Machine Precision is not enough precision for large K (probably due to summation):
p1 = Plot[f[7, 0, K], {K, 0, 1000}, WorkingPrecision -> 32, PlotStyle -> Red];
p2 = Plot[f[7, 0, K], {K, 0, 1000}];
Show[{p1, p2, p1}]
You see that for K > 250 the error grows considerably.
So for small K, the results are roughly the same, but the larger the K, the larger the error.
Of course you can do series expansions around infinity:
Series[function,{x,Infinity,order}]
But getting any kind of precision might be tricky, you might need a lot of terms!
Nice explanation!
I have used N[] in Manipulate[] to decrease the time for graph manipulation!
Hi,
Series[function,{x,Infinity,order}]
Have you tried this for the present function before writing? This does not seem to work. There are
some messages about singularities or indeterminate results. But (although I cannot yet prove it), I have
reasons to believe that there should exist an asymptotic expansion, probably as a series in powers
of 1/Sqrt[K].
Leslaw
No but when I do, I get a result:
out=Integrate[BesselJ[2n+1/2,L]BesselJ[2*m+1/2,L]/Sqrt[L^2+K],{L,0,\[Infinity]},Assumptions->n\[Element]Integers&&n>=0&&m\[Element]Integers&&m>=0&&K\[Element]Reals&&K>0]
Series[out,{K,\[Infinity],4}]
outputs a very large expression. Note that you can not substitute values directly into the output, first you have to apply Normal to it. See the detailed documentation of Series.