I am trying to prove that ∫∞−∞∫∞−∞sin(x2+y2)dxdy diverges and I did it like this:
x=rcosθ, y=rsinθ, θ∈[0,2π], r∈[0,∞], and the jacobian is r:
∫∞−∞∫∞−∞sin(x2+y2)dxdy=∫2π0∫∞0rsin(r2)drdθ=2π∫∞0rsin(r2)dr
Use the variable change v=r2, dv=2rdr and so:
2π∫∞0rsin(r2)dr=2π∫∞0rsin(v)dv2r=π∫∞0sin(v)dv
I checked wolfram, and it says that ∫∞0sin(x)dx diverges, so I must conclude that the original integral diverges as well.
But how do I actually prove that ∫∞0sin(x)dx diverges?
Answer
We can compute the limiting integral by writing ∫∞0sin(x)dx=lim We know that the limit \lim_{b\to\infty} \cos(b) diverges. This means the integral diverges as well.
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