# What is the exact value of $\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx$

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.

• Hmm this integral calculator shows that it is divergent? – aleden Nov 6 '17 at 15:39
• @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 $\mathbb{R}^+$. – Jack D'Aurizio Nov 6 '17 at 15:43
• @JackD'Aurizio Ah I see – aleden Nov 6 '17 at 15:44
• @JackD'Aurizio I hate it when people say “this integral calculator evaluates your integral to be ...” – Crescendo Nov 6 '17 at 17:29
• me too. with calculator every thing becomes banal – Guy Fsone Nov 6 '17 at 17:39

$$\frac{2}{3}\int_{0}^{+\infty}\frac{\sin(2x)}{x^{3/2}}\,dx =\frac{2\sqrt{2}}{3}\int_{0}^{+\infty}\frac{\sin x}{x^{3/2}}\,dx\stackrel{\mathcal{L},\mathcal{L}^{-1}}{=}\frac{4\sqrt{2}}{3\sqrt{\pi}}\int_{0}^{+\infty}\frac{\sqrt{s}}{s^2+1}\,ds$$ 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 $\frac{4}{3}\sqrt{\pi}$.
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 $$I = \Gamma\left(-\frac{3}{2}\right) = \frac{4\sqrt{\pi}}{3}$$
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, $$\int_0^\infty \cos(x^2)dx= \sqrt\frac\pi8$$