Solving Integral of cos(x^2): Finding the Result

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In summary: Once you have verified that the integrals converge and that the real and imaginary parts of the final answer are correctly obtained, you can proceed to applying various techniques for solving the integral.
  • #1
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Problem originally posted in a technical math section, so missing the template.
So i am trying to find the ##\int_{0}^{\infty} cos(x^2) dx##. I used Eulers identity to get ##\int_{0}^{\infty} cos(x^2) - isin(x^2) dx = \int_{0}^{\infty} e^{-i(x^2)} dx##. I squared this integral, changed to polar and evaluated and at the end of this process i got the result of ##\frac{1}{2} \sqrt{\frac{\pi}{i}}##. Where do i go from here? Am i even on the right track? Thanks for your help.
 
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  • #2
occh said:
So i am trying to find the ##\int_{0}^{\infty} cos(x^2) dx##. I used Eulers identity to get ##\int_{0}^{\infty} cos(x^2) - isin(x^2) dx = \int_{0}^{\infty} e^{-i(x^2)} dx##.
I see where your second equation comes from, but how do you get back to ##\int_0^{\infty}\cos^2(x)dx##?
IOW, how is ##\int_0^{\infty}\cos^2(x)dx = \int_{0}^{\infty} \cos(x^2) - i\sin(x^2) dx##?
occh said:
I squared this integral, changed to polar and evaluated and at the end of this process i got the result of ##\frac{1}{2} \sqrt{\frac{\pi}{i}}##. Where do i go from here? Am i even on the right track? Thanks for your help.
 
  • #3
well this problem gives you the integrals: ##\int_0^{\infty} cos(x^2) dx ## and ##\int_0^{\infty} sin(x^2) dx## and suggested proceeding in this manner to find the values of both of them by extracting the real and imaginary parts at the end after you solve the gaussian integral. This is where I am lost. I've viewed other threads on this topic this result appears there however they go through other steps to get the final result that i wasn't able to follow.
 
  • #4
What do you mean, "gives you the integrals"? It might be helpful to see the problem as stated.

At any rate, this problem belongs in the homework section, so I'm moving it there.
 
  • #5
This is an unusual integral, because usually, convergent integrals from [itex]0[/itex] to [itex]\infty[/itex] involve a function that goes to zero as [itex]x \rightarrow \infty[/itex]. In the case of [itex]cos(x^2)[/itex], it doesn't go to zero, but instead oscillates faster and faster, so the contribution from large values of [itex]x[/itex] tend to cancel out.

The way I would go about it is to use:

[itex]cos(x^2) = Re(e^{i x^2})[/itex]

where [itex]Re[/itex] means the real part. So if the integral converges, then we can write:

[itex]\int_0^\infty cos(x^2) dx = Re(\int_0^\infty e^{-i x^2} dx)[/itex]

How do you evaluate the right-hand side? I'm a little shaky about how to do it rigorously, but we know that for any [itex]\lambda[/itex] with a positive real part,

[itex]\int_0^\infty e^{-\lambda x^2} dx = \frac{\sqrt{\pi}}{2 \sqrt{\lambda}}[/itex]

If we assume (and this is where it gets a little hand-wavy) that this holds even when [itex]\lambda = i[/itex], then we would have:
[itex]\int_0^\infty e^{-i x^2} dx = \frac{\sqrt{\pi}}{2 \sqrt{i}}[/itex]

[itex]\frac{1}{\sqrt{i}} =\pm \frac{1-i}{\sqrt{2}}[/itex]

So plug that in, take the real part, and (hopefully) you have your answer. You have to choose the [itex]+[/itex] sign so that your integral is positive.
 
  • #6
Okay this yields the correct answer, for both the integrals, with the real part being for ##cos(x^2)## and (the same thing) for ##sin(x^2)## in the imaginary. Thank you very much! I guess i just needed to brush up on my manipulation of complex numbers.
 
  • #7
occh said:
Okay this yields the correct answer, for both the integrals, with the real part being for ##cos(x^2)## and (the same thing) for ##sin(x^2)## in the imaginary. Thank you very much! I guess i just needed to brush up on my manipulation of complex numbers.

There is an important issue you have "swept under the rug", namely: does the integral converge, and do your formal manipulations actually lead to a correct answer? This is trickier than usual in the present case because the integrand does not go to zero, but oscillates with increasing rapidity as ##x \to \infty## (as pointed out by stevendaryl).

You can justify what you have done by looking for properties of "Fresnel functions"; see, eg,
http://en.wikipedia.org/wiki/Fresnel_integral.
Basically, the function ##C(x) = \int_0^x \cos(t^2) \, dt## is a widely-used function in Optics and other physics fields, and its properties are well-studied. In particular, the behavior of ##C(x)## for large, positive ##x## is known and is given (without proof) in the article cited. The limit ##C(\infty)## is the value you obtained.
 
Last edited:

What is an integral and why is it important?

An integral is a mathematical concept that represents the area under a curve in a graph. It is important because it allows us to solve problems related to finding the total amount or accumulation of a quantity over a given interval.

How do you solve an integral of cos(x^2)?

To solve an integral of cos(x^2), you can use the substitution method. This involves substituting x^2 with u and then using the chain rule to solve the integral. The final result will involve the error function, which can be evaluated using a calculator or mathematical software.

What is the error function and how is it used to solve the integral of cos(x^2)?

The error function, also known as erf(x), is a special function in mathematics that is used to solve integrals involving the Gaussian function, such as cos(x^2). It is defined as the integral of the Gaussian function from 0 to x. In solving the integral of cos(x^2), we use the error function to evaluate the final result after using the substitution method.

Can the integral of cos(x^2) be solved without using the error function?

Yes, there are other methods for solving the integral of cos(x^2) without using the error function. These include using the Taylor series expansion or the Gauss-Hermite quadrature method. However, the substitution method using the error function is the most commonly used and efficient method for solving this integral.

What are some practical applications of solving integrals of cos(x^2)?

Solving integrals of cos(x^2) has many practical applications in physics, engineering, and statistics. It is used to solve problems related to calculating the area under a Gaussian curve, such as in probability and statistics. It is also used in signal processing and image reconstruction techniques. In addition, it has applications in quantum mechanics, where the wavefunction of a particle can be represented as a Gaussian function.

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