A very fickle function

We are tasked to find the area under the curve of $f(x)$ between $0$ and $1$ provided that $f(x)$ is above the $x$ - axis.

We first find the antiderivative of $f(x) = \sin ( \ln x )$ .

Let $I = \displaystyle \int \sin (\ln x) \mathrm{d}x$

By integration by parts, using $u= \sin (\ln x)$ , and $\mathrm{d}v = \mathrm{d}x$ , we get

$I = x \sin(\ln x) - \displaystyle \int x \frac { \cos (\ln x)}{x} \mathrm{d}x$

$I = x \sin(\ln x) - \displaystyle \int \cos(\ln x) \mathrm{d}x$

Then again, we use the substitutions $u= \cos(\ln x)$ and $\mathrm{d}v = \mathrm{d}x$ to get

$I = x \sin(\ln x) - [ x \cos (\ln x) + \displaystyle \int \sin (\ln x) \mathrm{d}x ]$

$I = x \sin(\ln x) - x \cos( \ln x) - I$

$2I = x \sin(\ln x) - x\cos (\ln x)$

$I = \frac{x}{2} [ \sin (\ln x) - \cos (\ln x) ] + C$

Now, we check the behavior of the function. Note that $f(x) = 0$ when $x = e^{ \pi n}$ for all integers. Additionally, note that $1 = e^0$ . This implies that we are dealing with zeroes of $f(x)$ of the form $e^{\pi k}$ where $k \in \mathbb{Z}, k \leq 0$ .

Since we can't find first zero of $f(x)$ around the origin, we inspect its behavior around $x=1$ and proceed to the left.

Notice that $\displaystyle \int_{e^{- \pi}}^{0} f(x) \mathrm{d}x = - \frac {1}{2} ( 1 + e^{\pi})e^{-\pi}$ . Since we dealt with two adjacent zeroes, seeing this area as negative implies that this part of the graph occurs below the $x$ - axis, and will not be considered.

By a similar manner, we can get the few succeeding zeroes, and get a pattern

$\displaystyle\int_{e^{-2 \pi}}^{e^{- \pi}} f(x) \mathrm{d}x = \frac {1}{2} (1 + e^{\pi}) e^{- 2\pi}$

$\displaystyle\int_{e^{-3 \pi}}^{e^{- 2\pi}} f(x) \mathrm{d}x = - \frac {1}{2} (1 + e^{\pi}) e^{- 3\pi}$

$\displaystyle\int_{e^{-4 \pi}}^{e^{- 3\pi}} f(x) \mathrm{d}x = \frac {1}{2} (1 + e^{\pi}) e^{- 4\pi}$

In fact, we can see that the area alternates every time the graph crosses the $x$ -axis. This will give us the sum for the areas above the $x$ -axis under the curve.

$A = \frac{1}{2}(1+e^{\pi})(e^{-2 \pi} + e^{-4 \pi} + e^{-6 \pi} +...)$

$A = \frac{1}{2} \cdot \frac { (1+e^{\pi})(e^{-2\pi})}{1 - e^{-2\pi}}$

$A = \frac{1}{2} \cdot \frac { (1+e^{\pi})(e^{-2\pi})}{1 - e^{-2\pi}} \cdot \frac {e^{2\pi}}{e^{2\pi}}$

$A = \frac {1}{2} \cdot \frac { 1+ e^{\pi}}{e^{2\pi} - 1}$

$A = \frac{1}{2} \cdot \frac { 1+e^{\pi}}{ (e^{\pi} + 1)(e^{- \pi} - 1)}$

$A = \frac{1}{2} \cdot \frac{1}{e^{\pi} - 1}$

$A \approx 0.02258285$

so $\lfloor 10^5 A \rfloor = 2258$ .