Evaluate $\int_{\pi}^{\infty}\left(x^{2}-\sin\left(x\right)-1\right)^{-1}dx=?$ It's an integral which seems simple but I confess I cannot evaluate this :
$$\int_{\pi}^{\infty}\left(x^{2}-\sin\left(x\right)-1\right)^{-1}dx=?$$
I can evaluate another integral where I start from :
$$\int_{\pi}^{\infty}\left(x^{2}-x-1\right)^{-1}dx=\frac{2\coth^{-1}\left(\frac{2\pi-1}{\sqrt{5}}\right)}{\sqrt{5}}$$
Show the convergence is not hard using bound for $\sin(x)$.
Question :
Can we hope to find a closed form ?
Thanks .

Here is an attempt at an antiderivative. You can consider it a comment.
Attempt 1:
Here is a series expansion for the antiderivative using geometric series which includes the $[\pi,\infty)$ interval of convergence:
$$\int \frac{dx}{x^2-\sin(x)-1}=-\int \sum_{n=0}^\infty  \left(i\frac{e^{-ix}-e^{ix}}{2}\right)^n(x^2-1)^{-n-1}dx$$
Which cannot be integrated in closed form. Let’s also use a binomial theorem expansions which have an infinite radius of convergence since they are truncated.
$$-\int \sum_{n=0}^\infty  \left(i\frac{e^{-ix}-e^{ix}}{2}\right)^n(x^2-1)^{-n-1}dx = -\int \sum_{n=0}^\infty i^n2^{-n} \left(e^{-ix}-e^{ix}\right)^n\sum_{k_1=0}^n\frac{n!}{(n-k_1)!k_1!}e^{-ix(n-k_1)}e^{ixk_1}\sum_{k_2=0}^n(x^2-1)^{-n-1}dx $$
Then use a Binomial Series which would constrict the series expansion.
Please let me know if there is a simpler series expansion.
Attempt 2:
$$\int \frac{dx}{x^2-\sin(x)-1} =\sum_{n=1}^\infty \frac{\frac{d^{n-1}}{dx^{n-1}}\frac1{x^2-\sin(x)-1}\big|_{x=a}}{n!}(x-a)^n$$
With the nth derivative and Gauss Hypergeometric function for a convergence interval:
$$\int \frac{dx}{x^2-\sin(x)-1}  =\sum_{n=1}^\infty\left((-2)^{n-1} a^{n-1} (n-1)!(a^2-\sin(a)-1)^{-(n-1)-1}\,_2\text F_1\left(\frac{1-(n-1)}2,-\frac {n-1}2;-(n-1);1-\frac{\sin(a)+1}{a^2}\right)\right) \frac{(x-a)^n}{n!}= \sum_{n=0}^\infty\left((-2)^n a^n n!(a^2-\sin(a)-1)^{-n-1}\,_2\text F_1\left(\frac{1-n}2,-\frac n2;-n;1-\frac{\sin(a)+1}{a^2}\right)\right) \frac{(x-a)^{n+1}}{(n+1)!}= \sum_{n=0}^\infty\,_2\text F_1\left(\frac{1-n}2,-\frac n2;-n;1-\frac{\sin(a)+1}{a^2}\right) \frac{(-2a)^n(x-a)^{n+1}}{(n+1) (a^2-\sin(a)-1)^{n+1}} $$
Even with the $-n$ in the hypergeometric function, the sum terms exist.
It can be shown that:
$$_2\text F_1\left(\frac{1-n}2,-\frac n2;-n;1-\frac{\sin(a)+1}{a^2}\right) =2^{-n-1 } \left(\sqrt{\frac{\sin(x) + 1}{x^2}} + 1\right)^n + 2^{-n-1} \frac{\left(\sqrt{\frac{\sin(x) + 1}{x^2}} + 1\right)^n}{\sqrt{\frac{\sin(x) + 1}{x^2}}}= 2^{-n-1 } \left(\sqrt{\frac{\sin(x) + 1}{x^2}} + 1\right)^n\left(1+ \frac1{\sqrt{\frac{\sin(x) + 1}{x^2}}}\right)$$
Therefore:
$$\int \frac{dx}{x^2-\sin(x)-1} =\sum_{n=0}^\infty 2^{-n-1 } \left(\sqrt{\frac{\sin(a) + 1}{a^2}} + 1\right)^n\left(1+ \frac1{\sqrt{\frac{\sin(a) + 1}{a^2}}}\right) \frac{(-2a)^n(x-a)^{n+1}}{(n+1) (a^2-\sin(a)-1)^{n+1}} =C+ \frac12 \left( \frac1{\sqrt{\frac{\sin(a) + 1}{a^2}}}+1\right)\sum_{n=0}^\infty \left(\sqrt{\frac{\sin(a) + 1}{a^2}} + 1\right)^n\frac{(-a)^n(x-a)^{n+1}}{(n+1) (a^2-\sin(a)-1)^{n+1}}  $$
This result is based on this result and this computation. Please correct me and give me feedback!
