Solving Integral Identity: Gradstein & Ryzhik

In summary, the conversation is about a person trying to understand an identity from a paper on dimensional regularization. The identity involves a k-integral and after performing it, it becomes a derivative of a function with respect to s. The person is unsure about the validity of the expression and is asking for clarification. They also mention a paper without references and ask if it is possibly from a specific source. A link to the paper in question is provided.
  • #1
romeo6
54
0
Hey folks!

I'm trying to figure out an identity from a paper on dimensional regularization.

Here's the identity:

[tex]-\frac{1}{2}\frac{d}{ds}|_{s=0}\int_0^\infty \frac{d^4k}{(2\pi)^4}(k^2+m^2)^{-s}[/tex]

after performing the k-integral this becomes:

[tex]=-\frac{1}{32\pi^2}\frac{d}{ds}|_{s=-2}\frac{1}{s(s+1)}m^{-2s}[/tex]

I found this in a paper with no references. Is this perhaps something out of Gradstein and Ryzhik?
 
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  • #2
There's a problem with the expression you've written. s appears on the left hand side, but on the right hand side it looks like after taking the derivative you're setting all the s's to 2. If s appears as a variable/parameter on the left hand side it has to appear on the right hand side, so presumably at least one of the s's on the right hand side is the same as on the left hand side. (or the s on the left hand side should be a 2). If you could clarify this it would be of some help.
 
  • #3
Hi Mute - I think I was modifying the expression while you were kindly looking at it. This one's good.
 
  • #4
By "k-integral" do you mean this:
[tex]-\frac{1}{2}\frac{d}{ds}|_{s=0}\int_0^\infty \frac{d^4k}{(2\pi)^4}(k^2+m^2)^{-s} dk[/tex]
?
I suppose this makes more sense:
[tex]-\frac{1}{2}\frac{d}{ds}|_{s=0}\int_0^\infty \frac{d^4k}{(2\pi)^4}(k^2+m^2)^{-s} ds[/tex]
 
  • #6
...anyone?
 

FAQ: Solving Integral Identity: Gradstein & Ryzhik

What is Gradstein & Ryzhik's Integral Identity?

Gradstein & Ryzhik's Integral Identity is a mathematical formula used in solving integrals. It is also known as the "Gradstein & Ryzhik formula" or the "Euler-Maclaurin formula".

How is the Integral Identity used in solving integrals?

The Integral Identity is used to transform complicated integrals into simpler ones that can be easily solved. It involves finding a series expansion of the integrand and using the Integral Identity to simplify the integral into a sum of terms that can be easily integrated.

What are the benefits of using the Integral Identity?

The Integral Identity is a powerful tool in solving integrals, as it can simplify complicated integrals and make them more manageable. It also allows for the evaluation of integrals that may not have a closed-form solution.

Are there any limitations to using the Integral Identity?

One limitation of using the Integral Identity is that it may not work for all types of integrals. It is most effective for integrals with known series expansions, and may not be applicable to integrals with complicated or unknown functions.

Who were Gradstein & Ryzhik and why is their Integral Identity important?

Gradstein & Ryzhik were two Russian mathematicians who published a comprehensive handbook of integrals, series, and products in 1923. Their Integral Identity has become an essential tool in solving integrals and is widely used by mathematicians, physicists, and engineers.

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