Area under period of $\mathrm{a^{csc(x)}}$. I am trying to find this graph’s area which was inspired by similar integrals of trigonometric function to the power of trigonometric functions on Math Stack Exchange. I have found the solution with some machine help and this is what cropped up:
$$\mathrm{C(a)=\int_{-\pi}^0 a^{csc(x)}dx=\int_{-\pi}^0 e^{csc(x)ln(a)}dx=2\sum_{n=0}^\infty\frac{ln^n(a)}{n!}\int_{-\frac\pi2}^0csc^n(x)dx}$$
Although the expression can be integrated like this, the substitution of the bounds fails as for even $n\ge1$, the integral diverges to $\infty$ as an improper integral.
Here is an example area representation of C(e):

Here is a plot of C(a) with a limit defined maximum at (1,$\pi$) as the integrand for the a=1 case turns, from a limit point of view, into a box like shape with height 1 and length $\pi$. The end behavior is ($\infty$,0) :
I am most interested in the cases of C(2) and C(e). I wish to have an integral with an exact solution, which means write C(a) without approximations nor an integral. This looks like it could use Bessel functions or  Struve functions as their main integral representation is similar to C(a). Here is why: $$\int_0^1 e^{\cos(\pi x)}dx=1.26606…=I_0(1), \int_0^1 e^{\sin(\pi x)}dx=\text L_0(1)+I_0(1)=1.976309… $$ Integral of exp(tan(x)) here. If I could figure out the integral of exp(csc(x)), then I can evaluate this improper integral. Please correct me and give me feedback!
 A: For $\lambda>0$ we have $${\rm C}({\rm e}^\lambda)=\int_{-\pi}^0{\rm e}^{\lambda\csc x}\,{\rm d}x\underset{t=-\csc x}{\phantom{\big[}=\phantom{\big]}}2\int_1^\infty\frac{{\rm e}^{-\lambda t}\,{\rm d}t}{t\sqrt{t^2-1}}=2f(\lambda)$$ and $f'(\lambda)=-K_0(\lambda)$ from an integral representation of the modified Bessel function $K_0$.
Hence, using this formula when doing $\int_\lambda^\infty$, one obtains $${\rm C}({\rm e}^\lambda)=\pi-\pi\lambda\big(K_0(\lambda)\mathbf{L}_{-1}(\lambda)+K_1(\lambda)\mathbf{L}_0(\lambda)\big).$$
Here is a PARI/GP session for numerical verification (with the default precision):
gp > struvel(a,x)=(x/2)^(a+1)*suminf(n=0,(x/2)^(2*n)/gamma(n+3/2)/gamma(n+a+3/2));
gp > foo(a)=Pi*(1-a*(besselk(0,a)*struvel(-1,a)+besselk(1,a)*struvel(0,a)));
gp > goo(a)=foo(a)-2*intnum(x=[1,-1/2],[+oo,a],exp(-a*x)/x/sqrt(x^2-1));
gp > goo(1)
%4 = 1.4987552972984165727 E-37
gp > goo(log(2))
%5 = 1.6750794499217596989 E-37
gp > foo(log(2))
%6 = 0.98349909539743160556395206700695146505

A: I have just come across the Sievert Integral:
$$\text S(x,\theta)\mathop=^\text{def}\int_0^\theta e^{-x\sec(t)}dt$$
Since $x$ is a constant variable, we can make $x\to\ln(a)$:
$$\text S(\ln(a),\theta)= \int_0^\theta e^{-\ln(a)\sec(t)}dt=\int_0^\theta a^{-\sec(t)} dt$$
Notice that $$\sec\left(\frac\pi 2-t\right)=\csc(t)\implies t=\frac \pi 2-y\implies dy=-dt,y_1=\frac\pi 2-0=\frac\pi 2,y_2=\frac\pi 2-\theta$$
Therefore:
$$\text S(\ln(a),\theta)= \int_0^\theta a^{-\sec(t)}dt=\int_{\frac\pi 2-\theta}^{\frac\pi 2} \left(\frac1a\right)^{\csc(y)} dy$$
The $\frac1a$ will just shift the function vertically. Our made up $\text C(a)$ function can then be expressed as:
$$\text C\left(\frac1a\right)=2\text S\left(\ln(a),\frac\pi 2\right)=2\int_0^\frac\pi 2 \left(\frac1a\right)^{\csc(y)} dy =2\int_0^\frac\pi 2 \left(\frac1a\right)^{\sec(y)}dy $$
Therefore:
$$\text C(a)=\text S\left(-\ln(a),\frac\pi2\right)$$
Please correct me and give me feedback!
