8.1: The Arctangent Function

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Definition

For any \(x \in \mathbb{R},\) we call

\[\arctan (x)=\int_{0}^{x} \frac{1}{1+t^{2}} d t\]

the arctangent of \(x .\)

Proposition \(\PageIndex{1}\)

The arctangent function is differentiable at every \(x \in \mathbb{R} .\)

Moreover, if \(f(x)=\arctan (x),\) then

\[f^{\prime}(x)=\frac{1}{1+x^{2}}.\]

Proof: The result follows immediately from Theorem \(7.5 .4 .\) \(\quad\) Q.E.D.

Proposition \(\PageIndex{2}\)

The arctangent is increasing on \(\mathbb{R}\).

Proof

The result follows immediately from the previous proposition and the fact that

\[\frac{1}{1+x^{2}}>0\]

for every \(x \in \mathbb{R}\). \(\quad\) Q.E.D.

Definition

\(\pi=2 \lim _{x \rightarrow+\infty} \arctan (x)=2 \int_{0}^{+\infty} \frac{1}{1+t^{2}} d t\)

Note that \(0<\pi<4\) by Example \(7.7 .1 .\)

The following proposition says that the arctangent function is an odd function.

Theorem \(\PageIndex{3}\)

For any \(x \in \mathbb{R},\) arctan \((x)=-\arctan (-x)\).

Proof

Using the substitution \(t=-u,\) we have

\[\begin{aligned} \arctan (x)=\int_{0}^{x} \frac{1}{1+t^{2}} d t=-\int_{0}^{-x} \frac{1}{1+u^{2}} d u=-\arctan (-x). & \end{aligned}\]

Q.E.D.

It now follows that

\[\lim _{x \rightarrow-\infty} \arctan (x)=-\lim _{x \rightarrow-\infty} \arctan (-x)=-\frac{\pi}{2}.\]

Hence the range of the arctangent function is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).

Proposition \(\PageIndex{4}\)

If \(x>0,\) then

\[\arctan (x)+\arctan \left(\frac{1}{x}\right)=\frac{\pi}{2}.\]

Proof

Using the substitution \(t=\frac{1}{u},\) we have

\[\begin{aligned} \arctan \left(\frac{1}{x}\right) &=\int_{0}^{\frac{1}{x}} \frac{1}{1+t^{2}} d t \\ &=\int_{+\infty}^{x} \frac{1}{1+\frac{1}{u^{2}}}\left(-\frac{1}{u^{2}}\right) d u \\ &=-\int_{+\infty}^{x} \frac{1}{1+u^{2}} d u \\ &=\int_{x}^{+\infty} \frac{1}{1+u^{2}} d u \\ &=\frac{\pi}{2}-\int_{0}^{x} \frac{1}{1+u^{2}} d u \\ &=\frac{\pi}{2}-\arctan (x). \end{aligned}\]

Q.E.D.

Proposition \(\PageIndex{5}\)

If \(x<0\), then

\[\arctan (x)+\arctan \left(\frac{1}{x}\right)=-\frac{\pi}{2}.\]

Proof: The result follows immediately from the preceding proposition and the fact that arctangent is an odd function.

Exercise \(\PageIndex{1}\)

Show that arctan \((1)=\frac{\pi}{4}\) and \(\arctan (-1)=-\frac{\pi}{4}\).