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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.

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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 $\frac{y}{x}$:


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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 –  apnorton 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) –  apnorton Dec 9 '12 at 2:36
Great picture! :-) –  000 Dec 9 '12 at 2:38

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

as you found.

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+1 Neat and sweet. –  amWhy Jan 6 '13 at 1:28

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}

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@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
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
@Argon Because $$\text{overkill}\cap \text{mathematics}\ne \emptyset.$$ I found this answer very entertaining and interesting. –  000 Dec 8 '12 at 20:33
Why would you need to use residues even if the integral was from $0$ to $\infty$? –  yunone Dec 8 '12 at 20:35
@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.)

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BAD PUN! :) ${}$ –  Argon Dec 8 '12 at 20:32

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