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I recently came into the following definite integral

$$ I = \int_{0}^{\infty} \sin\left(x^2\right)\:dx$$

To solve this, I used a composition of both the Feynman Trick with Laplace Transforms as given below. I was curious as to what other methods exist in solving this integral?

My method:

Let

$$ I(t) = \int_{0}^{\infty} \sin\left(tx^2\right)\:dx$$

Take the Laplace transform

\begin{align} \mathscr{L}\left[I(t) \right] &= \int_{0}^{\infty} \mathscr{L}\left[\sin\left(tx^2\right)\right]\:dx \\ &= \int_{0}^{\infty}\frac{x^2}{x^4 + s^2}\:dx \end{align}

Applying a change of variable $x = u\sqrt{s}$

\begin{align} \mathscr{L}\left[I(t) \right] &= \frac{1}{\sqrt{s}}\int_{0}^{\infty}\frac{u^2}{u^4 + 1}\:dx \\ &= \frac{1}{\sqrt{s}}\left[\frac{\ln\left|x^2 -\sqrt{2}x + 1 \right| - \ln\left|x^2 +\sqrt{2}x + 1 \right| -2\arctan\left(1 - \sqrt{2}x\right) + 2\arctan\left(\sqrt{2}x + 1 \right)}{4\sqrt{2}}\right]_{0}^{\infty} \\ &= \frac{1}{4\sqrt{2}\sqrt{s}}\left[\ln\left|\frac{x^2 -\sqrt{2}x + 1}{x^2 +\sqrt{2}x + 1} \right| -2\arctan\left(1 - \sqrt{2}x\right) + 2\arctan\left(\sqrt{2}x + 1 \right)\right]_{0}^{\infty} \\ &=\frac{1}{4\sqrt{2}\sqrt{s}}\left[-2\cdot\frac{-\pi}{2} + 2\cdot\frac{\pi}{2}\right] \\ &=\frac{\pi}{2\sqrt{2}\sqrt{s}} \end{align}

Hence,

$$I(t) = \mathscr{L}^{-1}\left[\frac{\pi}{2\sqrt{2}\sqrt{s}} \right] = \frac{\pi}{2\sqrt{2}}\mathscr{L}^{-1}\left[\frac{\pi}{1\sqrt{s}} \right] = \frac{\pi}{2\sqrt{2}} \frac{1}{\sqrt{t\pi}} = \frac{\sqrt{\pi}}{2\sqrt{2}\sqrt{t}}$$

Thus,

$$ I = I(1) = \frac{\sqrt{\pi}}{2\sqrt{2}\sqrt{1}} = \frac{\sqrt{\pi}}{2\sqrt{2}}$$

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    $\begingroup$ Possible duplicate of this question since you are asking for other methods on this integral. $\endgroup$ – Kemono Chen Nov 7 '18 at 5:02
  • $\begingroup$ @KemonoChen - Thanks for pointing out. I'm not sure how to proceed - do I wait for admin/moderators to speak to the issue? $\endgroup$ – user150203 Nov 7 '18 at 5:12
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    $\begingroup$ Some high-rep users or moderators will close it as duplicate. No need to worry about this. $\endgroup$ – Kemono Chen Nov 7 '18 at 5:24
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You can first substitute $u=x^2$, use a useful property of the Laplace Transform and the Beta function and then utilize the reflection formula for the Gamma function

\begin{align} \int_0^\infty \sin(x^2)\,dx &= \frac{1}{2}\int_0^\infty \frac{\sin u}{\sqrt{u}}\,du \\ &=\frac{1}{2\sqrt{\pi}} \int_0^\infty \frac{ds}{\sqrt{s}(s^2+1)}\, \\ &=\frac{1}{2\sqrt{\pi}} \int_0^\infty \frac{1}{\nu^{1/4}(1+\nu)}\frac{d\nu}{2\sqrt{\nu}} \\ &=\frac{1}{2\sqrt{\pi}} \int_0^\infty \frac{d\nu}{\nu^{3/4}(1+\nu)} \\ &=\frac{1}{2\sqrt{\pi}} B(1/4,3/4) \\ &=\frac{1}{2\sqrt{\pi}} \frac{\Gamma(1/4)\Gamma(3/4)}{\Gamma(1)} \\ &=\frac{1}{2\sqrt{\pi}} \frac{\pi}{\sqrt{2}} \\ &=\sqrt{\frac{\pi}{8}} \end{align}

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