Geometric Proof for Improper Integral Equals Pi?

In summary, the integral is equal to pi because the area under the graph of the arctan function is pi.
  • #1
themadhatter1
140
0
Hi, I came across this interesting integral

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

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?
 
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  • #2
If you think of it as a Stieltjes integral, it is one cycle of the tangent function (-π/2,π/2).
 
  • #3
Oops. Sorry, that lower bound is supposed to be [tex]-\infty[/tex] I didn't have the -
 
  • #4
I understand that if you [tex]\lim_{n\rightarrow\infty}\arctan x = \frac{\pi}{2}[/tex] or [tex]\lim_{n\rightarrow-\infty}\arctan x = \frac{-\pi}{2}[/tex]

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.
 
  • #5
It's just the fundamental theorem of calculus. For any function f whatsoever:

[tex]\int_{a}^{b} f'(x)\ dx = f(b) - f(a)[/tex]

So:

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

I'm not being very rigorous with the limits, of course, but this gives you the general idea.
 
  • #6
Citan Uzuki said:
It's just the fundamental theorem of calculus. For any function f whatsoever:

[tex]\int_{a}^{b} f'(x)\ dx = f(b) - f(a)[/tex]

So:

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

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?
 
  • #7
Could you be more precise as to what you have in mind for "geometrically why is this true"?
 
  • #8
mathman said:
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.
 
  • #9
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.
 
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What is an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite or where the integrand has an infinite discontinuity within the interval of integration.

Why do we need to prove the convergence of an improper integral?

We need to prove the convergence of an improper integral because it ensures that the integral has a well-defined value and that the function being integrated is well-behaved within the limits of integration.

What is the process for proving the convergence of an improper integral?

The process for proving the convergence of an improper integral involves evaluating the integral as a limit of a proper integral, and then using various techniques such as comparison tests, limit comparison tests, and the Cauchy criterion to determine if the limit exists.

What are some common techniques used in improper integral proofs?

Some common techniques used in improper integral proofs include the comparison test, where the integral is compared to a known convergent or divergent integral, and the limit comparison test, where the ratio of the integrand to a known function is taken to determine convergence. Other techniques include the Cauchy criterion, integration by parts, and substitution.

Are there any special cases in improper integral proofs?

Yes, there are special cases in improper integral proofs, such as when the integrand is unbounded or when the limits of integration are infinite or approach each other. These cases may require additional techniques or approaches to prove convergence.

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