Why does the integral of sine of x^2 from - infinity to + infinity diverge?

In summary, the conversation discusses the evaluation of the integral of sine of x^2 from -infinity to +infinity, which converges to sqrt(pi/2). The conversation also explores different methods of evaluating the integral, including using line integrals in the complex plane and applying Fubini's theorem. However, there is a mistake in one of the steps due to the integrand not being absolutely integrable.
  • #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?
 
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  • #2
I delete my message.
[EDIT]
Going forward your way of
[tex]A^2=i\pi(e^{iR^2}-1), R\rightarrow \infty[/tex]
[tex]=-\pi \sin R^2 - i\pi(1-\cos R^2)=\pi \sqrt{2} \sqrt{1-\cos R^2}e^{i\theta}[/tex]
where
[tex]\theta = \arctan \frac{1-\cos R^2}{\sin R^2}+\pi[/tex]
[tex]A=\sqrt{\pi \sqrt{2} \sqrt{1-\cos R^2}}e^{i\theta/2}[/tex]
[tex]I=Im\ A= \sqrt{\pi \sqrt{2} \sqrt{1-\cos R^2}} sin (\theta/2)[/tex]
I let Wolfram plot the graph. I draw red line y=##1/\sqrt{2}## of the expected limit.
plot integral 211226.png
 
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  • #3
This might be how to do it properly, using line integrals in the complex plane.

 
  • #4
An informal way to do it would be to look at the integral: $$\int_0^\infty \sin (x^2) dx = \lim_{n \rightarrow \infty}\int_0^{\sqrt{n\pi}}\sin(x^2)dx$$Then follow your technique to get
$$A^2 = \lim_{n \rightarrow \infty}\int_0^{2\pi} \int_{-2n\pi}^{2n\pi}e^{ir^2}r dr d\theta$$Etc.
 
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  • #5
I think there is an error in ##\int_{-\infty}^{+\infty} e^{ix^2} dx \int_{-\infty}^{+\infty} e^{iy^2} dy =\int_{0}^{2\pi} \int_{0}^{\infty} e^{ir^2} r dr d\theta. ##

It looks like you're trying to apply Fubini's theorem, but that requires the integrand to be absolutely integrable, which ##e^{i(x^2+y^2)}## is not. If you try to write out the argument formally with limits, you should see what goes wrong- your integral for ##A^2## would be over a large rectangle, but you would want it to be over a disk for polar coordinates to be useful. This distinction wouldn't matter in the case that your function is absolutely integrable, because then the difference integrals over the two domains would tend to zero.
 
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Related to Why does the integral of sine of x^2 from - infinity to + infinity diverge?

What is the integral of sine of x^2?

The integral of sine of x^2 is an indefinite integral, which means it does not have a specific numerical value. Instead, it is represented by the antiderivative of the function, which is written as ∫sin(x^2)dx.

How do you solve the integral of sine of x^2?

There is no simple formula for solving the integral of sine of x^2. However, it can be solved using integration techniques such as substitution or integration by parts. It is also possible to use numerical methods, such as Simpson's rule, to approximate the value of the integral.

What is the significance of the integral of sine of x^2 in mathematics?

The integral of sine of x^2 is a special type of integral known as an improper integral. It is often used in the study of Fourier series and in solving differential equations. It also has applications in physics, particularly in the study of wave phenomena.

Is the integral of sine of x^2 a periodic function?

No, the integral of sine of x^2 is not a periodic function. Unlike the sine function, which repeats itself at regular intervals, the integral of sine of x^2 does not have a repeating pattern. It can, however, exhibit certain symmetries depending on the limits of integration.

What is the relationship between the integral of sine of x^2 and the Fresnel integral?

The Fresnel integral is a special case of the integral of sine of x^2, with the limits of integration being 0 and a constant value. The Fresnel integral is used in optics and has applications in calculating the diffraction of light. It is also closely related to the error function, which is used in statistics and probability.

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