# Value of this integral and how to solve it?

## ${\int}_{0}^{\propto} \left(1 - {x}^{2} \sin \left(\frac{1}{x} ^ 2\right)\right) \mathrm{dx}$

Apr 16, 2018

$I = \frac{\sqrt{2 \pi}}{3}$

#### Explanation:

We want to solve

$I = {\int}_{0}^{\infty} \left(1 - {x}^{2} \sin \left(\frac{1}{x} ^ 2\right)\right) \mathrm{dx}$

Later on, we will use the fact

color(blue)(int_0^oo sin(x^2)dx=sqrt(pi/8)

Make a substitution color(red)(u=1/x^2=>du=-2/x^3dx and color(red)(x=1/sqrt(u)

$I = - \frac{1}{2} {\int}_{\infty}^{0} \frac{u - \sin \left(u\right)}{{u}^{\frac{5}{2}}} \mathrm{du}$

$\textcolor{w h i t e}{I} = \frac{1}{2} {\int}_{0}^{\infty} \frac{u - \sin \left(u\right)}{{u}^{\frac{5}{2}}} \mathrm{du}$

Use IBP

$I = - \frac{1}{3} {\left[\frac{u - \sin \left(u\right)}{{u}^{\frac{3}{2}}}\right]}_{0}^{\infty} + \frac{1}{3} {\int}_{0}^{\infty} \frac{1 - \cos \left(u\right)}{u} ^ \left(\frac{3}{2}\right) \mathrm{du}$

But

$\textcolor{red}{- \frac{1}{3} {\left[\frac{u - \sin \left(u\right)}{{u}^{\frac{3}{2}}}\right]}_{0}^{\infty} = 0} \leftarrow \text{Use LHS}$

Thus

$I = \frac{1}{3} {\int}_{0}^{\infty} \frac{1 - \cos \left(u\right)}{u} ^ \left(\frac{3}{2}\right) \mathrm{du}$

Use IBP

$I = - \frac{2}{3} {\left[\frac{1 - \cos \left(u\right)}{\sqrt{u}}\right]}_{0}^{\infty} + \frac{2}{3} {\int}_{0}^{\infty} \sin \frac{u}{u} ^ \left(\frac{1}{2}\right) \mathrm{du}$

But

$\textcolor{red}{- \frac{2}{3} {\left[\frac{1 - \cos \left(u\right)}{\sqrt{u}}\right]}_{0}^{\infty} = 0} \leftarrow \text{Use LHS}$

Thus

$I = \frac{2}{3} {\int}_{0}^{\infty} \sin \frac{u}{u} ^ \left(\frac{1}{2}\right) \mathrm{du}$

Make a substitution $s = \sqrt{u} \implies \mathrm{ds} = \frac{1}{2 s} \mathrm{du}$

$I = \frac{4}{3} {\int}_{0}^{\infty} \sin \left({s}^{2}\right) \mathrm{ds} = \frac{4}{3} \sqrt{\frac{\pi}{8}} = \frac{\sqrt{2 \pi}}{3}$