Consider the following integral: int sin(x^n)/x^n dx
In [8]: x = Symbol("x")
n = Symbol("n")
In [9]: integrate(sin(x**n)/x**n, x, meijerg=True)
Out[9]: x*gamma(1/(2*n))*hyper((1/(2*n),), (3/2, 1 + 1/(2*n)),
-x**(2*n)/4)/(2*n*gamma(1 + 1/(2*n)))
Btw, w/o meijerg=True this hangs.
In [10]: combsimp(_)
Out[10]: x*hyper((1/(2*n),), (3/2, 1 + 1/(2*n)), -x**(2*n)/4)
In case n=1 we get the well known Si(x) function:
In [12]: _.subs(n,1)
Out[12]: x*hyper((1/2,), (3/2, 3/2), -x**2/4)
In [14]: hyperexpand(_)
Out[14]: Si(x)
However, the general case n>1 has also a "nice" solution
in terms of incomplete Gamma functions.
We can find it for example here:
http://integrals.wolfram.com/index.jsp?expr=Sin[x^n]%2Fx^n&random=false
The problem is that we can not simplify the 1F2 returned
by our integrator. I tried to find a rule we could add
to the simplification routines but found none.
Similar issues arise for cis, sinh and cosh.
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