The antiderivative of 1/(x^4+1): Part 8

In the course of evaluating the antiderivative

\displaystyle \int \frac{1}{x^4 + 1} dx,

I’ve accidentally stumbled on a very curious looking trigonometric identity:

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi if x < -1,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) if -1 < x < 1,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi if x> 1.

The extra -\pi and \pi are important. Without them, the graphs of the left-hand side and right-hand sides are clearly different if x < -1 or x > 1:

TwoArctangents1

However, they match when those constants are included:

TwoArctangents2

Let’s see if I can explain why this trigonometric identity occurs without resorting to the graphs.

Since \tan^{-1} x assumes values between -\pi/2 and \pi/2, I know that

-\displaystyle \frac{\pi}{2} < \tan^{-1} ( x\sqrt{2} - 1 ) < \frac{\pi}{2},

-\displaystyle \frac{\pi}{2} < \tan^{-1} ( x\sqrt{2} + 1 ) < \frac{\pi}{2},

and so

-\pi< \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 ) < \pi.

However,

-\displaystyle \frac{\pi}{2} < \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) < \frac{\pi}{2},

and so \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 ) and \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) must differ if \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1) is in the interval [-\pi,-\pi/2] or in the interval [\pi/2,\pi].

I also notice that

-\pi< \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 ) < \pi,

-\displaystyle \frac{\pi}{2} < -\tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) < \frac{\pi}{2},

and so

-\displaystyle \frac{3\pi}{2} < \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 )-\tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) < \frac{3\pi}{2}.

However, this difference can only be equal to a multiple of \pi, and there are only three multiples of \pi in the interval \displaystyle \left( -\frac{3\pi}{2}, \frac{3\pi}{2} \right), namely -\pi, 0, and \pi.

To determine the values of x where this happens, I also note that f_1(x) = x \sqrt{2} - 1, f_2(x) = x \sqrt{2} + 1, and f_3(x) = \tan^{-1} x are increasing functions, and so f(x) = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) must also be an increasing function. Therefore, to determine where f(x) lies in the interval [\pi/2,\pi],it suffices to determine the unique value x_1 so that f(x_1) = \pi/2. Likewise, to determine where f(x) lies in the interval [-\pi,-\pi/2],it suffices to determine the unique value x_2 so that f(x_2) = -\pi/2.

In summary, I have shown so far that

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi if x < x_2,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) if x_2 < x < x_1,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi if x> x_1,

where x_1 and x_2 are the unique values so that

\tan^{-1} ( x_1\sqrt{2} - 1 ) + \tan^{-1}( x_1 \sqrt{2} + 1) = \displaystyle \frac{\pi}{2},

\tan^{-1} ( x_2\sqrt{2} - 1 ) + \tan^{-1}( x_2 \sqrt{2} + 1) = -\displaystyle \frac{\pi}{2}.

So, to complete the proof of the trigonometric identity, I need to show that x_1 = 1 and x_2 = -1. I will do this in tomorrow’s post.

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1 Comment

  1. The antiderivative of 1/(x^4+1): Index | Mean Green Math

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