# Using Laplace Transforms to solve int_0^(oo) sin(x^2)dx

Using Laplace Transforms to solve ${\int }_{0}^{\mathrm{\infty }}\mathrm{sin}\left({x}^{2}\right)\phantom{\rule{mediummathspace}{0ex}}dx$
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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
$\begin{array}{rl}{\int }_{0}^{\mathrm{\infty }}\mathrm{sin}\left({x}^{2}\right)\phantom{\rule{thinmathspace}{0ex}}dx& =\frac{1}{2}{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}u}{\sqrt{u}}\phantom{\rule{thinmathspace}{0ex}}du\\ & =\frac{1}{2\sqrt{\pi }}{\int }_{0}^{\mathrm{\infty }}\frac{ds}{\sqrt{s}\left({s}^{2}+1\right)}\phantom{\rule{thinmathspace}{0ex}}\\ & =\frac{1}{2\sqrt{\pi }}{\int }_{0}^{\mathrm{\infty }}\frac{1}{{\nu }^{1/4}\left(1+\nu \right)}\frac{d\nu }{2\sqrt{\nu }}\\ & =\frac{1}{2\sqrt{\pi }}{\int }_{0}^{\mathrm{\infty }}\frac{d\nu }{{\nu }^{3/4}\left(1+\nu \right)}\\ & =\frac{1}{2\sqrt{\pi }}B\left(1/4,3/4\right)\\ & =\frac{1}{2\sqrt{\pi }}\frac{\mathrm{\Gamma }\left(1/4\right)\mathrm{\Gamma }\left(3/4\right)}{\mathrm{\Gamma }\left(1\right)}\\ & =\frac{1}{2\sqrt{\pi }}\frac{\pi }{\sqrt{2}}\\ & =\sqrt{\frac{\pi }{8}}\end{array}$