Differentiation/Integration Disagreement

[ChaoticLlama]

Consider the function \(\displaystyle \L\ y = \arctan \left( {\frac{1}{x}} \right)\).

To the best of my knowledge;
\(\displaystyle \L\begin{array}{c}
\frac{{dy}}{{dx}} = \frac{1}{{1 + \frac{1}{{x^2 }}}}\left( {\frac{{ - 1}}{{x^2 }}} \right) \\
= \frac{{ - 1}}{{x^2 + 1}} \\
\end{array}\)

and if you integrate the new function, you get;
\(\displaystyle \L\int {\frac{{ - dx}}{{x^2 + 1}}} = - \arctan (x) + C\)

Now unless I'm missing something here; I think I broke Calculus.

 

[soroban]

Hello, ChaoticLlama!

A fascinating "contradiction" . . . thank you for pointing it out!
It is going immediately into my "Mathematical Curiosities" file.

Consider the function \(\displaystyle \L\ y = \arctan \left( {\frac{1}{x}} \right)\).

To the best of my knowledge;
\(\displaystyle \L\;\;\; \frac{{dy}}{{dx}} \:= \:\frac{1}{{1 + \frac{1}{{x^2 }}}}\left( {\frac{{ - 1}}{{x^2 }}} \right) \;= \;\frac{{ - 1}}{{x^2 + 1}}\)

and if you integrate the new function, you get;
\(\displaystyle \L\;\;\;\int \frac{- dx}{x^2 + 1} \; =\; - \arctan (x) + C\)

Now unless I'm missing something here, I think I broke Calculus.

We will find that the two answers are equal (for a particular value of $\displaystyle C$).

$\displaystyle \;\;$That is: $\displaystyle \,\arctan\left(\frac{1}{x}\right) \;= \;-\arctan(x)\,+\,C\;\;\text{ for }C\, =\, \frac{\pi}{2}$


Let $\displaystyle \alpha\,=\,\arctan\left(\frac{1}{x}\right)$

Then $\displaystyle \alpha$ is an acute angle in a right triangle with: $\displaystyle opp\,=\,1,\;adj\,=\,x.$

                  *
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            *     * 1
         * α      *
      * * * * * * *
            x

Let $\displaystyle \beta\,=\,\arctan(x)$

Then $\displaystyle \beta$ is an acute angle in a right triangle with: $\displaystyle opp\,=\,x,\;adj\,=\,1$


Get it? . . . $\displaystyle \,\beta$ is the "other acute angle": $\displaystyle \:\alpha\,+\,\beta\:=\:\frac{\pi}{2}$


Hence: $\displaystyle \,\arctan\left(\frac{1}{x}\right)\,+\,\arctan(x)\;=\;\frac{\pi}{2}\;\;\Rightarrow\;\;\arctan\left(\frac{1}{x}\right)\;=\;-\arctan(x)\,+\,\frac{\pi}{2}$

 

[pka]

Hence: $\displaystyle \,\arctan\left(\frac{1}{x}\right)\,+\,\arctan(x)\;=\;\frac{\pi}{2}\;\;\Rightarrow\;\;\arctan\left(\frac{1}{x}\right)\;=\;-\arctan(x)\,+\,\frac{\pi}{2}$

That is true only if x>0!

 

[ChaoticLlama]

In some way the functions are still equal, but this is a strange situation indeed. Both functions have the same derivative, but are only equal themselves for a single constant. Is there any reason why? Or maybe it's just another oddity?

(I still like to think that I broke Calculus :wink: )