- #1

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How can I evaluate the following integral please?

##\int_{-\infty}^{+\infty}\frac{e^{i A x}}{\sqrt{x^2+a^2}}dx##

Thank you

- Thread starter newgate
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- #1

- 13

- 0

How can I evaluate the following integral please?

##\int_{-\infty}^{+\infty}\frac{e^{i A x}}{\sqrt{x^2+a^2}}dx##

Thank you

- #2

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- 2,381

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

How can I evaluate the following integral please?

##\int_{-\infty}^{+\infty}\frac{e^{i A x}}{\sqrt{x^2+a^2}}dx##

Thank you

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.

- #3

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Thank you Dr.CourtneyUse 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.

- #4

mathman

Science Advisor

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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.Thank you Dr.Courtney

- #5

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How? I'm curious :DI 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.

- #6

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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.##How? I'm curious :D

- #7

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Ok thank you very much.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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