Strange equation

**wonderboy1953** · January 18th 2011, 01:27 PM

Here's an equation I derived a long time ago:

$2\tan^{-1}x + C = ln\frac{(x - i)}{(x + i)}$ where C is the constant of integration and i is the imaginary unit (you'll have to imagine the vertical lines for the absolute value of the fraction on the right-hand side of the equation).

The strangest thing I find about this equation is that i shows up on the right-hand side and not on the left-hand side (to check out the validity of this equation, differentiate both sides). So there's a type of crossover from the realm of real numbers to the realm of complex numbers.

Does anyone have an explanation as to what's really going on? That's my puzzle.

**Bruno J.** · January 18th 2011, 04:18 PM

All inverse trigonometric functions can be expressed in similar ways. I'm not sure what you mean about the "constant of integration", if not just that the logarithm is multiple-valued, so there's an ambiguity of $2\pi$ on both sides. You can't replace $C$ by anything you like, though.

Trigonometric functions are just exponential functions, so it's no surprise the inverse trigonometric functions can be written down using the logarithm.

**TheCoffeeMachine** · January 18th 2011, 05:35 PM

Originally Posted by wonderboy1953

The strangest thing I find about this equation is that i shows up on the right-hand side and not on the left-hand side (to check out the validity of this equation, differentiate both sides). So there's a type of crossover from the realm of real numbers to the realm of complex numbers.

Well, $i$ does appear on the left-hand side (here) -- it's just that you missed it. :P
We must have $i$ as a multiple of the LHS (or $\frac{1}{i}$ on the RHS) for it to make sense.
Either way, it's just due to the integral definition of the arctangent function as
$\arctan{x} = \int_{0}^{x}\frac{1}{x^2+1}\;{dx}$ and the fact that $x^2+1$ is reducible over $\mathbb{C}$ .

**Soroban** · January 18th 2011, 07:01 PM

Hello, wonderboy1953!

TheCoffeeMachine is right . . . The left side needs an $\,i.$

$2i\tan^{-1}x \:=\:\ln\left|\dfrac{x - i}{x + i}\right|+ C$

I derived this, too, many many years ago.
Here's how I did it.

$\displaystyle \text{We know that: }\:\int\frac{dx}{x^2+1} \;=\;\tan^{-1}x + C_1$ .[1]

$\text{Since }\:x^2+1 \;=\;x^2 - (\text{-}1) \;=\;x^2 - i^2 \;=\;(x-i)(x+i)$

. . $\displaystyle \text{then: }\;\int\frac{dx}{x^2+1} \;=\;\int\frac{dx}{(x-i)(x+i)} \;=\;\frac{1}{2i}\ln\left|\frac{x-i}{x+i}\right| + C_2$ .[2]

Equate [1] and [2]: . $\displaystyle \tan^{-1}x + C_1 \;=\;\frac{1}{2i}\ln\left|\frac{x-i}{x+i}\right| + C_2$

$\displaystyle\text{Therefore: }\;2i\tan^{-1}x \;=\;\ln\left|\frac{x-i}{x+i}\right| + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle \text{Another identity: }\;i\ln\left|\frac{x-1}{x+1}\right| \;=\;2\tan^{1}\left(\frac{x}{i}\right) + C$

$\displaystyle \text{This comes from: }\:\int\frac{dx}{x^2-1} \;=\;\int\frac{dx}{x^2+i^2}$

**wonderboy1953** · January 20th 2011, 11:10 AM

Originally Posted by Soroban

Hello, wonderboy1953!

TheCoffeeMachine is right . . . The left side needs an $\,i.$

I derived this, too, many many years ago.
Here's how I did it.

$\displaystyle \text{We know that: }\:\int\frac{dx}{x^2+1} \;=\;\tan^{-1}x + C_1$ .[1]

$\text{Since }\:x^2+1 \;=\;x^2 - (\text{-}1) \;=\;x^2 - i^2 \;=\;(x-i)(x+i)$

. . $\displaystyle \text{then: }\;\int\frac{dx}{x^2+1} \;=\;\int\frac{dx}{(x-i)(x+i)} \;=\;\frac{1}{2i}\ln\left|\frac{x-i}{x+i}\right| + C_2$ .[2]

Equate [1] and [2]: . $\displaystyle \tan^{-1}x + C_1 \;=\;\frac{1}{2i}\ln\left|\frac{x-i}{x+i}\right| + C_2$

$\displaystyle\text{Therefore: }\;2i\tan^{-1}x \;=\;\ln\left|\frac{x-i}{x+i}\right| + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle \text{Another identity: }\;i\ln\left|\frac{x-1}{x+1}\right| \;=\;2\tan^{1}\left(\frac{x}{i}\right) + C$

$\displaystyle \text{This comes from: }\:\int\frac{dx}{x^2-1} \;=\;\int\frac{dx}{x^2+i^2}$

I see said the blind man.

**Bruno J.** · January 20th 2011, 01:37 PM

Originally Posted by wonderboy1953

I see said the blind man.

What's that supposed to mean?

Oh and please see this.

**wonderboy1953** · January 22nd 2011, 09:28 AM

Originally Posted by Bruno J.

What's that supposed to mean?

Oh and please see this.

"I see said the blind man" simply means that Soroban alerted me to my algebra mistake.

When you leave $2i$ on the right side of the equation, you have an interesting situation whereby you're extending the realm of the real numbers on left side of the equation to the realm of the complex numbers on the right side.

I'm sure the average person would find this situation to be unexpected if that person never saw it before. I want to mention that I find calculus to be very interesting, for various reasons, and I like how algebra and calculus interact.

Thank you for the link to Inverse trigonometric functions - Wikipedia, the free encyclopedia as I was just thinking about inverse trig functions.

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