Taylor series of $\sin\ln x$ at $x = 1$ I tried to discover the Taylor series for and get the answer from Wolfram Alpha, but I couldn't prove the expression provided.
The expression is:

$$\sin\ln x=\sum_{n=0}^{\infty}\frac{1}{2}i(x-1)^n\Bigg(\binom{-i}{n}-\binom{i}{n}\Bigg)$$

How can I prove this relation?
 A: A solution based on differential equations seems made to order.
Render $y=\sin(\ln x)$.  Differentiate using the chain and quotient rules:
$\dfrac{dy}{dx}=\dfrac{\cos(\ln x)}{x}$
$\dfrac{d^2y}{dx^2}=\dfrac{[x(-\sin(\ln x)/x)]-[\cos(\ln x)]}{x^2}=-\dfrac{\sin(\ln x)+\cos(\ln x)}{x^2}$
Now take the linear combination $x(\dfrac{d^2y}{dx^2})+\dfrac{dy}{dx}$ which eliminates the cosine term, and put in $y=\sin(\ln x)$ to get rid of the sine function.  After clearing fractions and collecting terms you have the Cauchy-Euler equation
$x^2\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}+y=0$
whose methods of solution are described in the link.  You basically assume solutions having the form $x^\lambda$ and derive an algebraic equation for $\lambda$ that fits the differential equation, which in this case turns out to be $\lambda^2+1=0$.  Thus we get a general solution
$y=Ax^i+Bx^{-i}$
To render $A$ and $B$ that fit the specific function $y=\sin(\ln x)$, use the function value, and the first derivative given above, evaluated at $x=1$ to render
$A+B=0, A-B=-i$
Thereby $A=-i/2,B=i/2$ and then
$\sin(\ln x)=(-i/2)x^i+(i/2)x^{-i}$
Now render $x^i=(1+y)^i$ with $y=x-1$ and expand with the Binomial Theorem.  Do the same with $x^{-i}$ and combine the results to get the claimed expansion.
A: The function $f(x)=\sin\log x$ is $C^{\infty}$ on $(0,\infty)$ and analytic in a neighborhood of $1$, as the composition of two analytic functions. We will compute $f^{(n)}(x)$ to derive the Taylor series at $x=1$.
The derivative of $a_n\dfrac{\cos\log x}{x^n}+b_n\dfrac{\sin\log x}{x^n}$ is
$$-na_n\dfrac{\cos\log x}{x^{n+1}}-a_n\dfrac{\sin\log x}{x^{n+1}}-nb_n\dfrac{\sin\log x}{x^{n+1}}+b_n\dfrac{\cos\log x}{x^{n+1}}\\=(-na_n+b_n)\dfrac{\cos\log x}{x^{n+1}}+(-a_n-nb_n)\dfrac{\sin\log x}{x^{n+1}}$$
And it's easy to verify this formula is still valid for $n=0$. So, with $a_0=0, b_0=1$, the $n$th derivative of $f(x)=\sin\log x$ (the $0$th derivative is simply $f$) is:
$$f^{(n)}(x)=a_n\dfrac{\cos\log x}{x^n}+b_n\dfrac{\sin\log x}{x^n}$$
With
$$\left\{\begin{eqnarray}
a_{n+1}&=-na_n+b_n \\
b_{n+1}&=-a_n-nb_n
\end{eqnarray}\right.$$
We are only interested in $a_n$ since the $b_n$ part will contribute $0$ to $f^{(n)}(1)$. Rewrite $b_n=a_{n+1}+na_n$ from the first relation, then the second can be written:
$$a_{n+2}+(n+1)a_{n+1}=-a_n-n(a_{n+1}+na_n)$$
That is
$$a_{n+2}+(2n+1)a_{n+1}+(n^2+1)a_n=0$$
With $a_0=0$ and $a_1=1$.
Then the Taylor series at $x=1$ is:
$$f(x)=\sum_{n=0}^\infty a_n\frac{(x-1)^n}{n!}$$
We need two solutions of the recurrence equation. Let $a_n=(s)_n=s(s-1)\cdots(s-n+1)$ for some $s$. Then
$$a_{n+2}+(2n+1)a_{n+1}+(n^2+1)a_n=(s)_n(s-n)(s-n-1)+(s)_n(s-n)(2n+1)+(n^2+1)(s)_n\\
=(s)_n\left[(s-n)(s-n-1)+(s-n)(2n+1)+(n^2+1)\right]$$
We want
$$(s-n)(s-n-1)+(s-n)(2n+1)+(n^2+1)=0$$
$$s^2+(-2n-1+2n+1)s+[n(n+1)-n(2n+1)+n^2+1]=0$$
$$s^2+1=0$$
With $s=\pm i$, we get two linearly independent solutions, hence the general solution of the recurrence equation is:
$$a_n=\alpha (i)_n+\beta (-i)_n$$
Now use the initial conditions:
$$\left\{\begin{eqnarray}
a_0&=&0=\alpha+\beta \\
a_1&=&1=i(\alpha-\beta)\end{eqnarray}\right.$$
Hence $\alpha=-i/2$ and $\beta=i/2$.
Finally:
$$\sin\log x=\sum_{n=0}^\infty \frac{i}{2}\frac{(x-1)^n}{n!}\left[(-i)_n-(i)_n\right]$$
$$\sin\log x=\sum_{n=0}^\infty \frac{i}{2}(x-1)^n\left[\binom{-i}{n}-\binom{i}{n}\right]$$
Now to complete the proof, we still need the radius of convergence. It's easy to see that $\binom{\pm i}{n}$ is bounded, so the radius of convergence must be at least $1$. It can't be larger because the function has a singularity at $x=0$.
Write:
$$\binom{i}{n}=\frac{i}{1}\cdot\frac{i-1}{2}\cdots\frac{i-n+1}{n}$$
And for $k\ge1$,
$$\left|\frac{i-k+1}{k}\right|^2=\frac{1+(k-1)^2}{k^2}=1-2\frac{k-1}{k^2}\le1$$
Hence
$$\left|\binom{i}{n}\right|\le1$$
Likewise,
$$\left|\binom{-i}{n}\right|\le1$$
