Recent content by wxstall

Thought I would also share that the Mathematica Online Integrator comes up with this disaster of a solution: http://integrals.wolfram.com/index.jsp?expr=Exp%5B-kt%28x%5E2%29%5D*x*Sin%5Bax%5D&random=false

That's what I was wondering :) So from here: ∂I/∂x = -(e-ktw2/2kt) sin(wx) + (x/2kt)I Evaluating the first term at the limits [0,∞] will give 0. For infinity, the negative exponential becomes 0, and at 0, sin(0)=0. So we should just be left with ∂I/∂x = (x/2kt)I correct? From there...

I have made it to: ∂I/∂x = -(e-ktw2/2kt) sin(wx) + (x/2kt)I Solving for I gives: I = (2kt/x) ∂I/∂x + -(e-ktw2/x) sin(wx) Does this need to be solved further? I have done as the problem instructed, but is it acceptable to leave the expression in terms of ∂I/∂x?

Evaluating a "Fourier Transform" Integral Homework Statement Evaluate I = ∫[0,∞] e-ktw2 cos(wx) dw in the following way: Determine ∂I/∂x, then integrate by parts. Homework Equations Possibly? The Attempt at a Solution Since integral limits do not depend on x, the partial with respect...

So as I suspected, it's some rule that I was unfamiliar with. Thanks for the help!

Homework Statement Show that y(t) = (1/w) ∫[0,t] f(s)*sin(w(t-s)) ds is a particular solution to y'' +w2 y = f(t)where w is a constant. The Attempt at a Solution After wasting several pages of paper I have made virtually no progress. Obviously, substitution suggests you plug in y(t)...