$\require{begingroup}\begingroup\renewcommand{\dd}[1]{\,\mathrm{d}#1}$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.$\endgroup$
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1As $I(t)=tI(1)$, I suspect you'll just go round in circles with this method. – Angina Seng May 07 '19 at 04:11
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Is there any other form of substitution that can be made? In the past, I’ve made substitutions in the arguments of the trig functions (as seen above), but I’ve also done a substitution where I put the parameter $t$ as the exponent. – ItIsLastThursday May 07 '19 at 04:13
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5You say that you know how to integrate $\int_{0}^{\infty}\frac{sin(x)}{x}dx$ , so why don't you stop after $I'$ ? – user619894 May 07 '19 at 05:28
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Because at $I’(t)$, its $\frac{sin(2tx)}{x}$, and unless there’s a good u-sub for that, there’s no way to get it to be $\frac{sin(x)}{x}$ – ItIsLastThursday May 07 '19 at 09:47
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$\int_{0}^{\infty}\frac{sin(2tx)}{x}dx= \int_{0}^{\infty}\frac{sin(2tx)}{x(2t)}d(2tx)$ – user619894 May 07 '19 at 11:09
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I’m not understanding the notation that you are using for the differential. Also, is $x$ now a function of $t$? Also, how does this help? – ItIsLastThursday May 07 '19 at 11:16
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Treat $t$ as a fixed parameter. – user619894 May 07 '19 at 12:05
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So then is the denominator equivalent to $2tx$? And what does $d(2tx)$ mean – ItIsLastThursday May 07 '19 at 12:37
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Let us continue this discussion in chat. – ItIsLastThursday May 07 '19 at 12:38
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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, $$ \begin{align} \int_0^\infty\frac{\sin^2(x)}{x^2}\,\mathrm{d}x &=I(1)\\ &=I(0)+\int_0^1\frac\pi2\,\mathrm{d}a\\[6pt] &=\frac\pi2 \end{align} $$
robjohn
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Nice (+1)! Isn't easier to integrate by parts?:) BTW, how is it going lately? – user 1591719 May 12 '19 at 09:24
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It might be a bit simpler, but not much. However, they asked for this approach.The question I have is how they they handled $\int_0^\infty\frac{\sin(x)}x,\mathrm{d}x$ with this approach. In this answer, I use Fourier series to compute this integral (avoiding contour integration). – robjohn May 12 '19 at 09:40
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