Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods? I have seen the Fresnel integral
$$\int_0^\infty \sin x^2\, dx = \sqrt{\frac{\pi}{8}}$$
evaluated by contour integration and other complex analysis methods, and I have found these methods to be the standard way to evaluate this integral.  I was wondering, however, does anyone know a real analysis method to evaluate this integral?
 A: 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 $$F(1)=\int_0^{\infty}\sin (x^2)\ dx$$ 
$$\int_0^{\infty}\sin (x^2)\ dx=\frac{\pi\sqrt{2}}{4\sqrt{\pi\times 1}}=\sqrt{\frac{\pi}{8}}$$
A: 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. 
A: Let $u=x^2$, then
$$
   \int_0^\infty \sin(u) \frac{\mathrm{d} u}{2 \sqrt{u}}
$$
The real analysis way of evaluating this integral is to consider a parametric family:
$$\begin{eqnarray}
    I(\epsilon) &=& \int_0^\infty \frac{\sin(u)}{2 \sqrt{u}} \mathrm{e}^{-\epsilon u} \mathrm{d} u = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\int_0^\infty u^{2n+\frac{1}{2}} \mathrm{e}^{-\epsilon u} \mathrm{d} u \\ &=& \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \Gamma\left(2n+\frac{3}{2}\right) \epsilon^{-\frac{3}{2}-2n} \\
  &=& \frac{1}{2 \epsilon^{3/2}} \sum_{n=0}^\infty \left(-\frac{1}{\epsilon^2}\right)^n\frac{\Gamma\left(2n+\frac{3}{2}\right)}{\Gamma\left(2n+2\right)} \\
  &\stackrel{\Gamma-\text{duplication}}{=}&\frac{1}{2 \epsilon^{3/2}} \sum_{n=0}^\infty \left(-\frac{1}{\epsilon^2}\right)^n\frac{\Gamma\left(n+\frac{3}{4}\right)\Gamma\left(n+\frac{5}{4}\right)}{\sqrt{2} n! \Gamma\left(n+\frac{3}{2}\right)} \\
   &=& \frac{1}{(2 \epsilon)^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{2}\right)} {}_2F_1\left(\frac{3}{4}, \frac{5}{4}; \frac{3}{2}; -\frac{1}{\epsilon^2}\right) \\
  &\stackrel{\text{Euler integral}}{=}& \frac{1}{(2 \epsilon)^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{2}\right)} \frac{1}{\operatorname{B}\left(\frac{5}{4}, \frac{3}{2}-\frac{5}{4}\right)} \int_0^1 x^{\frac{5}{4}-1} (1-x)^{\frac{3}{2}-\frac{5}{4} -1} \left(1+\frac{x}{\epsilon^2}\right)^{-3/4} \mathrm{d} x \\
  &=& \frac{1}{2^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{2}\right)} \frac{\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\frac{1}{4}\right)} \int_0^1 x^{\frac{5}{4}-1} (1-x)^{\frac{1}{4} -1} \left(\epsilon^2+x\right)^{-3/4} \mathrm{d} x
\end{eqnarray}
$$
Now we are ready to compute $\lim_{\epsilon \to 0} I(\epsilon)$:
$$\begin{eqnarray}
  \lim_{\epsilon \to 0} I(\epsilon) &=& \frac{1}{2^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)} \int_0^1 x^{\frac{1}{2}-1} \left(1-x\right)^{\frac{1}{4}-1} \mathrm{d} x = \frac{1}{2^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)} \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)} \\ &=& \frac{1}{2^{3/2}} \Gamma\left(\frac{1}{2}\right) = \frac{1}{2} \sqrt{\frac{\pi}{2}}
\end{eqnarray}
$$
A: 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*}
$$
J
=
\sqrt{\,{\rm e}^{{\rm i}\pi/2}\,{\pi \over 4}\,}
=
{\rm e}^{{\rm i}\pi/4}\,\sqrt{\,\pi \over 4\,}
$$
$$
\int_{0}^{\infty}\sin\left(x^{2}\right)\,{\rm d}x
=
\Im J
=
\overbrace{\quad\sin\left(\pi \over 4\right)\quad}^{=\ 1/\sqrt{\,2\,}}
\sqrt{\,\pi \over 4\,}
=
\sqrt{\,\pi \over 8\,}
$$
A: 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
$$ \int_{0}^{\infty} \cos\left(x^2\right) \; dx = \sqrt{\frac{\pi}{8}}. $$
A: 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 $\cos(x^2)$.
A: Short
ELet 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 present of the 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}


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:How to prove only by Transformation that: $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $
A: 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 $n=2$,
$$\int_0^\infty \sin(x^2)\,\mathrm{d}x=\Gamma\left(\frac{3}{2}\right)\sin\frac{\pi}{4}=\frac{\sqrt{\pi}}{2\sqrt{2}}$$
