# Geometric Interpretation for Definite Integrals with $\pi$ in the result [duplicate]

What is the geometric interpretation for the following integral?

What is a nice geometric interpretation for the following integral (possibly in relationship to a circle) that emphasizes why we get the result of π in the right hand side?

$$\int^\infty_{-\infty} \frac{1}{1+x^2}dx = \pi$$

I do know, of course that the indefinite integral for the integrand is $\tan^{-1} x$.

In the answers, I'd also appreciate examples of other integrals with a geometric interpretation.

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## marked as duplicate by Qiaochu YuanApr 1 '13 at 22:36

You forgot $dx$ :) –  Harold Apr 1 '13 at 9:55
The primitive of $1/(1+x^2)$ is $\tan$ which has a geometric interpretation/ –  Damien L Apr 1 '13 at 9:57
@Harold, thanks! :) Damien - yes, but I'm wondering whether we could go directly to the geometric interpretation. –  Vincent Tjeng Apr 1 '13 at 10:03
" The area between the positive function $\,\frac{1}{1+x^2}\,$ and the $\,x-$axis over the whole real line equals $\,\pi\,$ " . When a function is positive over some real interval the geometric interpretation of the definite integral is pretty much the same usually studied in high school: geometric area. Here's the same even if we have an improper integral (improper because of the infinite interval, only that) –  DonAntonio Apr 1 '13 at 10:10
@Damien, the primitive of $\,\frac{1}{1+x^2},$ is $\,\arctan x\,$ , not $\,\tan x\,$ ... –  DonAntonio Apr 1 '13 at 10:11

Consider the stereographic projection of the line onto the circle.

$\hspace{1cm}$

Let $x$ be the distance from the point of tangency and $r$ be the distance from the center of the circle. Using similar triangles in the right inset, we get that $\mathrm{d}x$ on the tangent line is projected onto $\frac{\mathrm{d}x}{r}$ on the piece parallel to the circle. Then, again by similar triangles in the left inset, that is reduced to $\frac{\mathrm{d}x}{r^2}$. Thus, the projection of $\mathrm{d}x$ onto the circle yields $$\frac{\mathrm{d}x}{r^2}=\frac{\mathrm{d}x}{1+x^2}\tag{1}$$ Integrating $(1)$ over the entire real line will give the length of half the circle, $\pi$. That is, $$\int_{-\infty}^\infty\frac{\mathrm{d}x}{1+x^2}=\pi\tag{2}$$

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This I think explains the value of the integral. +1 –  coffeemath Apr 1 '13 at 17:07

I guess this will be a bit to long for a comment, althought it should be more a comment I post it as an answer.

You have to be careful not to give a meaningful geometric interpretation which doesn't have one. In your example you have $\int_{-\infty}^\infty \frac{1}{1+x^2} \; \mathrm{d}x$ Using $x=\tan(u)$ this integral is the same as $\int_{-\infty}^\infty \frac{1}{1+x^2} \; \mathrm{d}x= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \; \mathrm{d}u=\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)=\pi$

One can say that $\tan$ is a diffemorphism which makes from the intervall $(-\frac{\pi}{2},\frac{\pi}{2})$ the Real numbers.

But for sure you can get from $\int_{-\frac{\pi}{2}}^\frac{\pi}{2} 1 \; \mathrm{d}u = \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \sqrt{\sin^2(u)+\cos^2(u)} \; \mathrm{d}u$ Ok now we make a 2 D transformation to polar coordinates and see that our integral is just the same as $\int_{-\frac{\pi}{2}}^\frac{\pi}{2} \sqrt{\sin^2(u)+\cos^2(u)} \; \mathrm{d}u= \frac{1}{2} \cdot \int_\gamma \| r\| \; \mathrm{d}r$ where $\gamma$ is the unit circle, so we just measure the half circumference of the unit circle.

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One should be careful not to deny a meaningful geometric interpretation where one might exist. –  robjohn Apr 1 '13 at 22:29
@robjohn that why it was orginally meant as a comment, i didn't say there isn't a meaningful interpretation, I just say that when there isn't one you shouldn't try to give it, but our solutions are the same or ? –  Dominic Michaelis Apr 2 '13 at 8:20
If you mean that we arrive at the same answer, then yes. However, I wouldn't say that our method of solution is the same. –  robjohn Apr 2 '13 at 8:46

Suppose there is a light source at $(0,0)$ which shines on the right semicircle $x^2+y^2=1,\ x\ge 0.$ And suppose the light intensity is proportional to arclength along the circle, so that the total light given off is $\pi$, the length of the semicircle. Now as we all know, light dissipates according to an inverse square law. So if we now compute the total amount of light which hits the line $x=1$, that total should also be $\pi$. The squared distance from $(0,0)$ to $(1,y)$ being $1+y^2$, we would, in finding the total light hitting the line, be computing $$\int_{-\infty}^{\infty}\frac{dx}{1+y^2}.$$ I admit this is a bit fanciful, but thought I'd add it. Maybe someone can make the argument rigorous.

EDIT: (No it wasn't an April fool's joke.) A major flaw in the above is that in a linear situation, light intensity would fall of as the inverse first power of distance, not inverse square. But there is an adjusting factor of $1/\sqrt{1+x^2}$ due to the fact that the line $x=1$ is on a varying slant with respect to the direction of the light rays. So I think the "april fool" argument could be salvaged, but it's likely not worth it in view of robjohn's clear argument (not depending on physics, light, etc).

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Not sure if you are serious or it shall be an april fool –  Dominic Michaelis Apr 1 '13 at 10:40
@coffeemath perhaps we could use the idea of Gaussian Surfaces and the Divergence Theorem, but I think that your answer is already pretty intuitive. –  Vincent Tjeng Apr 1 '13 at 10:53
I think it's not $x \geq 0$ but $y \geq 0$. Anyway I like it! +1 –  Taro Apr 1 '13 at 11:02
@DominicMichaelis, I'm strongly suspecting it is the second option. Now we just need an explanation including little flying fairies along some cute, flowered semi-circle, together with the great news that along the centuries we've been writing the number $\,5\,$ upside down and, thus, all of mathematics is worthless... –  DonAntonio Apr 1 '13 at 11:06

By definition :

$$a,b\in\Bbb R\;,\;a,b>0\;:\;\;\int\limits_{-\infty}^\infty\frac{dx}{1+x^2}=\lim_{a,b\to\infty}\int\limits_{-a}^b\frac{dx}{1+x^2} =\left.\lim_{a,b\to\infty}\arctan x\right|_{-a}^b=$$

$$=\lim_{a,b\to\infty}\left(\arctan b-\arctan(-a)\right)=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi$$

The idea of a geometric interpretation is in the comments above.

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perhaps my question isn't so nicely put. what I'm looking for is, can you change your viewpoint so the integral becomes the area of the unit circle, or perhaps the circumference of a unit semi-circle? –  Vincent Tjeng Apr 1 '13 at 10:19
No, I can't. Perhaps somebody else has some idea, and yes: if this is what you meant then it wasn't nicely put. –  DonAntonio Apr 1 '13 at 10:30
@VincentTjeng maybe mine answer is more what you mean –  Dominic Michaelis Apr 1 '13 at 10:34
@DonAntonio sorry about it, and thanks for your answer! –  Vincent Tjeng Apr 1 '13 at 10:36

Geometric interpretation of a definite integral is the signed area between the graph of integrant and x-axis.
Now let's have a look at the graph of your function. Having $\int^\infty_{-\infty} \frac{1}{1+x^2}dx = \pi$ means that if you take a rectangle with side lengths $1$ and $\pi$, you can cut it into infinitely small pieces and fill the region between the curve and x-axis.

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