Here's a way to get an exact symbolic expression for the integral.
Use the notation fi[m,n,a]
to denote Integrate[(Exp[-a x](x^2 - 1)^(m/2))/x^n, {x, 1, Infinity}]
.
Define integration by parts, where the boundary term vanishes.
fi[m_, n_, a_] /; m > 0 && n > 1 :=
(m fi[-2 + m, -2 + n, a] - a fi[m, -1 + n, a])/(-1 + n);
See how we're doing so far.
fi[13, 10, a] // Expand
(*
143/3 fi[3, 0, a] + 9295/384 a^2 fi[5, 0, a] -
3003/128 a fi[5, 1, a] + (3289 a^4 fi[7, 0, a])/2688 -
143/32 a^3 fi[7, 1, a] + (143 a^6 fi[9, 0, a])/10368 -
143/960 a^5 fi[9, 1, a] + (13 a^8 fi[11, 0, a])/362880 - (
13 a^7 fi[11, 1, a])/10080 - (a^9 fi[13, 1, a])/362880
*)
Define how the fi[m_, 0, a_]
cases evaluate.
fi[m_, 0, a_] /; m > 0 =
Assuming[Re[m] > 0 && Re[a] > 0, Integrate[Exp[-a x] (x^2 - 1)^(m/2), {x, 1, ∞}]]
(* (2^((1 + m)/2) a^(1/2 (-1 - m)) BesselK[(1 + m)/2, a] Gamma[1 + m/2])/Sqrt[π] *)
See how we are doing so far.
fi[13, 10, a] // Expand
(*
(143 BesselK[2, a])/a^2 + (46475 BesselK[3, a])/(128 a) +
16445/128 BesselK[4, a] + 5005/384 a BesselK[5, a] +
143/384 a^2 BesselK[6, a] - 3003/128 a fi[5, 1, a] -
143/32 a^3 fi[7, 1, a] - 143/960 a^5 fi[9, 1, a] - (
13 a^7 fi[11, 1, a])/10080 - (a^9 fi[13, 1, a])/362880
*)
Define how the fi[m_, 1, a_]
cases evaluate, using differentiation w.r.t. a
under the integral to get rid of the 1/x
factor, then integrating w.r.t. a
afterwards.
fi[m_, 1, a_] /; m > 0 =
Assuming[Re[m] > 0 && Re[a] > 0,
Integrate[-Exp[-a x] (x^2 - 1)^(m/2), {x, 1, ∞}] //
Integrate[#, a] & //
Simplify //
(# - (Limit[#, a -> ∞]//FullSimplify)&)]
(*
-(1/2) π Csc[(m π)/2] +
1/4 a^-m Sqrt[π]
Gamma[1 + m/2] (a^(1 + m) Sqrt[π]
HypergeometricPFQRegularized[{1/2}, {(3 + m)/2, 3/2}, a^2/4]
- 2^(1 + m) Gamma[-(m/2)] HypergeometricPFQRegularized[{-(m/2)}, {1/2 - m/2, 1 - m/2}, a^2/4]) Sec[(m π)/2]
*)
The last step fixes the constant of integration to ensure that the result goes to 0 (as it should) as a
goes to infinity.
However, if m
is an odd integer then the Sec[(m π)/2]
factor blows up, and the factor containing the difference between the two HypergeometricPFQRegularized
terms goes to zero, but these factors have a well-behaved product. So we need to manually intervene to make the expression manifestly finite when m
is an odd integer.
It is sufficient to do a series expansion about odd integer values of m
to extract the relevant coefficients. Here is how I did it.
factor1 = Assuming[m0 ∈ Integers && m0 >= 0,
SeriesCoefficient[
a^(1 + m) Sqrt[π]
HypergeometricPFQRegularized[{1/2}, {(3 + m)/2, 3/2}, a^2/4] -
2^(1 + m) Gamma[-(m/2)] HypergeometricPFQRegularized[{-(m/2)}, {1/2 - m/2,
1 - m/2}, a^2/4], {m, 2 m0 + 1, 1}] // FullSimplify];
factor2 = Assuming[m0 ∈ Integers && m0 >= 0,
SeriesCoefficient[Sec[(m π)/2], {m, 2 m0 + 1, -1}]];
Gather the pieces together, and rewrite the definition for the fi[m_, 1, a_]
cases.
fi[m_, 1, a_] /; m > 0 =
-(1/2) π Csc[(m π)/2] + 1/4 a^-m Sqrt[π] Gamma[1 + m/2] (factor1 factor2 /. m0 -> (m - 1)/2)
(*
(-(1/2))*Pi*Csc[(m*Pi)/2] - (Gamma[1 + m/2]*Sec[(1/2)*(-1 + m)*Pi]*
(a^(1 + m)*(2*
HypergeometricPFQ[{1/2}, {3/2, 3/2 + m/2},
a^2/4]*(-Log[4] + 2*Log[a] +
PolyGamma[0, -(1/2) + (1 - m)/2]) +
Sqrt[Pi]*Gamma[2 + (1/2)*(-1 + m)]*
Derivative[{0}, {1, 0}, 0][
HypergeometricPFQRegularized][{1/2}, {2 + (1/2)*(-1 + m),
3/2}, a^2/4]) +
4^(1 + (1/2)*(-1 + m))*Gamma[-(1/2) + (1 - m)/2]*
Gamma[2 + (1/2)*(-1 + m)]*
(Derivative[{0}, {0, 1}, 0][
HypergeometricPFQRegularized][{-(1/2) + (1 - m)/
2}, {(1 - m)/2, 1/2 + (1 - m)/2},
a^2/4] +
Derivative[{0}, {1, 0}, 0][
HypergeometricPFQRegularized][{-(1/2) + (1 - m)/2},
{(1 - m)/2, 1/2 + (1 - m)/2}, a^2/4] +
Derivative[{1}, {0, 0}, 0][HypergeometricPFQRegularized][
{-(1/2) + (1 - m)/2}, {(1 - m)/2, 1/2 + (1 - m)/2},
a^2/4])))/(a^m*(4*Sqrt[Pi]*Gamma[2 + (1/2)*(-1 + m)]))
*)
Numerically verify that this gives you the right answer.
With[{m = 13, n = 10, a = RandomReal[{10^-6, 3}]},
{NIntegrate[(Exp[-a x] (x^2 - 1)^(m/2))/x^n, {x, 1, ∞}], fi[m, n, a]}]
The required integral is then
fi[13, 10, a] // Expand
(* Lots of HypergeometricPFQ and HypergeometricPFQRegularized *)
The symbolic result looks incredibly messy to me, so maybe there are some further simplifications that could compress it down somewhat ...
Infinity
with a large number to use these schemes. $\endgroup$-1
and1
are the branch points, thus approaching1
you should know how to tackle this integral. As a reference examine this answer How to calculate contour integrals with Mathematica? $\endgroup$