Even MIT fears the wrath of FEYNMAN

For more on Laplace Transforms here ya go:
• Laplace Transforms
Triple angle sine:
• Triple Angle Formula sine
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[@MikeMagTech]

It's interesting that differentiating under the integral sign is given only very brief coverage in Woods' Advanced Calculus, but Feynman used it to great effect and made it quite popular. Feynman also had a differentiation technique he called the "dispatch method", but that is not nearly so well known.

[@owl3math]

Oooh I don’t know the dispatch method. And yes it’s an interesting history of the Feynman technique! It goes from basically never used to sometimes used to use it all the time for everything. 🤣

[@adandap]

Seems like it has been a while since the answer was pi/4. Now I feel all nostalgic!

[@owl3math]

@@adandap exactly! 🤣🤣🤣

[@hamdamoverali]

I used lobacheskiy's formula Integral of f(x)*sin(x)/x from 0 to infinity=integral of f(x) from 0 to pi/2 where f(x) is both continuous and has a period of pi

[@slavinojunepri7648]

@@hamdamoverali good stuff right there

[@slavinojunepri7648]

We can use the triple angle formula for sin and Dirichlet's integral to arrive quickly at the answer. The exponential parametrization in this video was used to derive the Dirichlet integral in the first place, but repeating it provides good review into Laplace Transform for rusty viewers. Good stuff 👌

[@owl3math]

Good point! thanks :)

[@harrydiv321]

if you wanna make f(infinity)=0 more rigorous i think you can show that the sequence of functions e^(-nx) sin^3(x)/x converges uniformly to 0 on any closed interval [epsilon, t] for 0 < epsilon < t so the limit of the integral as s -> infty is 0 by the integral limit theorem

[@owl3math]

Yep makes sense. Thanks!

[@celkat]

I was going to ask about this: It's clear (i.e. intuitive) that the integrand converges to zero as s->oo, but it's a lot less clear to me that the entire integral therefore converges to zero.

[@owl3math]

Hi @@celkat yes see what you mean. Because even though the integral of 0 is 0 it's not as clear when it's a limit and we are integrating to infinity. You could look at the whole integral as a Laplace transform. The laplace transform of just sin(ax) = a/(s^2 + a^2) which approaches 0 as s approaches infinity. I use that just because sin(ax)/x approaches 0 even quicker then the problem. or Laplace transform of sin(x)/x = pi/2 - arctan(s) which again is 0 as s approaches infinity.

[@theelk801]

it’s probably faster to use power reduction here since you already had the previous value

[@owl3math]

Hi Evan. Power reduction on sin^3 x?

[@theelk801]

@@owl3math yeah

[@owl3math]

@@theelk801 I think the triple angle formula is easier? Maybe I'm misunderstanding

[@theelk801]

@@owl3math I’m saying you can turn sin^3(x) into (3sin(x)-sin(3x))/4 and then use u=3x on the second term and the whole thing becomes a scalar multiple of the integral of sin(x)/x

[@owl3math]

@@theelk801 ah makes sense & thanks for clarifying :) It would save some time do it that way 👍👍