The integral of the function \(\int_{-\infty}^{+\infty}\frac{e^{i A x}}{\sqrt{x^2+a^2}}dx\) can be evaluated using Euler's formula to separate it into cosine and sine components. The sine component integrates to zero due to its odd nature, while the cosine component evaluates to \(2K_0(aA)\), where \(K_0\) is the modified Bessel function of the second kind. Tools such as Mathematica or Wolfram Alpha Pro can assist in computing this integral, especially when substituting constants to simplify the expression.
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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
Thank you Dr.CourtneyDr. 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.
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.newgate said:Thank you Dr.Courtney
How? I'm curious :Dmathman 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.
newgate said:How? I'm curious :D
Ok thank you very much.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.##