This time it's an inverse

If you realize that $\arctan(x)$ is never more than $\frac{\pi}{2}\approx 1.5707...$ , this problem is very easy.

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So the integral is just the area of the box, which has a length $100 - \tan(1)$ and height $1$ . So the area is just $100-\tan(1)$ , so $\alpha + \beta$ is $\boxed{101}$ .

We will split the integral into two parts :

$I=\int _{ 0 }^{ tan(1) }{ \left\lfloor { tan }^{ -1 }(x) \right\rfloor dx } +\int _{ tan(1) }^{ 100 }{ \left\lfloor { tan }^{ -1 }(x) \right\rfloor dx }$

Now for $0 \le x \le tan(1)$

$\Rightarrow 0 \le {tan}^{-1}(x) \le 1$

$\Rightarrow \left\lfloor {tan}^{-1}(x) \right\rfloor = 0$

Also for $tan(1) \le x \le 100$

$\Rightarrow 1 \le {tan}^{-1}(x) \le {tan}^{-1}(100) \le 2$

$\Rightarrow \left\lfloor {tan}^{-1}(x) \right\rfloor = 1$

$\Rightarrow I=\int _{ 0 }^{ tan(1) }{ 0dx } +\int _{ tan(1) }^{ 100 }{ 1dx }$

$I=100-tan(1)$

$\Rightarrow \alpha=100, \beta=1$

$\boxed{\alpha+\beta=101}$

Just integrate it directly, as follows

$\displaystyle \int _{ 0 }^{ 100 }{ \left( -\dfrac { 1 }{ 2 } +ArcTan\left( x \right) -\dfrac { 1 }{ 2\pi } iLog\left( -{ e }^{\displaystyle -2i\pi ArcTan\left( x \right) } \right) \right) } dx=$

$100-Tan\left( 1 \right)$

ta-da!

$\begin{aligned}\displaystyle \int\limits_0^{100} \left \lfloor \tan^{-1} x\right \rfloor \mathrm{d}x&=\int\limits_0^{\tan 1} \left \lfloor \tan^{-1} x\right \rfloor \mathrm{d}x + \int\limits_{\tan 1}^{100} \left \lfloor \tan^{-1} x\right \rfloor \mathrm{d}x \\&=\int\limits_0^{\tan 1} 0 \mathrm{d}x + \int\limits_{\tan 1}^{100} 1 \mathrm{d}x\\ &= 100 - \tan 1\\&\approx 98.44259\end{aligned}\\\huge \boxed{\displaystyle\int\limits_0^{100} \left \lfloor \tan^{-1} x\right \rfloor \mathrm{d}x = 100 - \tan 1^c}$