Integral of 1/sqrt(x)exp(-ix) dx

In summary, the conversation discusses finding the integral of 1/sqrt(x)exp(ix)dx from minus infinity to infinity. The two suggested solutions are to use the substitution x=u^2 or to use Jordan's Lemma. However, neither solution seems to be applicable. The conversation then suggests splitting the integral into two parts and setting u=sqrt(x), but there are concerns about the limits. The conversation concludes with a clarification on the limits for u when x is positive or negative.
  • #1
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Hi,

I am just doing this out of curiousity.

Homework Statement



I want to integrate 1/sqrt(x)exp(ix) dx from minus infinity to infinity.


Homework Equations





The Attempt at a Solution



I had a couple of ideas one was to substitute x=u^2
but then you mess up the limits and you get minus imaginary infinity.

The other idea was to use Jordan's Lemma. But as far as I know 1/sqrt(x) doesn't have a residue so it can't be applied.

How do you solve this integral then?

thank you
 
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  • #2
That substitution looks like a good idea, if you split up your integral into two parts first.
 
  • #3
Hey,

Thanks for the hint. I have done the splitting, but I am not sure about the limits.

If I set u=sqrt(x) then the lower limit is plus or minus imaginary infinity and moreover the upper limit can take two values: plus infinity or minus infinity.

I don't know how to integrate that. I am aware that cos(x^2)dx and sin(x^2)dx are the well known fresnel integrals.

thanks for your help.
 
  • #4
For x from 0 to plus (real) infinity, u goes from 0 to plus (real) infinity.
For negative x, you can transform the integral to the integral for positive x.
 

What is the integral of 1/sqrt(x)exp(-ix) dx?

The integral of 1/sqrt(x)exp(-ix) dx is equal to -2i*sqrt(pi/x)exp(-ix). This can be solved by using the substitution method, where u = -ix and du = -idx. This leads to the integral becoming -2i*sqrt(pi/x)exp(u) du, which can then be solved using integration by parts.

What is the significance of 1/sqrt(x)exp(-ix) in science?

1/sqrt(x)exp(-ix) is known as the Gaussian wave packet, and it has many applications in science. It is commonly used in quantum mechanics to describe the wave function of a particle in a given energy state. It is also used in signal processing to analyze and filter signals.

How is the integral of 1/sqrt(x)exp(-ix) dx related to the error function?

The integral of 1/sqrt(x)exp(-ix) dx is directly related to the error function, erf(x). In fact, it can be written as erf(x/sqrt(2)) + C, where C is a constant. This relationship is useful in solving various integrals and in statistics, where the error function is used to calculate probabilities in normal distributions.

Can the integral of 1/sqrt(x)exp(-ix) dx be solved using other methods?

Yes, the integral of 1/sqrt(x)exp(-ix) dx can also be solved using contour integration. This involves integrating over a closed contour in the complex plane and using Cauchy's integral formula. This method is often used in complex analysis and can lead to simpler solutions for certain integrals.

Are there any real-world applications of the integral of 1/sqrt(x)exp(-ix) dx?

Yes, there are many real-world applications of the integral of 1/sqrt(x)exp(-ix) dx. One example is in the calculation of electric fields in electrostatics, where the electric potential can be represented by the integral of 1/sqrt(x)exp(-ix) dx. It is also used in optics to calculate the intensity of light passing through a slit or diffraction grating.

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