Can the Integration of 1/sin(x)e^(csc^2(x)) be Solved?

[custer]

I'm told that this integration can't be solved, can it ?

ʃ dx/(sin (x) . exp((csc x)^2)

 

[HallsofIvy]

Since "almost all" elementary expressions cannot be integrated in terms of elementary expressions, the question really is, "do you have any reason to think it can be?"

 

[squidsoft]

I'm told that this integration can't be solved, can it ?

ʃ dx/(sin (x) . exp((csc x)^2)

Make it so. Suppose that was all you were told. What would you do? Numerically counts as "solved" in my book. How about a power series? Can I use that (in principle) and integrate over it's radius of convergence? Ain't that a "solution"? Find a way . . . try. That's what math is about. :)

 

[jbunniii]

Make it so. Suppose that was all you were told. What would you do? Numerically counts as "solved" in my book. How about a power series? Can I use that (in principle) and integrate over it's radius of convergence? Ain't that a "solution"? Find a way . . . try. That's what math is about. :)

And if all else fails, and it's a useful enough integral, then give it a name, e.g.

"For all real x, we define

cus(x) = \int_{-\infty}^x \frac{dx}{\sin(x) \exp((\csc x)^2)}"

Presto, solved! Now all you need to do is lobby for a calculator button. Oh, and maybe establish some facts about the function. If people like it enough, they might assign your name to it. It worked for Bessel.

 

[squidsoft]

If the integrand is:

f(x)=\frac{1}{\sin(x)e^{\sec^2(x)}}

then I'd suggest we define:

\text{cus}(x)=\int_0^x \frac{1}{\sin(x)e^{\sec^2(x)}}

and I'd like to establish the first property of this \text{cus}(x) (why? just be-cus):

\text{cus}(-x)=-\text{cus}(x)

Also, just to get some empirical data about \text{cus} (why, just . . . ok, I'll stop it), I calculated the first 25 terms of the Taylor series about the point x=\pi/2 and then calculated the antiderivative of those terms, substituted the limits \pi/4 and 3\pi/4 then compared it to a direct numerical integration of the integrand. The results were accurate to 6 places:

f[x_] := 1/(Sin[x]*Exp[Csc[x]^2]); 
tay[x_] := Normal[Series[f[x], 
    {x, Pi/2, 25}]]
cus[x_] = Integrate[tay[x], x]
N[cus[3*(Pi/4)] - cus[Pi/4]]
NIntegrate[f[x], {x, Pi/4, 3*(Pi/4)}]