Value of $a$ for which two integrals are equal

Question

Determine $a$ such that $$\int _0 ^a ([\arctan \sqrt {x}]dx=\int _0 ^a [\frac {\pi}{2}-\arctan\sqrt {x}]dx$$ where $[.] $ is greatest integer function so i first took the first integral to get integral as $a [\arctan (a)] $ as I think derivative of integer function is 0 as its always an integer. But I dont know how to proceed.

Answer 1

$\tan^{-1} \sqrt{x}$ is defined for $x\ge 0$, so we need to consider $a\ge 0$ only.

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• $\tan^{-1} \sqrt{x}$ is an increasing function and

$$0\le \tan^{-1} \sqrt{x} \le \frac{\pi}{2} < 2$$

\begin{align*} \left \lfloor \tan^{-1} \sqrt{x} \right \rfloor &= \left \{ \begin{array}{lr} 0 \, , & 0\le x < \tan^2 1 \\ 1 \, , & x \ge \tan^2 1 \end{array} \right. \\[5pt] \int_0^a \left \lfloor \tan^{-1} \sqrt{x} \right \rfloor \, dx &= \left \{ \begin{array}{lr} 0 \, , & 0\le a < \tan^2 1 \\ a-\tan^2 1 \, , & a \ge \tan^2 1 \end{array} \right. \end{align*}

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• $\dfrac{\pi}{2}-\tan^{-1} \sqrt{x}$ is a decreasing function and

$$0\le \frac{\pi}{2}-\tan^{-1} \sqrt{x} \le \frac{\pi}{2} < 2$$

\begin{align*} \left \lfloor \frac{\pi}{2}-\tan^{-1} \sqrt{x} \right \rfloor &= \left \{ \begin{array}{lr} 1 \, , & 0\le x < \cot^2 1 \\ 0 \, , & x \ge \cot^2 1 \end{array} \right. \\[5pt] \int_0^a \left \lfloor \frac{\pi}{2}-\tan^{-1} \sqrt{x} \right \rfloor \, dx &= \left \{ \begin{array}{lr} a \, , & 0\le a < \cot^2 1 \\ \cot^2 1 \, , & a \ge \cot^2 1 \end{array} \right. \end{align*}

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Note that $\cot^2 1 <\tan^2 1$,

$$\int_0^a \left \lfloor \tan^{-1} \sqrt{x} \right \rfloor \, dx = \int_0^a \left \lfloor \frac{\pi}{2}-\tan^{-1} \sqrt{x} \right \rfloor \, dx$$

$$\implies a=0 \quad \text{or} \quad a=\tan^2 1+\cot^2 1$$