Table of Integrals: Solving \int_0^{\infty}x^A\,(x^2+x)^{B/2}\,e^{-Cx}\,K_{(B)}

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SUMMARY

The integral under discussion is defined as \(\int_0^{\infty}x^A\,(x^2+x)^{B/2}\,e^{-Cx}\,K_{(B)}\left(2\,\alpha\,\sqrt{x^2+x}\right)\,dx\), where \(A\), \(B\), and \(C\) are positive integers, and \(K_{(B)}\) represents the B-th order modified Bessel function of the second kind. The forum participants suggest that this integral may be suitable for graduate-level homework help, indicating its complexity and relevance in advanced mathematics. The discussion emphasizes the need for resources such as a table of integrals to find equivalent forms of this expression.

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EngWiPy
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Hello,
During my derivation, I am faced with the following integral:

\int_0^{\infty}x^A\,(x^2+x)^{B/2}\,e^{-Cx}\,K_{(B)}\left(2\,\alpha\,\sqrt{x^2+x}\right)\,dx

where A, B, and C are positive integers, K_{(B)} is the B^{th} order modified bessel function of the second kind. I am intending to find an equivalent integral in the table of integrals. Can anyone help me, please?
 
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saeddawoud said:
Hello,
During my derivation, I am faced with the following integral:

\int_0^{\infty}x^A\,(x^2+x)^{B/2}\,e^{-Cx}\,K_{(B)}\left(2\,\alpha\,\sqrt{x^2+x}\right)\,dx

where A, B, and C are positive integers, K_{(B)} is the B^{th} order modified bessel function of the second kind. I am intending to find an equivalent integral in the table of integrals. Can anyone help me, please?

I think this forum is for general discussions rather than homework help, you should post this in homework help section,
here
https://www.physicsforums.com/forumdisplay.php?f=152"
 
Last edited by a moderator:
aaryan0077 said:
I think this forum is for general discussions rather than homework help

Graduate-level homework is acceptable here. This looks like it might fit -- I certainly didn't do hairy integrals with Bessel functions as an undegrad.
 
CRGreathouse said:
Graduate-level homework is acceptable here. This looks like it might fit -- I certainly didn't do hairy integrals with Bessel functions as an undegrad.

Okay, whatever you say.
 

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