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
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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 =...