# Other ways to evaluate the integral $\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} \, dx$?

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

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

• Fourier inversion? – Angina Seng Jul 7 '17 at 7:27
• you can use the primitve of $1/(x^2+1)$ :) – tired Jul 7 '17 at 7:28
• You can apply the result on this question, since $\frac{1}{x^2+1}$ is an even function, therefore: $$\int_{-\infty}^{\infty} \frac{1}{x^2+1}~dx=2\int_0^{\infty} \frac{1}{x^2+1}~dx$$ – projectilemotion Jul 7 '17 at 7:38
• 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? – Sangchul Lee Jul 7 '17 at 7:38
• nice @SangchulLee – user440024 Jul 7 '17 at 7:40

Here is a solution using trigonometry. Consider the following situation:

$\hspace{5em}$

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.

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 $\arctan x$.

• Thanks @ChrisTaylor , i did the same using complex numbers – user440024 Jul 7 '17 at 7:42

\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

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 $\omega=0$ $$\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{x^2+1}=\pi$$

• thanks i don't know fourier transforms though – user440024 Jul 7 '17 at 7:34

What about Euler's Beta function? We have

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


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}

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

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

• sir i knew this approach , i asked something other than this one – user440024 Jul 7 '17 at 7:33

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.