Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am practicing for an exam and am having trouble with this problem. Find the integral from 1 to infinity of $\frac{1}{1+x^2}$. I believe the integral's anti-derivative is $\arctan(x)$ which would make this answer $\arctan(\infty)-\arctan(1)$ but from here I'm lost. I did find out that this comes out to $\pi/4$ but I don't know why.

share|improve this question

4 Answers 4

up vote 6 down vote accepted

Essentially, your question appears to be "Why does $\lim_{x\to\infty}\arctan{x} = \frac{\pi}{2}$?"

Remember that $$\theta = \arctan\left(\frac{y}{x}\right)$$ When will $\frac{y}{x}\to\infty$? That's when $x \to 0$.

Think of a right triangle with height $y$ and base $x$. $\theta$ is the angle between $x$ and the hypotenuse. As $x$ gets smaller and smaller, what does $\theta$ get close to? Well, $\theta$ approaches 90 degrees or $\frac{\pi}{2}$.

EDIT: As requested, here's a picture to illustrate this idea. The angle (in degrees) is displayed inside the tangent function, alongside the fraction $\frac{y}{x}$:

Illustration

share|improve this answer
    
This is such awesome intuition. Can you make a picture to illustrate it and relate it explicitly to the unit circle? –  000 Dec 8 '12 at 20:36
    
@Limitless Not quite sure what you mean by explicitly relate it to the unit circle... I'll see about creating a graphic –  anorton Dec 9 '12 at 1:35
    
I envision the triangle as part of the unit circle. –  000 Dec 9 '12 at 1:37
    
Picture has been added (@Limitless) –  anorton Dec 9 '12 at 2:36
1  
Great picture! :-) –  000 Dec 9 '12 at 2:38

With contour integration we will calculate $\int_{0}^{\infty}\frac{dx}{1+x^2}$:

First note that by symmetry \begin{equation}\int_{0}^{+\infty}\frac{1}{1+x^2}\, dx=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{1}{1+x^2}\, dx=\lim_{R\to +\infty}\frac{1}{2}\int_{-R}^{+R}\frac{1}{1+x^2}\, dx\end{equation} The associated complex function is $f(z)=\frac{1}{1+z^2}$. We shall integrate it over the contour $C=[-R,R]\cup \gamma_{R}$ where $\gamma_{R}$ is an anticlockwise semicircle arc centered at $0$ with radius $R>0$. $f$ has a simple poles at $\pm i$ but only $i$ is in the region bounded by $C$. By the Residue Theorem, \begin{equation}\oint_{C} f(z)\, dz=2\pi \text{Res}_{i}(f)=2\pi \lim_{z\to i}\frac{z-i}{(z-i)(z+i)}=\pi\end{equation} Observe that \begin{equation}\oint_{C}f(z)\, dz=\int_{[-R,R]}f(z)\, dz+\int_{\gamma_{R}}f(z)\, dz\Rightarrow \int_{[-R,R]}f(z)\, dz=\pi -\int_{\gamma_{R}}f(z)\, dz \end{equation} We wish to show that the integral in the right hand side converges to $0$ as $R\to +\infty$.

Indeed, \begin{equation}\lim_{R\to +\infty}\int_{\gamma_R}f(z)\, dz= \lim_{R\to +\infty}\int_{0}^{\pi}\frac{Re^{it}}{1+R^2e^{2it}}\, dt=\int_{0}^{\pi}\lim_{R\to +\infty}\frac{1}{R}\frac{e^{it}}{\frac{1}{R^2}+e^{2it}}\, dt=0 \end{equation} which yields that \begin{equation}\lim_{R\to +\infty}\int_{-R}^{+R}\frac{1}{1+x^2}\, dx=\lim_{R\to +\infty}\int_{[-R,R]}f(z)\, dz=\pi -\lim_{R\to +\infty}\int_{\gamma_{R}}f(z)\, dz=\pi \end{equation} Therefore, \begin{equation}\int_{0}^{+\infty}\frac{1}{1+x^2}\, dx=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{1}{1+x^2}\, dx=\frac{\pi}{2}\end{equation}

We will now calculate $\int_{0}^{1}\frac{1}{1+x^2}\, dx$

\begin{gather}\int\limits_{0}^{1}\frac{dx}{1+x^2}=\int\limits_{0}^{1}\sum\limits_{k=0}^{n}(-1)^k x^{2k}\, dt+\int\limits_{0}^{1}\frac{(-1)^{n+1}x^{2n+2}}{1+x^2}\, dt=\\ \sum\limits_{k=0}^{n}\frac{(-1)^k}{2k+1}+(-1)^{n+1}\int\limits_{0}^{1}\frac{x^{2n+2}}{1+x^2} dt\end{gather} Since \begin{equation}\left|\int\limits_{0}^{1}(-1)^{n+1}\frac{x^{2n+2}}{1+x^2}\, dt\right|= \int\limits_{0}^{1}\frac{x^{2n+2}}{1+x^2}\, dt\le \int\limits_{0}^{1}x^{2n+2}\, dt=\frac{1}{2n+3} \to 0\end{equation} and $$\sum\limits_{k=0}^{+\infty}\frac{(-1)^k}{2k+1}=\frac{\pi}{4}$$ it follows that $$\int\limits_{0}^{1}\frac{dt}{1+x^2}=\frac{\pi}{4}$$

The required integral is \begin{equation}\int_{1}^{+\infty}\frac{1}{1+x^2}\, dx=\int_{0}^{+\infty}\frac{1}{1+x^2}\, dx-\int_{0}^{1}\frac{1}{1+x^2}\, dx=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}\end{equation}

share|improve this answer
2  
@Nameless It would've manically depressed me if you deleted this answer on the sole basis of not noting the bounds of integration. +1 –  000 Dec 8 '12 at 20:30
2  
Why use residues at all if you know the antiderivatve is $\arctan$ and you even use it to find a definite integral of $\frac{1}{x^2+1}$, (albeit with different bound) anyways? –  Argon Dec 8 '12 at 20:31
3  
@Argon Because $$\text{overkill}\cap \text{mathematics}\ne \emptyset.$$ I found this answer very entertaining and interesting. –  000 Dec 8 '12 at 20:33
2  
Why would you need to use residues even if the integral was from $0$ to $\infty$? –  yunone Dec 8 '12 at 20:35
2  
@Limitless Residues are not just overkill - they aren't necessary here at all. If the antiderivative is required and found anyways, why use residues? –  Argon Dec 8 '12 at 20:37

The key point is that you do not evaluate the antiderivative at $\infty$ since most functions aren't defined for $x=\infty$. You take the limit: $$\int_a^{\infty}f(x)dx=\lim_{m \to \infty}\int_a^{m}f(x)dx=\lim_{m \to \infty}F(m)-F(a).$$

(It's punny because I'm Limitless.)

share|improve this answer
1  
BAD PUN! :) ${}$ –  Argon Dec 8 '12 at 20:32

$$\int_1^\infty \frac{dx}{x^2+1} = \arctan x\bigg|_1^\infty = \lim_{x \to \infty}\arctan x - \arctan 1 = \frac{\pi}{2} -\frac{\pi}{4} = \frac{\pi}{4}$$

as you found.

share|improve this answer
    
+1 Neat and sweet. –  amWhy Jan 6 '13 at 1:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.