Recent content by iggyonphysics

Got it, thanks so much!

Right, so I have I'(α) = 1/\sqrt{α}\int_0^∞ e^{-x^2 - u^2} du = 1/\sqrt{α}\int_0^∞ e^{-a/u^2 - u^2} du , which looks just like the l(a)=\int_0^\infty e^{-x^2- a/x^2}dx I don't understand what to do next.

Yes, with dx. Sorry! (Also, how do you format equations?) For this problem, I am supposed to find I(1) using u = sqrt(x)/a and the answer is e-2sqrt(π)/2 .

I(α) = 0∞∫e-(x2+α/x2) dx Differentiating under the integral sign leads to: I(α) = 0∞∫-e-(x2+α/x2)/x2 dx Here I am supposed to let u = sqrt(a)/x, but the -x2 doesn't cancel out, Wolfram-Alpha tells me the answer is: e(-2 sqrt(α) sqrt(π))/(2 sqrt(α)). I understand where the sqrt(π))/(2)sqrt(α)...

Great, so that just comes to i, correct? So, F^2=π/4 eiπ/2 --> F = ei/4sqrt(π/4) How do I get to S and C from here? (I know if I evaluate the product of sqrt(π/4) and sin(π/4)2 it spits out sqrt(π/8). Why does this work?)

Homework Statement Evaluate the following integrals C = 0inf∫cos(x2) dx and S = 0inf∫sin(x2) dx Homework Equations [/B] Hint: use Euler formula to write the integral for F = C + iS. Square the integral and evaluate it in polar coordinates. Temporary add a convergence factor. Answer: C = S =...