- #1

jaumzaum

- 434

- 33

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?

$$ I = \int_{-\infty}^{+\infty} sin(x^2) dx = Im(\int_{-\infty}^{+\infty} e^{ix^2} dx) $$

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?

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