Integral of exponential over square root

[newgate]

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

 

[Dr. Courtney]

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.

 

[newgate]

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

 

[mathman]

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.

 

[newgate]

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

 

[micromass]

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 $\int_{-\infty}^{+\infty} \frac{e^{Cix}}{\sqrt{x^2 + 1}}dx.$

 

[newgate]

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.