Integral of exponential over square root

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  • #1
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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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  • #2
Dr. Courtney
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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
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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  • #3
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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
 
  • #4
mathman
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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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  • #5
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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
 
  • #6
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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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  • #7
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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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