integral of sin(x)/x from 0 to inf by Feynman's Technique
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it when someone wants to evaluate the sin(x)/x integral without complex analysis or Laplace transforms, or wants to follow the Feynman technique step by step.
Recommended by 2 people across 2 different months, 2017 to 2020.
What it is
blackpenredpen works through the Dirichlet integral, the integral of sin(x)/x from 0 to infinity, using Feynman's technique of differentiation under the integral sign. He introduces a parameter with an exponential damping factor, differentiates, solves for the constant, and then sends the parameter back to 0. It is a roughly 23-minute whiteboard-style walkthrough.
What people said
Recommends Rick · 2017-12 · Mathematics Stack Exchange · on “Is it possible to compute $\int_{-\infty}^\infty {x \sin x\over x^4+1}dx$ without using complex analysis?”
@R_Berger I would have considered a function $f(t) = \int_{0}^{\infty} e^{-tx}\frac{\sin x }{x} \text{d}x$, from that 2f(0) = I and evaluate f'(t).. Something like this [this video]
Recommends Andrei · 2018-08 · Mathematics Stack Exchange · on “why do we use laplace transform to solve the integral $\int _{0}^{\infty}\frac{\sin x}{x}dx$ ? why is it insolvable if we go the normal way?”
You can solve it using Feynman's technique. But it boils down to providing the same calculation steps, with an intuitive explanation. See this video [this video] for example
Cites newbie105 · 2020-11 · Mathematics Stack Exchange · on “$\int_0^\infty \frac{\sin x}{x}dx$ using Feynman's technique”
While watching this [this video] video, he says that $$\lim_{b\to\infty}\int_0^\infty \frac{\sin x}{x}e^{-bx}dx\to 0$$ since the integrand tends to $0$ . However I think that it must be an indeterminate form, since $\infty\times0$ may not always be $0$ Am I right in saying this? If yes, how else can we evaluate the integration constant? Reference picture: [LINK]