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$$\int_{-\infty}^{\infty}\frac{1}{x^2+1}\,dx=\pi $$

I can do it with the substitution $x= \tan u$ or complex analysis. Are there any other ways to evaluate this?

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    $\begingroup$ Fourier inversion? $\endgroup$ – Lord Shark the Unknown Jul 7 '17 at 7:27
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    $\begingroup$ you can use the primitve of $1/(x^2+1)$ :) $\endgroup$ – tired Jul 7 '17 at 7:28
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    $\begingroup$ You can apply the result on this question, since $\frac{1}{x^2+1}$ is an even function, therefore: $$\int_{-\infty}^{\infty} \frac{1}{x^2+1}~dx=2\int_0^{\infty} \frac{1}{x^2+1}~dx$$ $\endgroup$ – projectilemotion Jul 7 '17 at 7:38
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    $\begingroup$ Does $$\int_{-\infty}^{\infty} \frac{dx}{x^2+1} = 4\int_{0}^{1} \frac{dx}{x^2+1} = 4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} = 4\cdot\frac{\pi}{4} = \pi $$ count? $\endgroup$ – Sangchul Lee Jul 7 '17 at 7:38
  • $\begingroup$ nice @SangchulLee $\endgroup$ – JohnnySinns Jul 7 '17 at 7:40
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Here is a solution using trigonometry. Consider the following situation:

$\hspace{5em}$enter image description here

Since triangles $\triangle CP_1P_2$, $\triangle CQ_1Q_2$ and $\triangle CR_1R_2$ are congruent with ratio

$$1 \ : \ \frac{1}{\sqrt{1+t^2}} \ : \ \frac{1}{\sqrt{1+(t+\Delta t)^2}},$$

it follows that the area of the wedge $CQ_1R_2$, which equals $\frac{1}{2}\angle Q_1 C R_2$, is bounded between

$$ \frac{\Delta t}{2(1+(t+\Delta t)^2)} = \mathrm{Area}(\triangle CR_1R_2) \leq \frac{1}{2}\angle Q_1 C R_2 \leq \mathrm{Area}(\triangle CQ_1Q_2) = \frac{\Delta t}{2(1+t^2)}. $$

Hence for any $\theta \in (0,\frac{\pi}{2})$ and for any partition $\Pi = \{0 = t_0 < t_1 < \cdots < t_n = \tan\theta \}$ we have

$$ \sum_{i=1}^{n} \frac{\Delta t_i}{1+t_i^2} \leq \theta \leq \sum_{i=1}^{n} \frac{\Delta t_i}{1 + t_{i-1}^2}, \qquad (\Delta t_i = t_i - t_{i-1}). $$

Taking the limit $\|\Pi\|\to 0$, the squeezing lemma tells

$$ \int_{0}^{\tan\theta} \frac{dt}{1+t^2} = \theta. $$

Then taking $\theta \uparrow \frac{\pi}{2}$ proves the desired identity through symmetry.

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You can use partial fractions:

$$ \begin{align} \int_{-\infty}^\infty \frac{dx}{1+x^2} & = \int_{-\infty}^\infty \frac{1}{2} \left( \frac{1}{1+ix} + \frac{1}{1-ix} \right) dx \\ & = \frac{1}{2i} \bigg[\log(1+ix) - \log(1-ix)\bigg]_{-\infty}^\infty \\ & = \frac{1}{2i} \left[ \lim_{x\to\infty} \log\left( \frac{1+ix}{1-ix} \right) - \lim_{x\to-\infty} \log\left(\frac{1+ix}{1-ix}\right)\right] \\ & = \frac{1}{2i}\bigg[ i\pi - (-i\pi)\bigg] = \pi \end{align} $$

Note that the expression on the second line obviously must be equal to

$$ \bigg[ \arctan x\bigg]_{-\infty}^\infty $$

because we know the antiderivative of $1/(1+x^2)$ is $\arctan x$.

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  • $\begingroup$ Thanks @ChrisTaylor , i did the same using complex numbers $\endgroup$ – JohnnySinns Jul 7 '17 at 7:42
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\begin{align} \int_{-\infty}^{\infty}\frac{dx}{1+x^2}&=2\int_{0}^{\infty}\frac{dx}{1+x^2}\\ &=2\int_{0}^{1}\frac{dx}{1+x^2}+2\int_{1}^{\infty}\frac{dx}{1+x^2}\\ &=4\int_{0}^{1}\frac{dx}{1+x^2}\\ &=4\int_{0}^{1}\sum_{j=0}^{\infty}\left(-x^2\right)^jdx\\ &=4\sum_{j=0}^{\infty}\left(-1\right)^j\int_{0}^{1}\left(x^2\right)^jdx\\ &=4\sum_{j=0}^{\infty}\frac{(-1)^j}{2j+1}\\ &=4\times \frac{\pi}{4} \end{align}

Note

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I was learning some Fourier analysis recently , so here is a little overkill

Using the well known FT of $e^{-a|t|}$ $$\mathcal{F} \left\{e^{-a|t|}\right\}=\frac{2a}{a^2+\omega^2}$$

It's nothing but evaluation of this integral $$\mathcal{F} \left\{e^{-a|t|}\right\}=\int_{-\infty}^{\infty}e^{-j\omega t}\cdot e^{-a|t|}\,dt$$ Then using Duality

$$x(t) \Leftrightarrow X(\omega) \implies X(t)\Leftrightarrow2\pi \cdot x(-\omega) $$

$$e^{-a|t|} \Leftrightarrow \frac{2a}{a^2+\omega^2} $$ $$ \frac{2a}{a^2+t^2}\Leftrightarrow 2\pi \cdot e^{-a|\omega|} $$

$$\frac{1}{a^2+t^2} \Leftrightarrow \left(\frac{\pi}{a}\right)\cdot e^{-a|\omega|} $$

$$\frac{1}{1+t^2} \Leftrightarrow \pi \cdot e^{-|\omega|} $$ So your integral evaluates at $a=1$ and $\omega=0$ $$\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{x^2+1}=\pi$$

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  • $\begingroup$ thanks i don't know fourier transforms though $\endgroup$ – JohnnySinns Jul 7 '17 at 7:34
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What about Euler's Beta function? We have

$$\begin{eqnarray*} \int_{-\infty}^{+\infty}\frac{dt}{t^2+1}&\stackrel{\text{parity}}{=}&2\int_{0}^{+\infty}\frac{dt}{t^2+1}\\&\stackrel{\frac{1}{t^2+1}\mapsto u}{=}&\int_{0}^{1}u^{-1/2}(1-u)^{-1/2}\,du\\&=&\Gamma\left(\frac{1}{2}\right)^2=\color{red}{\pi}.\end{eqnarray*} $$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{-\infty}^{\infty}{\dd x \over 1 + x^{2}} & = 2\int_{0}^{\infty}{\dd x \over 1 + x^{2}} \\[5mm] & = 2\lim_{\substack{R \to \infty\\ \epsilon \to 0^{+}}}\left[% -\,\Re\int_{0}^{\pi/2}{R\expo{\ic\theta}\ic\,\dd\theta \over 1 + R^{2}\expo{2\ic\theta}} - \Re\int_{\infty}^{1 + \epsilon}{\ic\,\dd y \over 1 - y^{2}}\right. \\[2mm] & \phantom{2\lim_{\substack{R \to \infty\\ \epsilon \to 0^{+}}}\left[AA\,\right.} \left.-\Re\int_{\pi/2}^{-\pi/2}{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over 1 + \pars{\ic^{2} + 2\ic\epsilon\expo{\ic\theta} + \epsilon^{2}\expo{2\ic\theta}}} - \Re\int_{1 - \epsilon}^{0}{\ic\,\dd y \over 1 - y^{2}}\right\rbrack \\[1mm] & = 2\lim_{\epsilon \to 0^{+}}\Re\int_{-\pi/2}^{\pi/2}{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over 2\ic\epsilon\expo{\ic\theta}} = \bbx{\pi} \end{align}

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It's a bit of overkill, but since $f(x) = \frac{1}{1+x^{2}}$ is a monotonically decreasing function on $(0, \infty)$, the Riemann-like sum $$\frac{1}{n}\sum_{k=0}^{{\color{red}{\infty}}} f \left(\frac{k}{n} \right)= \frac{1}{n} \sum_{k=0}^{\infty} \frac{1}{1+ \left(\frac{k}{n} \right)^{2}} = n\sum_{k=0}^{\infty} \frac{1}{n^{2}+k^{2}} $$ converges to the value of $\int_{0}^{\infty} \frac{dx}{1+x^{2}}$ as $n \to \infty$.

And the partial fraction expansion of $\coth (\pi z)$ is $$\coth(\pi z) = \frac{1}{\pi z} + \frac{2z}{\pi} \sum_{k=1}^{\infty} \frac{1}{z^{2}+k^{2}} = - \frac{1}{\pi z} + \frac{2z}{\pi} \sum_{k=0}^{\infty} \frac{1}{z^{2}+k^{2}} . $$

Therefore, $$ \begin{align}\int_{-\infty}^{\infty} \frac{dx}{1+x^{2}} &= 2 \int_{0}^{\infty} \frac{dx}{1+x^{2}} \\ &= 2 \lim_{n \to \infty} \left(\frac{1}{2n} + \frac{\pi}{2} \, \coth(\pi n) \right) \\ &= 2 \left(0+ \frac{\pi}{2}(1) \right) \\ &= \pi. \end{align}$$

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An antiderivative of $\frac{1}{x^2+1}$ is $\arctan(x)$, hence

$\int_{-\infty}^{\infty}\frac{1}{x^2+1}\,dx=2 \int_{0}^{\infty}\frac{1}{x^2+1}\,dx=2 \lim_{t \to \infty} \arctan(t)= \pi$

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  • $\begingroup$ sir i knew this approach , i asked something other than this one $\endgroup$ – JohnnySinns Jul 7 '17 at 7:33
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Derivation using series the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{\pi}{4}$.

The integrand is an even function & break the interval at $1$ and we have \begin{eqnarray*} \int_{-\infty}^{\infty} \frac{dx}{1+x^2} =2 \int_{0}^{\infty} \frac{dx}{1+x^2} =2 \int_{0}^{1} \frac{dx}{1+x^2}+2 \underbrace{\int_{1}^{\infty} \frac{dx}{1+x^2}}_{x \rightarrow \frac{1}{x}}=4 \int_{0}^{1} \frac{dx}{1+x^2} \end{eqnarray*} Now geometrically expand the integrand \begin{eqnarray*} \frac{1}{1+x^2} =\sum_{i=0}^{\infty} (-1)^n x^{2n} \end{eqnarray*} interchange the order of the integration & sum ... & then perform the each integration.

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