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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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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

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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?2012-12-08
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    @Limitless Not quite sure what you mean by explicitly relate it to the unit circle... I'll see about creating a graphic2012-12-09
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    I envision the triangle as part of the unit circle.2012-12-09
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    Picture has been added (@Limitless)2012-12-09
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    Great picture! :-)2012-12-09
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$$\int_1^\infty \frac{dx}{x^2+1} = \arctan x\bigg|_1^\infty = \lim_{x \to \infty}\arctan x - \arctan 1 = \frac{\pi}{2} -\frac{\pi}{4} = \frac{\pi}{4}$$

as you found.

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    Why ingnore the period?2018-03-05
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    @DeepNorth What period? If you mean that the general inverse tan has a $\pi n$ term, it cancels out from when taking the difference between the two arctans.2018-03-05
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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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    The integral is from $1$ to $\infty$.2012-12-08
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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. +12012-12-08
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    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?2012-12-08
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    @Argon I agree. I thought the integral in question was from $0$ to $\infty$ and that's why I used residues. When I was informed by Antonio that the integral is from $1$ to $\infty$ I tried to modify it.2012-12-08
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    @Argon Because $$\text{overkill}\cap \text{mathematics}\ne \emptyset.$$ I found this answer very entertaining and interesting.2012-12-08
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    Why would you need to use residues even if the integral was from $0$ to $\infty$?2012-12-08
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    @Limitless Residues are not just overkill - they aren't necessary here at all. If the antiderivative is required and found anyways, why use residues?2012-12-08
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    @Argon That is a very fair point. I guess the only justification at this point would be stating that Nameless wanted to translate this problem to complex analysis, albeit it would appear there is nothing profound or notable about $$\int \frac{1}{1+x^2}dx=\arctan(x)+C$$ in terms of complex analysis techniques. I may very well be wrong, however. Actually: It is notable that $$\begin{equation}\int_{0}^{+\infty}\frac{1}{1+x^2}\, dx=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{1}{1+x^2}\, dx=\frac{\pi}{2}\end{equation},$$ which was shown via complex analysis. However, it requires an (cont below)2012-12-08
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    unnecessary use of (I forget the official term, rewriting bounds?). The part where Nameless states $\int_1^{+\infty}=\int_{0}^{+\infty}-\int_0^{1}.$2012-12-08
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    @Argon Yes, I see that and I agree with you. Note my edits above. We should challenge Nameless to use purely complex analysis to evaluate this integral! :-)2012-12-08
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    @Limitless I removed the need of $\arctan$ completely (although the series for $\frac{\pi}{4}$ is a consequence of $\arctan(1)=\frac{\pi}{4}$)2012-12-08
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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! :) ${}$2012-12-08