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 .
 A: 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!
A: This is a typical integral for a numerical solution and in the mathematical field of tables of integrals, it is famous and well known.
Elsewhere on the internet, this is a calculation of seconds.
Some approximation that is nicely:
$\int_{\pi}^{\infty} (\frac{1}{x^2-sin(x)-1}-\frac{1}{x^2-1})dx$
This second term can be calculated as Your approximation.
The value is
$ -0.004998176$.
$\int_{\pi}^{\infty} \frac{dx}{x^2-sin(x)-1}\approx0.3247671$
There is no solution in terms of standard mathematical functions.
It is astonishing that the difference to the function without $sin(x)$ is so big. This is probably due to the large extent of the interval over which is integrated.
So my quick approximation reduction is much closer.
That can be seen from really examining the denominator alone. For very large values of $x$. $x$ dominates very much over $sin(x)$ and that is important for the integral approximation on the given function.
The above difference can be rewritten as
$\frac{sin(x)}{(x^2-1)(x^2-sin(x)-1)}$
This again shows how to approximate this recursively. So this is not standard as expected but a convergent approximation that gets really close to the value. It makes meaningful use of the value $\pi$. But this integral is infinite and even more difficult to calculate other than numerical.
On the other hand an approximation up to $\frac{1}{200}$ is close.
This gives an infinite series in the matter started. This should be the most natural representation of the integral despite it is unusual. There were times that this was developed and popular.
Mind the greatest zero is that of the first term. $\{-0.636733,1.40962\}$. So convergence is sure in $\{\pi,\infty \}$.
So in the next step
$\int_{\pi}^{\infty} (\frac{1}{x^2-sin(x)-1}-\frac{1}{x^2-1}-\frac{sin(x)}{(x^2-1)^2})dx$
and so on.
