Integral Using Feynman's Method

Question

$\require{begingroup}\begingroup\renewcommand{\dd}[1]{\,\mathrm{d}#1}$I was trying to solve this integral using Feynman's Method: $$\int_{0}^{\infty}\frac{ \sin^{2}(x)}{x^{2}} \dd{x}$$ I thought I'd use Feynman's Method because I was able to solve this integral using the same method: $$\int_{0}^{\infty}\frac{ \sin(x)}{x} \dd{x}$$ Here are my steps: $$I=\int_{0}^{\infty}\frac{ \sin^{2}(x)}{x^{2}} \dd{x} \\ I(t)=\int_{0}^{\infty}\frac{ \sin^{2}(tx)}{x^{2}}dx ~, \qquad I(1)=I \\ \begin{aligned} I'(t) = \frac{\dd{}}{\dd{t}} \int_{0}^{\infty}\frac{ \sin^{2}(tx)}{x^{2}} \dd{x} &=\int_{0}^{\infty}\frac{\partial}{\partial t}\frac{ \sin^{2}(tx)}{x^2} \dd{x} \\ &=\int_{0}^{\infty}\frac{2 \sin(tx)\cos(tx)}{x} \dd{x} \\ &=\int_{0}^{\infty}\frac{ \sin(2tx)}{x} \dd{x} \end{aligned} \\ \begin{aligned} I''(t) = \frac{\dd{}}{\dd{t}} \int_{0}^{\infty}\frac{ \sin(2tx)}x \dd{x} &=\int_{0}^{\infty}\frac{\partial}{\partial t}\frac{ \sin(2tx)}x \dd{x} \\ &=\int_{0}^{\infty}2 \sin(2tx) \cos(2tx) \dd{x} \\ &=\int_{0}^{\infty} \sin(4tx) \dd{x} \end{aligned} \\ u=4tx \implies \dd{u} = 4t \dd{x} \implies \frac1{4t} \dd{u} = \dd{x} \\ \begin{aligned}I''(t) &= \frac1{4t}\int_{0}^{\infty} \sin(u) \dd{u} \\ &=\frac{1}{4t}\bigl[-\cos(u) \bigr]_{0}^{\infty} \end{aligned} $$ But $-\cos(\infty)$ is not defined, so this is where I got stuck. With whatever hint or answer you provide, it would be greatly appreciated if you explained what lead you to this solution and not just why its correct.$\endgroup$

Answer 1

Let $$ I(a)=\int_0^\infty\frac{\sin^2(ax)}{x^2}\,\mathrm{d}x $$ Note that $I(0)=0$ and $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}a}I(a) &=\int_0^\infty\frac{2\sin(ax)\cos(ax)}{x}\,\mathrm{d}x\\ &=\int_0^\infty\frac{\sin(2ax)}{x}\,\mathrm{d}x\\[3pt] &=\frac\pi2 \end{align} $$ Therefore, $$ \begin{align} \int_0^\infty\frac{\sin^2(x)}{x^2}\,\mathrm{d}x &=I(1)\\ &=I(0)+\int_0^1\frac\pi2\,\mathrm{d}a\\[6pt] &=\frac\pi2 \end{align} $$