# Mathematica and special functions

How is it possible that Mathematica doesn't recognize it's own definitions of special functions ? I tried as input:

Integrate[Cosh[I2t]/E^(6 Cosh[t]), {t, 0, Infinity}]


The above should give $K_{i2}(6)$ according to this. Unfortunately, it just happily reproduces my input, and leaves me ignorant of how it should have been phrased.

Does someone know how I could get Mathematica recognize to evaluate such input to something useful?

Moreover, when I plot

Plot[BesselK[2 I, z], {z, 0, 7}]


I get a graph, but when I evaluate

BesselK[2 I, 6]


again, Mathematica repeats my input.

I'm not really used to Mathematica, so sometimes I get this sort of frustrating output, Mathematica not understanding what I'm trying to say.

• 1. I2t is treated as a single variable; put spaces in between variables that you intend to multiply. 2. BesselK[2 I, 6] is an exact number without an apparently simpler closed form; to see an approximation, use N[]: N[BesselK[2 I, 6]]. – J. M.'s ennui Jun 3 '15 at 0:17
• What makes you think this transformation rule is in Mathematica? While related, the Wolfram Functions site is not a source for Mathematica transformation rules per se... – ciao Jun 3 '15 at 0:35
• @Guesswhoitis. You'll become the first portable computerless computer language interpreter – Dr. belisarius Jun 3 '15 at 2:03
• Not quite yet there, @bel. :) – J. M.'s ennui Jun 3 '15 at 2:14
• Closely related: Incorrect results for elementary integrals when using Integrate. Unfortunately the solution there (IntegrateInverseIntegrate) doesn't work in general. Also related: Why does Mathematica give the wrong answer when integrating?. – Jens Jun 3 '15 at 4:07

Mathematica does not recognize your integral is BesselK[2 I, 6] even though it is. All the identities shown on http://functions.wolfram.com are not necessarily incorporated into Mathematica. The numeric equivalence of BesselK[2 I, 6] and the integral can be demonstrated by
N[BesselK[2 I, 6], 16] ==

True
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