Christmas Streak 42/88: A Weird Function

$\begin{aligned} f(x) & = \sum_{k=1}^\infty (-1)^{k-1}\left(\frac {x^{2k-1}\sin x}{(2k-1)!} - \frac {x^{2k}\cos x}{(2k)!}\right) \\ & = \sum_{k=1}^\infty \frac{x^k\color{#3D99F6}\cos \left(\frac {k\pi}2 - x \right)}{k!} & \small \color{#3D99F6} \text{By Euler's formula: }e^{i\theta} = \cos \theta + i\sin \theta \\ & = \sum_{k=1}^\infty \frac{x^k\color{#3D99F6}\Re \left(e^{i\left(\frac {k\pi}2 - x\right)}\right)}{k!} & \small \color{#3D99F6} \text{where } \Re(z) \text{ is the real part of }z. \\ & = \Re \left(\frac 1{e^{ix}}\sum_{k=1}^\infty \frac{\left(xe^{\frac {i\pi} 2}\right)^k}{k!} \right) \\ & = \Re \left(\frac 1{e^{ix}}\sum_{k=1}^\infty \frac{\left(ix\right)^k}{k!} \right) \\ & = \Re \left(\frac 1{e^{ix}} \left(e^{ix}-1\right) \right) & \small \color{#3D99F6} \text{Multiply up and down by }e^{-ix} \\ & = \Re \left(1-e^{-ix} \right) \\ & = \Re \left(1-\cos x + i\sin x \right) \\ & = 1-\cos x \end{aligned}$

Therefore,

$\begin{aligned} I & = \int_0^1 f(x)^2 dx = \int_0^1 (1-\cos x)^2 dx = \int_0^1 (1-2\cos x + \cos^2 x) \ dx \\ & = x - 2\sin x + \frac x2 + \frac {\sin 2x}4 \bigg|_0^1 = \frac 32 - 2\sin 1 + \frac {\sin 2}4 \approx 0.044382387 \end{aligned}$

$\implies \lfloor 10000I\rfloor = \boxed{443}$

First, notice that $f(0)=0.$

Use integration by parts for infinite times.

$\begin{aligned}\int 1\cdot \sin x dx & = x\sin x - \int x(\cos x) dx \\ & = x\sin x - \frac{1}{2}x^2\cos x - \int \frac{1}{2}x^2\sin x dx \\ & = x\sin x - \frac{1}{2}x^2\cos x - \frac{1}{3!}x^3\sin x + \int \frac{1}{3!}x^2(\cos x) dx \\ & = x\sin x - \frac{1}{2!}x^2\cos x - \frac{1}{3!}x^3\sin x + \frac{1}{4!}x^4\cos x + \int \frac{1}{4!}x^4\sin x dx \\ & = \frac{1}{1!}x\sin x - \frac{1}{2!}x^2\cos x - \frac{1}{3!}x^3\sin x + \frac{1}{4!}x^4\cos x +\frac{1}{5!}x^5\sin x - \frac{1}{6!}x^6\cos x - \cdots \\ & = \sum_{n=1}^{\infty} \left\{(-1)^{n-1}\left(\frac{x^{2n-1}}{(2n-1)!}\cdot\sin x-\frac{x^{2n}}{(2n)!}\cdot\cos x\right)\right\} \\ & = f(x) \end{aligned}$

Therefore $f(x)=-\cos x + 1.~(\because f(0)=0)$

And so $\displaystyle \left\lfloor 10000\int_{0}^{1}f(x)^2dx\right\rfloor = \left\lfloor10000\cdot\dfrac{1}{4}(6-8\sin(1)+\sin(2))\right\rfloor = \boxed{443}.$

Just have to know the following identity:

$sinx = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...$ and $cosx = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...$

Then expand the sum carefully, you'll get:

$f(x) = sinx(sinx) + cosx(cosx-1) = 1-cosx$

Then do the integration, easily to get

$10000 \int_0^1 (1-cosx) dx = 443.8238709...$

Get the floor function, the answer is $\boxed{443}$ .

Noted that I first see that $x$ is in the range $[0,1]$ , so I can compute those two identity, otherwise there might be another way to solve this.