Solve this integration ?

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

ʃ dx/(sin (x) . exp((csc x)^2)

HallsofIvy
Homework Helper
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?"

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
Homework Helper
Gold Member
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.

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:

Code:
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)}]