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Mathematics
General Math
Why does the integral of sine of x^2 from - infinity to + infinity diverge?
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[QUOTE="jaumzaum, post: 6579939, member: 339620"] Hello guys. I was trying to evaluate the integral of sine of x^2 from - infinity to + infinity and ran into some inconsistencies. I know this integral converges to sqrt(pi/2). Can someone help me to figure out why I am getting a divergent answer? $$ I = \int_{-\infty}^{+\infty} sin(x^2) dx = Im(\int_{-\infty}^{+\infty} e^{ix^2} dx) $$ Now let's call: $$ A=\int_{-\infty}^{+\infty} e^{ix^2} dx $$ $$ A^2=\int_{-\infty}^{+\infty} e^{ix^2} dx \int_{-\infty}^{+\infty} e^{iy^2} dy $$ $$ A^2=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{i(x^2+y^2)} dx dy $$ $$ A^2=\int_{0}^{2\pi} \int_{0}^{+\infty} e^{ir^2}r dr d\theta $$ $$ A^2= -i\pi \left. e^{ir^2} \right|_0^{\infty} $$ $$ A^2= -i\pi (e^{i\infty}-1) $$ But we know ## lim_{M->\infty} e^{iM}## does not converge, so A^2 would not converge either! However, if we consider ##e^{i\infty}## as being zero, we get: $$A^2= i\pi$$ $$A=1/\sqrt{2}\pi(1+i)$$ $$I=\sqrt{\pi/2}$$ But why is ##e^{i\infty}## considered zero if ## lim_{M->\infty} e^{iM}## does not exist? [/QUOTE]
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Mathematics
General Math
Why does the integral of sine of x^2 from - infinity to + infinity diverge?
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