How may one show that

$$\int_{-\infty}^\infty \frac{\mathrm dx}{(1+x^2)(\pi+\cos x)} =\color{red}1\tag1$$

$x=\tan y$ then $\mathrm dx=\sec^2 y \, \mathrm dy$

$$\int_{-\pi/2}^{\pi/2} \sqrt{1+y^2} \cdot {\mathrm dy\over 1+\pi\sqrt{1+y^2}} \tag2$$

  • $\begingroup$ Your substitution looks wrong. Where did you find this integral, and why do you think it is equal to 1? $\endgroup$ – mickep Nov 5 '17 at 18:30
  • $\begingroup$ on my book and the answer one, but I am not it is 1 $\endgroup$ – user499615 Nov 5 '17 at 18:32
  • $\begingroup$ If you let $x=\tan y$, you will for sure have a term $\cos(\tan y)$ that looks difficult to get rid of. $\endgroup$ – mickep Nov 5 '17 at 18:32
  • $\begingroup$ Wolfram can't seem to compute it. $\endgroup$ – Ivo Terek Nov 5 '17 at 18:33
  • $\begingroup$ yeah you are right @mickep $\endgroup$ – user499615 Nov 5 '17 at 18:34

No, the integral doesn't integrate to $1$. The correct value is around $0.9352357$.

Since the derivation is a little bit messy, I will skip justification of most algebraic manipulations and test of convergence below. Furthermore, all sums of the form $\sum_n (\cdots)$ should be interpreted as $\lim_{N\to\infty} \sum_{n=-N}^N(\cdots)$.

Notice $\cos(x)$ is a periodic function of period $2\pi$, we have

$$\begin{align} \mathcal{I} \stackrel{def}{=} \int_{-\infty}^\infty \frac{dx}{(1+x^2)(\pi + \cos(x))} &= \left( \sum_n \int_{(2n-1)\pi}^{(2n+1)\pi}\right) \frac{dx}{(1+x^2)(\pi + \cos(x))}\\ &= \int_{-\pi}^\pi \left(\sum_n \frac{1}{(x+2n\pi)^2+1}\right)\frac{dx}{\pi + \cos(x)} \end{align}\tag{*1} $$

Let $t = \tan\frac{x}{2}$, $\alpha = \tanh\frac12$ and $\beta = \sqrt{\frac{\pi+1}{\pi-1}}$.

Start from the infinite product expansion of $\sin(x)$,

$$\sin(x) = x \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)$$ one can deduce $$\cot(x) = \sum_n\frac{1}{x+n\pi}\quad\implies\quad \frac12\cot\frac{x}{2} = \sum_n \frac{1}{x+2n\pi}$$ With a little bit of algebra, this leads to $$\begin{align}\sum_n \frac{1}{(x+2n\pi)^2+1} &= \frac{1}{4i}\left[\cot\left(\frac{x - i}{2}\right) - \cot\left(\frac{x+i}{2}\right)\right]\\ &= \frac{1}{4i}\left(\frac{1 + i \alpha t}{t - \alpha i} - \frac{1 - i\alpha t}{t + \alpha i}\right) = \frac{\alpha}{2}\left(\frac{1+t^2}{t^2+\alpha^2}\right) \end{align} $$ Substitute this into $(*1)$, use the fact both $\cos(x)$ and above function is an even function and change variable to $t$, we obtain

$$\begin{align}\mathcal{I} &= \alpha \int_0^{\pi}\frac{(1+t^2)}{(t^2+\alpha^2)(\pi + \cos x)}dx = \alpha \int_0^\infty \frac{(1+t^2)}{(t^2+\alpha^2)\left(\pi + \frac{1-t^2}{1+t^2}\right)}\frac{2dt}{1+t^2}\\ &= \frac{\alpha}{\pi-1} \int_{-\infty}^\infty \frac{1+t^2}{(t^2+\alpha^2)(t^2+\beta^2)}dt \end{align} $$ Notice $$\frac{1+t^2}{(t^2+\alpha^2)(t^2+\beta^2)} = \frac{1-\alpha^2}{(t^2+\alpha^2)(-\alpha^2+\beta^2)} + \frac{1-\beta^2}{(-\beta^2+\alpha^2)(t^2+\beta^2)}$$

We find $$\begin{align} \mathcal{I} &= \frac{\pi\alpha}{(\pi-1)(\beta^2-\alpha^2)}\left[\frac{1-\alpha^2}{\alpha} - \frac{1-\beta^2}{\beta}\right] = \frac{\pi}{(\pi - 1)\beta}\frac{1+\alpha\beta}{\beta+\alpha}\\ &= \frac{\pi}{\sqrt{\pi^2-1}}\frac{\sqrt{\pi-1} + \sqrt{\pi+1}\tanh\frac12} {\sqrt{\pi+1} + \sqrt{\pi-1}\tanh\frac12}\\ &\approx 0.93523570580939545599059373053102730519937591000461724... \end{align}\tag{*2} $$ As a double check, I have evaluated $\mathcal{I}$ numerically on WA using two equivalent form of the integral:

  • N[Int[1/((1+x^2)*(Pi+Cos[x])),{x,-Inf,Inf}],50] gives $0.935197907462057583934216094872870832745553392068185$

  • N[Int[1/(Cos[Tan[y]]+Pi),{y,-Pi/2,Pi/2}],50] gives $0.935221913498036727655246950920018882262473130195986$

One can see that our analytic result is compatible with the part where above two numerical methods agree ($\approx 0.9352$).


It turns out there is an alternate (simpler?) way to evaluate this integral.

Pick a $\gamma > 1$ such that $\pi = \frac{\gamma^2+1}{2\gamma}$ (i.e. set $\gamma = \pi + \sqrt{\pi^2-1}$ ), we have

$$\frac{1}{\pi + \cos(z)} = \frac{2\gamma}{1+\gamma^2 + 2\gamma\cos(z)} = \frac{2\gamma}{(\gamma + e^{iz})(\gamma + e^{-iz})} = \frac{\gamma}{\gamma^2-1}\left[\frac{\gamma - e^{iz}}{\gamma + e^{iz}} + \frac{\gamma-e^{-iz}}{\gamma + e^{-iz}}\right] $$ This leads to $$\mathcal{I} = \frac{\gamma}{\gamma^2-1}\int_{-\infty}^\infty \left[\frac{\gamma - e^{iz}}{\gamma + e^{iz}} + \frac{\gamma-e^{-iz}}{\gamma + e^{-iz}}\right] \frac{dz}{1+z^2}\tag{*3}$$ On upper complex plane,

$$\Im z \ge 0 \quad\implies\quad |e^{iz}| \le 1 \quad\implies\quad \frac{\gamma - e^{iz}}{\gamma + e^{iz}} \quad\text{ is bounded and analytic there.} $$ On lower complex plane, $$\Im z \le 0 \quad\implies\quad |e^{-iz}| \le 1 \quad\implies\quad \frac{\gamma - e^{-iz}}{\gamma + e^{-iz}} \quad\text{ is bounded and analytic there.} $$ We can evaluate the integral $(*3)$ by splitting what is in the square bracket into two pieces. The first piece involving $\frac{\gamma - e^{iz}}{\gamma + e^{iz}}$ can be computed by completing the contour with a infinite large circle in upper complex plane and then taking residue at pole $i$ of $\frac{1}{1+z^2}$. Doing essentially the same thing to the second piece but in lower complex plane and taking residue at pole $-i$, we obtain:

$$\begin{align}\mathcal{I} &= \frac{\gamma}{\gamma^2-1} \left[ \left(\frac{2\pi i}{2i}\right)\left(\frac{\gamma - e^{i(i)}}{\gamma + e^{i(i)}}\right) + \left(\frac{-2\pi i}{-2i}\right)\left(\frac{\gamma - e^{-i(-i)}}{\gamma + e^{-i(-i)}}\right) \right]\\ &= \frac{2\pi\gamma}{\gamma^2-1}\left(\frac{\gamma - e^{-1}}{\gamma + e^{-1}}\right) = \frac{\pi}{\sqrt{\pi^2-1}}\left(\frac{\gamma e - 1}{\gamma e + 1}\right) \end{align} $$ Notice $$2\gamma = (\sqrt{\pi + 1} + \sqrt{\pi - 1})^2 \quad\implies\quad \begin{cases} \sqrt{\pi+1} + \sqrt{\pi-1} = \sqrt{2\gamma}\\ \sqrt{\pi+1} - \sqrt{\pi-1} = \frac{2}{\sqrt{\pi+1} - \sqrt{\pi-1}} = \sqrt{\frac{2}{\gamma}}\end{cases}$$

With a little bit of algebra, we can transform above result to $$\mathcal{I} = \frac{\pi}{\sqrt{\pi^2-1}}\left[\frac{\sqrt{\pi-1}(e+1) + \sqrt{\pi+1}(e-1)}{ \sqrt{\pi+1}(e+1) + \sqrt{\pi-1}(e-1)}\right] = \frac{\pi}{\sqrt{\pi^2-1}} \left[\frac{\sqrt{\pi-1} + \sqrt{\pi+1}\tanh\frac12}{ \sqrt{\pi+1} + \sqrt{\pi-1}\tanh\frac12}\right] $$ Reproducing what has been derived by another method.


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