202109160920 Exercises 5.3.B (Q21)

Show that for all $x>0$ ,

$\displaystyle{\int_{0}^{x}\frac{\mathrm{d}t}{1+t^2}+\int_{0}^{1/x}\frac{\mathrm{d}t}{1+t^2}}$

is independent of $x$ .

Solution.

$\begin{aligned} & \qquad\int_{0}^{x}\frac{\mathrm{d}t}{1+t^2}+\int_{0}^{1/x}\frac{\mathrm{d}t}{1+t^2} \\ \dots &\enspace\textrm{as }\sin^2u +\cos^2u =1 \enspace\dots\\ \dots &\enspace\textrm{so }\tan^2u +1 = \sec^2u\enspace\dots \\ \dots &\enspace\textrm{let }t=\tan u \textrm{ then }\mathrm{d}t=\sec^2u\,\mathrm{d}u\enspace\dots \\ & = \int_{0}^{\arctan x}\frac{\sec^2u\,\mathrm{d}u}{1+\tan^2u} + \int_{0}^{\arctan (\frac{1}{x})}\frac{\sec^2u\,\mathrm{d}u}{1+\tan^2u} \\ & = \int_{0}^{\arctan x}\frac{\sec^2u\,\mathrm{d}u}{\sec^2u} + \int_{0}^{\arctan (\frac{1}{x})}\frac{\sec^2u\,\mathrm{d}u}{\sec^2u} \\ & = [u]|_{0}^{\arctan x} + [u]|_{0}^{\arctan (1/x)}\\ & = \arctan x + \arctan\bigg(\frac{1}{x}\bigg) \\ & \stackrel{(*)}{=} \frac{\pi}{2} \end{aligned}$

Lemma (*)

Let $y=\arctan x$ and $w=\arctan \displaystyle{\bigg(\frac{1}{x}\bigg)}$ ,

then $x=\tan y$ and $\displaystyle{\frac{1}{x}}=\tan w$ .

Thus,

$\begin{aligned} \tan y & = \frac{1}{\tan w} \\ \tan y & = \cot w \\ \tan y & = \tan \bigg(\frac{\pi}{2}-w\bigg) \\ y & = \frac{\pi}{2} - w \\ y + w & = \frac{\pi}{2} \\ \end{aligned}$

$\therefore\enspace \displaystyle{\arctan x + \arctan\bigg(\frac{1}{x}\bigg) = \frac{\pi}{2}}$

(independent of $x$ )

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