Lately, I'm trying to prove that $\lim \limits_{n\to\infty} \int_0^\infty \sin(xt)e^{-t}\sum_{k=0}^n \frac{t^k}{k!}dt =\frac{1}{x}$ (for $x\neq$ zeros of $\sin(x)$) by using mellin transform of $f(t)=\sin(xt)e^{-t}$, to be specific, $\int_0^\infty t^{s-1}\sin(xt)e^{-t}dt = \frac{\Gamma(s)\sin(s\tan^{-1}(x))}{(x^{2}+1)^{\frac{s}{2}}}$. After dividing $\Gamma(s)$, summing up on both sides and changing the variable, I got the following equation( needed to be proven): $\sum_{k=1}^\infty \frac{\sin(k\tan^{-1}(x))}{(x^2+1)^{\frac{k}{2}}}=\frac{1}{x}$. I noticed that $\tan^{-1}(x)=\cos^{-1}({(x^{2}+1)^{-\frac{1}{2}}})$, so after substitution, this is the equivalent equation: $\sum_{k=0}^\infty \sin((k+1)x)\cos(x)^k = \csc(x)$. I have worked a lot and I found some other equations (with the assumption that the equation above is true):
$\sum_{k=0}^\infty \cos((k+2)x)\cos(x)^{k}= -1$, $\sum_{k=0}^\infty \cos(kx)\cos(x)^{k}= 1$
By using those equations, with the formula $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$, I could form a geometric series which visually proves that my assumption is true. But the problem is I actually guessed the answer first, means that my whole work is just a circular reasoning without a proof.
I'm wondering if this identity was discovered and proven before because I can't find them in public discussions. By the way, could you help me prove this?
Any helps will be respected. Thanks in advance.
