Apply the product rule for differentiation to the solution. This gives you the integrand. If you recognize this then you just need to do the reverse to get the solution.
You can expand the exponential term on the rhs as a product of two series:
\exp\left(\frac{\sqrt{x^2-1}}{2}\left(t-\frac{1}{t}\right)\right)=\sum_{n=0}^{\infty}\frac{(\sqrt{x^2-1})^n}{2^nn!}t^n\sum_{k=0}^{\infty}\frac{(\sqrt{x^2-1})^k}{2^kk!}(-t)^{-k}
and so it becomes...
This website is an attempt to compile proofs of all integrals listed in the Gradshteyn and Ryzhik integral tables:
http://129.81.170.14/~vhm/Table.html" [Broken]
There are still lots of gaps to be filled in though!
you can use the identity
\sin N\theta = \sum_{k=0}^N \binom{N}{k} \cos^k \theta\,\sin^{N-k} \theta\,\sin\left(\frac{1}{2}(N-k)\pi\right)
to obtain a polynomial in \sin \theta which can then be solved for \theta . However, you will probably need to solve it numerically for N>3.
Are you sure that the only nonzero point is (0,\pi)? Depending on the values of a and b (e.g. if one of them is negative) the function g(x,y)=a(cos(x)-1)+b(cos(y)+1) could have a number of roots between the limits of integration.
Given the following identity for a single variable Dirac Delta...
Does anyone know how to prove the following identity:
\Sigma_{k=0}^{n}\left(\stackrel{n}{k}\right) H_{k}(x)H_{n-k}(y)=2^{n/2}H_{n}(2^{-1/2}(x+y))
where H_{i}(z)represents the Hermite polynomial?
Yes, I think you can assume that the angle is dependent on only one of the angles in a spherical polar co-ordinate system by assuming that the vector \vec{a} lies along one of the axes in the system similar to the approach in the following integral...
Hi Jason,
That is very nice, thanks! Do you know of any methods/tricks that I could try to reduce it from a double to a single summation?
Thanks for your help.
Hi Orbb,
I think the formula you are looking for is
\frac{\pi^{\frac{d}{2}}}{\alpha}\frac{1}{\Gamma(\frac{d}{2})}
It is straightforward to get this result in spherical co-ordinates and I don’t think you can find any easier way to do it.
Assume
|\vec{y}| = r =...
Hi JasonRF,
Thanks for all your help, it looks like you put in a lot of work! I am going to take sometime to go through it and see how it looks - I am still clinging to the hope that a neater analytical solution might pop out! Thanks again.
Hello,
There is a book called "Integral representation and the computation of combinatorial sums" by G. P. Egorychev that might be useful. The general idea is to convert each term of the series to a contour integral and then using some theorems from several complex variables to manipulate the...