Integral of exponential over square root

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Discussion Overview

The discussion centers around evaluating the integral ##\int_{-\infty}^{+\infty}\frac{e^{i A x}}{\sqrt{x^2+a^2}}dx##. Participants explore various methods for solving this integral, including the use of Euler's formula and substitutions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest using Euler's formula to expand the complex exponential into cosine and sine components.
  • It is proposed that the integral can be separated into two parts, with the sine integral being odd and thus equal to zero.
  • One participant mentions that the integral of the cosine part can be expressed as ##2*K0(a*A)##, where ##K0## is the modified Bessel function of the second kind.
  • There are suggestions that tools like Mathematica or Wolfram Alpha Pro could be used to evaluate the integral.
  • Some participants express uncertainty about using Wolfram Alpha and discuss modifying the integral to simplify it by eliminating one of the constants.
  • A substitution ##u = Ax## is proposed to transform the integral into a more manageable form.

Areas of Agreement / Disagreement

Participants generally agree on the methods of using Euler's formula and substitution, but there is no consensus on the best approach to evaluate the integral or on the use of specific computational tools.

Contextual Notes

Some participants note the challenge of evaluating the integral with symbolic constants and suggest that numerical examples may provide insights.

newgate
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Hello guys,
How can I evaluate the following integral please?
##\int_{-\infty}^{+\infty}\frac{e^{i A x}}{\sqrt{x^2+a^2}}dx##
Thank you
 
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newgate said:
Hello guys,
How can I evaluate the following integral please?
##\int_{-\infty}^{+\infty}\frac{e^{i A x}}{\sqrt{x^2+a^2}}dx##
Thank you

Use Euler to expand the complex exponential into cos() + i*sin().

Separate into two separate integrals. The integrand with sin() is odd, so its integral is ZERO.

Wolfram Alpha won't work the cos() integrand with a and A as symbols, but with a bit of plugging in numbers, one can quickly determine that the integral of the cos() integrand is 2*K0(a*A), where K0 is the modified Bessel function of the second kind. A good integral table would likely tell you that also.

Mathematica or Wolfram Alpha Pro would probably do the original integral.
 
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Dr. Courtney said:
Use Euler to expand the complex exponential into cos() + i*sin().

Separate into two separate integrals. The integrand with sin() is odd, so its integral is ZERO.

Wolfram Alpha won't work the cos() integrand with a and A as symbols, but with a bit of plugging in numbers, one can quickly determine that the integral of the cos() integrand is 2*K0(a*A), where K0 is the modified Bessel function of the second kind. A good integral table would likely tell you that also.

Mathematica or Wolfram Alpha Pro would probably do the original integral.
Thank you Dr.Courtney
 
newgate said:
Thank you Dr.Courtney
I am not familiar with Wolfram. However, it is easy enough to modify the integral to get rid of A or a, so that you have only one constant.
 
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mathman said:
I am not familiar with Wolfram. However, it is easy enough to modify the integral to get rid of A or a, so that you have only one constant.
How? I'm curious :D
 
newgate said:
How? I'm curious :D

Substitution ##u = Ax##. Then you just have to find an integral of the type ##\int_{-\infty}^{+\infty} \frac{e^{ix}}{\sqrt{x^2 + C}}dx.##. Likewise, a good substitution will leave you with an integral of the type ##\int_{-\infty}^{+\infty} \frac{e^{Cix}}{\sqrt{x^2 + 1}}dx.##
 
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micromass said:
Substitution ##u = Ax##. Then you just have to find an integral of the type ##\int_{-\infty}^{+\infty} \frac{e^{ix}}{\sqrt{x^2 + C}}dx.##. Likewise, a good substitution will leave you with an integral of the type ##\int_{-\infty}^{+\infty} \frac{e^{Cix}}{\sqrt{x^2 + 1}}dx.##
Ok thank you very much.
 

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