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It is well known that $$\int\frac{1}{1+x^2}\,\mathrm{d}x=\tan^{-1}x+C \tag{1}$$

However, I integrated this differently and got an unusual result.

Suppose we make the substitution $x=\sinh\theta$ and $\mathrm{d}x=\cosh\theta\,\mathrm{d}\theta$ so the integral becomes $$\int\frac{\cosh\theta}{\cosh^2\theta}\,\mathrm{d}\theta=\int\frac{1}{\cosh\theta}\,\mathrm{d}\theta \tag{2}$$

By the definition of $\cosh\theta$, we can rewrite this as $$\int\frac{2e^\theta}{e^{2\theta}+1}\,\mathrm{d}\theta=2\tan^{-1}e^\theta+C \tag{3}$$

Using the fact that $e^\theta=\cosh\theta+\sinh\theta$, we get $e^\theta=x+\sqrt{1+x^2}$, so the answer is then $$2\tan^{-1}\left(x+\sqrt{1+x^2}\right)+C \tag{4}$$

Equating $(4)$ with $(1)$, we have $$2\tan^{-1}\left(x+\sqrt{1+x^2}\right)+C=\tan^{-1}x \tag{5}$$

Plugging in $x=0$, we find $C=-\frac\pi2$. We now have the following strange relationship

$$\tan^{-1} x= 2\tan^{-1}\left(x+\sqrt{1+x^2}\right)-\frac\pi2 \tag{$\star$}$$ This leads me to wonder: Why is this true geometrically, and does this relationship extend into the complex plane?

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  • $\begingroup$ Relevant: en.wikipedia.org/wiki/Gudermannian_function $\endgroup$ – Chappers May 1 '18 at 3:36
  • $\begingroup$ This just looks false. When is this relationship suppose to hold? Let $x=1$ and evaluate this relationship. $\endgroup$ – Mason May 1 '18 at 3:53
  • $\begingroup$ @Mason : Or compute $\frac{\mathrm{d}}{\mathrm{d}x}\left( \tan^{-1}x - 2 \tan^{-1}(x+\sqrt{1+x^2}) - \frac{\pi}{2}\right)$ and get $0$... $\endgroup$ – Eric Towers May 1 '18 at 3:55
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The geometric interpretation could be found in the figure below.

We have $\alpha = \frac{\pi}{2} - \arctan x , ~\beta = \frac{\pi}{2} - \arctan \left(x + \sqrt{1 + x^2}\right)$, and $\alpha = 2 \beta$.

Combining these three equations would lead to your results.

For a formal proof, let $\theta = \arctan \left(x + \sqrt{1 + x^2}\right)$. Since

$$ x + \sqrt{1 + x^2} \geq \sqrt{1 + x^2} - |x| > 0,$$

we have $\theta \in (0, \frac{\pi}{2})$.

Also, note that

$$ \begin{align} \tan\left(2\theta - \frac{\pi}{2}\right)&= -\frac{1}{\tan(2\theta)}\\ &= -\frac{1 - \tan^2 \theta}{2 \tan \theta}\\ &= -\frac{1 - \left(x + \sqrt{1 + x^2}\right)^2}{2 \left(x + \sqrt{1 + x^2}\right)}\\ &= x. \end{align} $$ As $\left(2\theta - \frac{\pi}{2}\right) \in (-\frac{\pi}{2}, \frac{\pi}{2})$, i.e., $\left(2\theta - \frac{\pi}{2}\right)$ lies in the range of $\arctan(\cdot)$, we know that $$ \arctan x = 2\theta - \frac{\pi}{2} = 2\arctan \left(x + \sqrt{1 + x^2}\right) - \frac{\pi}{2}. $$

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  • $\begingroup$ Can we pin down exactly when this holds? $\endgroup$ – Mason May 1 '18 at 4:00
  • $\begingroup$ A formal proof could be done by using the formula of $\tan(2 \theta)$, but need to consider the range of $\arctan$. $\endgroup$ – Xiangxiang Xu May 1 '18 at 4:08
  • $\begingroup$ Very nice geometry. See my answer using the arctan addition formula. $\endgroup$ – marty cohen May 1 '18 at 5:15
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Since $\arctan(x)+\arctan(y) =\arctan(\frac{x+y}{1-xy}) $ with $k\pi$ thrown in as needed,

$\begin{array}\\ \arctan x- 2\arctan\left(x+\sqrt{1+x^2}\right) &=\arctan x- \arctan\left(\frac{2x+2\sqrt{1+x^2}}{1-(x+\sqrt{1+x^2})^2}\right)\\ &=\arctan x- \arctan\left(\frac{2x+2\sqrt{1+x^2}}{1-(x^2+2x\sqrt{1+x^2}+1+x^2)}\right)\\ &=\arctan x- \arctan\left(\frac{2x+2\sqrt{1+x^2}}{1-(2x^2+2x\sqrt{1+x^2}+1)}\right)\\ &=\arctan x- \arctan\left(\frac{2x+2\sqrt{1+x^2}}{-(2x^2+2x\sqrt{1+x^2})}\right)\\ &=\arctan x- \arctan\left(\frac{x+\sqrt{1+x^2}}{-x(x+\sqrt{1+x^2})}\right)\\ &=\arctan x- \arctan\left(\frac{1}{-x}\right)\\ &=\arctan x+ \arctan\left(\frac{1}{x}\right)\\ &=\arctan\left(\frac{x+\frac1{x}}{1-x\frac1{x}}\right)\\ &=\arctan\left(\frac{x+\frac1{x}}{0}\right)\\ &=\frac{\pi}{2}\\ \end{array} $

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We need to be very careful with principal values ,

Let arccot$(x)=2y\implies x=\cot2y$ and $0<2y<\pi$

$x+\sqrt{1+x^2}=\dfrac{\cos2y+1}{\sin2y}=\cot y$ as $y\ne\dfrac\pi2$

$\implies2\tan^{-1}(x+\sqrt{1+x^2})=2\left(\dfrac\pi2-\cot^{-1}(x+\sqrt{1+x^2})\right)=\pi-2y$

and $\tan^{-1}x=\dfrac\pi2-\cot^{-1}x=\dfrac\pi2-2y$

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