac{i\pi}{2n}}\right]=-\sin\left(\frac{\pi}{2n}\right)\right) \end{aligned}$$

We can then conclude that

$$\int_0^\infty \sin(x^n)\,\mathrm{d}x=\Gamma\left(1+\frac{1}{n}\right)\sin\left(\frac{\pi}{2n}\right), \quad n>1$$

With $n=2$,

$$\int_0^\infty \sin(x^2)\,\mathrm{d}x=\Gamma\left(\frac{3}{2}\right)\sin\frac{\pi}{4}=\frac{\sqrt{\pi}}{2\sqrt{2}}$$

Using $$ \int_0^\infty f(x)\mathcal L\{g(t)\}(x)\;dx=\int_0^\infty \mathcal \{f(x)\}(t)g(t)\;dt $$ one has $$\begin{eqnarray} I&=& \int_{0}^{\infty}\sin\left(x^2\right)\;dx\overset{x^2\to x}=\int_{0}^{\infty}\frac{\sin x}{2\sqrt{x}}\;dx\\ &=&\frac1{2\sqrt\pi}\int_0^\infty\sin x\,\mathcal L\{\frac1{\sqrt t}\}(x)\;dx=\frac1{2\sqrt\pi}\int_0^\infty\mathcal L\{\sin x\}(t)\frac1{\sqrt t}\;dt\\ &=&\frac1{2\sqrt\pi}\int_0^\infty\frac1{\sqrt t(t^2+1)}\;dt=\frac1{2\sqrt\pi}\cdot\frac{\pi}{\sqrt2}=\frac{\sqrt \pi}{2\sqrt2}. \end{eqnarray}$$