Geometric Proof for Improper Integral Equals Pi?

[themadhatter1]

Hi, I came across this interesting integral

\int_{-\infty}^{\infty}\frac{d}{dx}(\arctan x) dx=\pi

I can derive the solution analytically, but I cannot think of a geometric proof. Does anyone know geometrically why this integral is equal to pi?

 

[mathman]

If you think of it as a Stieltjes integral, it is one cycle of the tangent function (-π/2,π/2).

 

[themadhatter1]

Oops. Sorry, that lower bound is supposed to be -\infty I didn't have the -

 

[themadhatter1]

I understand that if you \lim_{n\rightarrow\infty}\arctan x = \frac{\pi}{2} or \lim_{n\rightarrow-\infty}\arctan x = \frac{-\pi}{2}

That's apparent because your ratio of opp./adj. side lengths is growing very quickly as θ approaches π/2. Until, theoretically you have a tiny adjacent side and a relatively huge opposite side. Then, theoretically you would have 2 angles of pi/2 and one angle of 0 if you take the limit as arctan x approaches infinity. Same can be applied for a negative opposite side, except it'd be -pi/2.

It facinates me how the area under the rate of change of the arctan function can be pi. Don't know why this is.

 

[Citan Uzuki]

It's just the fundamental theorem of calculus. For any function f whatsoever:

\int_{a}^{b} f'(x)\ dx = f(b) - f(a)

So:

\int_{-\infty}^{\infty} \frac{d}{dx}\arctan x\ dx = \arctan (\infty) \ -\ \arctan (-\infty) = \frac{\pi}{2}\ -\ \left( -\frac{\pi}{2} \right) = \pi

I'm not being very rigorous with the limits, of course, but this gives you the general idea.

 

[themadhatter1]

It's just the fundamental theorem of calculus. For any function f whatsoever:

\int_{a}^{b} f'(x)\ dx = f(b) - f(a)

So:

\int_{-\infty}^{\infty} \frac{d}{dx}\arctan x\ dx = \arctan (\infty) \ -\ \arctan (-\infty) = \frac{\pi}{2}\ -\ \left( -\frac{\pi}{2} \right) = \pi

I'm not being very rigorous with the limits, of course, but this gives you the general idea.

I understand this; what I don't understand is: geometrically why is this true?

 

[mathman]

Could you be more precise as to what you have in mind for "geometrically why is this true"?

 

[gb7nash]

Could you be more precise as to what you have in mind for "geometrically why is this true"?

I think he wants to know how pi could possibly be the total area between arctanx and the x-axis from -inf to inf. I've seen weirder questions, like how the integral containing a natural log contains pi's in it, and how an infinite sum of 1/n^2 has a pi in it. I certainly have no idea how you could figure it out geometrically.

 

[RandomMystery]

If you look at the bottom circle, tan is shown geometrically. If can some how find it's inverse, it might help.
http://en.wikipedia.org/wiki/Trigonometric_functions

What would help more was to find the equation for calculating trigonometric functions and pi or maybe know the nature of the estimations.

Actually, if you can solve for y in 1 = x^2 + y^2 and then do the absolute value of the integral with respect to dx, it should give you pi.

I hope this helps.