Using Laplace Transforms to solve ${\int}_{0}^{\mathrm{\infty}}\mathrm{sin}\left({x}^{2}\right)\phantom{\rule{mediummathspace}{0ex}}dx$

Altenbraknz
2022-09-20
Answered

Using Laplace Transforms to solve ${\int}_{0}^{\mathrm{\infty}}\mathrm{sin}\left({x}^{2}\right)\phantom{\rule{mediummathspace}{0ex}}dx$

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baselulaox

Answered 2022-09-21
Author has **8** answers

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

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

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

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