Putting it through twice

Relevant wiki: Taylor Series - Problem Solving

$\begin{aligned} f(x) & = \sum_{n=\color{#D61F06}{0}}^\infty \frac {\sin(nx)}{7^n} \quad \quad \small \color{#D61F06}{\text{Since }\sin 0 = 0} \\ & = \sum_{n=\color{#D61F06}{1}}^\infty \frac {\sin(nx)}{7^n} = \sum_{n=1}^\infty \frac {\Im \left( e^{inx}\right)}{7^n} = \Im \left( \sum_{n=1}^\infty \left(\frac {e^{ix}}7\right)^n \right) = \Im \left( \frac {\frac{e^{ix}}7}{1-\frac{e^{ix}}7} \right) \\ & = \Im \left( \frac {e^{ix}}{7-e^{ix}} \right) = \Im \left( \frac {\cos x + i \sin x}{7-\cos x - i \sin x} \right) \\ & = \Im \left( \frac {(\cos x + i \sin x)(7-\cos x + i \sin x)}{(7-\cos x - i \sin x)(7-\cos x + i \sin x)} \right) \\ & = \Im \left( \frac {7\cos x - \cos^2 x + i \sin x \cos x + 7i\sin x - i \sin x \cos x - \sin^2 x}{49-14\cos x + \cos^2 x + \sin^2 x} \right) \\ & = \Im \left( \frac {7\cos x - 1 + 7i\sin x }{50-14\cos x} \right) \end{aligned}$

$\begin{aligned} \implies f(x) & = \frac {7\sin x}{50-14\cos x} \end{aligned}$

Now, we have:

$\begin{aligned} a & = \int_0^\color{#3D99F6}{\pi} \frac {7\sin x}{50-14\cos x} dx & \small \color{#3D99F6}{P/2 = \pi} \\ & = \int_0^\pi \frac 12 \cdot \frac {14\sin x}{50-14\cos x} dx \\ & = \int_0^\pi \frac 12 \cdot \frac d{dx} \ln \left(50-14\cos x \right) dx \\ & = \frac 12 \left(\ln 64 - \ln 36 \right) \\ \implies a & = \ln \frac 43 \end{aligned}$

And finally,

$\begin{aligned} f \left(\pi e^{|a|} \right) & = f \left(\pi e^{\ln \frac 43} \right) = f \left(\frac 43 \pi \right) \\ & = \frac {7\sin \frac 43 \pi }{50-14\cos \frac 43 \pi} = \frac {-7\sqrt 3}{100+14} = - \frac {7\sqrt 3}{114} \end{aligned}$

$\implies \alpha + \beta + \gamma = 7+3+114 = \boxed{124}$

Relevant wiki: Taylor Series - Problem Solving

Since $\frac{\sin(x \cdot 0)}{7^0} = 0$ for all $x$ , let us first redefine $f(x) = \sum_{n=1}^\infty \frac{\sin(x \cdot n)}{c^n}$ , where $c$ is a constant that we will eventually replace with $7$ . Now we first find the closed form of $f(x)$ .

$\displaystyle \sum_{n=1}^\infty \frac{\sin(x \cdot n)}{c^n}$

$\displaystyle = \text{Im}\left\{\sum_{n=1}^\infty \frac{e^{ixn}}{c^n}\right\}$

$\displaystyle = \text{Im}\left\{\sum_{n=1}^\infty \left(\frac{e^{ix}}{c}\right)^n \right\}$

$\displaystyle = \text{Im}\left\{\frac{e^{ix}/c}{1-e^{ix}/c}\right\}$

$\displaystyle = \text{Im}\left\{\frac{\cos x + i\sin x}{c-\cos x - i\sin x}\right\}$

$\displaystyle = \text{Im}\left\{\frac{\cos x}{c-\cos x - i\sin x}\cdot \frac{c-\cos x + i\sin x}{c-\cos x + i\sin x} + \frac{i\sin x}{c-\cos x - i\sin x} \cdot \frac{c-\cos x + i\sin x}{c-\cos x + i\sin x}\right\}$

$\displaystyle = \text{Im}\left\{\frac{c\cos x - \cos^2 x + i\sin x\cos x - i\sin x\cos x - \sin^2 x + ia\sin x}{c^2-2c\cos x +\cos^2 x + \sin^2 x}\right\}$

$\displaystyle = \text{Im}\left\{\frac{c\cos x - 1 + ic\sin x}{c^2-2c\cos x + 1}\right\}$

$\displaystyle = \frac{c\sin x}{c^2-2c\cos x + 1}$

The area $a$ between $f(x)$ and the $x$ -axis over a half-period of $f(x)$ is $\displaystyle\int_0^\pi \frac{c\sin x}{c^2-2c\cos x + 1} \, \mathrm{d}x$ .

We perform the $u$ -substitution $u = -c\cos x$ , $\mathrm{d}u = c\sin x$ , giving

$\displaystyle \int_{-c}^{c} \frac{\mathrm{d}u}{c^2 + 2u + 1} = \frac{1}{2}\left( \ln\frac{c^2+2c+1}{2} - \ln\frac{c^2-2c+1}{2}\right) = \ln\frac{c+1}{c-1}$ .

Substituting in $c=7$ results in $a = \ln \frac{8}{6} = \ln \frac{4}{3}.$ Finally, the argument we pass into $f(x)$ is $\pi e^{\left|\ln 4/3\right|} = \frac{4\pi}{3}$ .

$\displaystyle f\left(\frac{4\pi}{3}\right) = \frac{-\sqrt{3}/2 \cdot 7}{7^2 + 2\cdot 7/2 + 1} = \frac{-7\sqrt{3}}{114}$ .