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I would like to get the exact value of the following integral. $$\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx$$

I was able to prove the convergence as well. But I don't how to compute its exact value. I tried with the Residue Theorem of the complex function

$$z\mapsto \frac{\sin^2 z}{z^{5/2}}$$ But I could not move further.

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  • $\begingroup$ Hmm this integral calculator shows that it is divergent? $\endgroup$ – aleden Nov 6 '17 at 15:39
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    $\begingroup$ @aleden: that integral calculator sucks. The given function behaves like $\frac{1}{\sqrt{x}}$ in a right neighbourhood of the origin and it is bounded by $\frac{1}{x^2}$ far from the origin, hence it is integrable over $\mathbb{R}^+$. $\endgroup$ – Jack D'Aurizio Nov 6 '17 at 15:43
  • $\begingroup$ @JackD'Aurizio Ah I see $\endgroup$ – aleden Nov 6 '17 at 15:44
  • $\begingroup$ @JackD'Aurizio I hate it when people say “this integral calculator evaluates your integral to be ...” $\endgroup$ – Crescendo Nov 6 '17 at 17:29
  • $\begingroup$ me too. with calculator every thing becomes banal $\endgroup$ – Guy Fsone Nov 6 '17 at 17:39
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With a step of integration by parts the problem boils down to computing

$$ \frac{2}{3}\int_{0}^{+\infty}\frac{\sin(2x)}{x^{3/2}}\,dx =\frac{2\sqrt{2}}{3}\int_{0}^{+\infty}\frac{\sin x}{x^{3/2}}\,dx\stackrel{\mathcal{L},\mathcal{L}^{-1}}{=}\frac{4\sqrt{2}}{3\sqrt{\pi}}\int_{0}^{+\infty}\frac{\sqrt{s}}{s^2+1}\,ds$$ and the last integral is elementary (just enforce the substitution $s\mapsto u^2$ and perform a partial fraction decomposition). By the properties of the Laplace transform, the final outcome is $\frac{4}{3}\sqrt{\pi}$.

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This integral $I$ can be interpreted as the Mellin transform of $\sin(x)^2$ evaluated at $s=-3/2$. $$ \mathcal{M}[\sin(x)^2] = -2^{-1-s}\cos\left(\frac{\pi s}{2}\right)\Gamma(s) $$ so $$ I = \Gamma\left(-\frac{3}{2}\right) = \frac{4\sqrt{\pi}}{3} $$

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Performing the change of variables $2u = x^2$ together with two integration by parts, we get, $$ \int_0^\infty \cos(x^2)dx = \frac{1}{\sqrt{2}}\int^\infty_0\frac{\cos(2x)}{\sqrt{x}}\,dx\\=\frac{1}{2\sqrt{2}}\underbrace{\left[\frac{\sin 2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{4\sqrt{2}} \int^\infty_0\frac{\sin 2 x}{x^{3/2}}\,dx\\= \frac{1}{4\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{3}{8\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx$$

Hence $$\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx = \frac{8\sqrt2}{3}\int_0^\infty \cos(x^2)dx = \frac{4\sqrt \pi}{3}$$ Since See Here, $$\int_0^\infty \cos(x^2)dx= \sqrt\frac\pi8$$

or How to prove only by Transformation that: $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $

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