Approximating $\pi$ and $\ln 2$ with $I_k=\int_0^\infty \left(\text{sech}x\tanh\tfrac12x\right)^k\,dx$ for integer $k$ Consider the following integral:
$$I_k=\int_0^\infty \left(\text{sech}x\tanh\tfrac12x\right)^k\,dx$$
where $k\in\Bbb N$.
If we evaluate $I_1$, $I_2$, $I_3$, etc. we get the following pattern:


*

*$I_1=\log(2)$

*$I_2=-3+\pi$

*$I_3=\frac 72-5\log(2)$

*$I_4=22-7\pi$

*$I_5=-\frac{341}{12}+41\log(2)$

*$I_6=-\frac{968}{5}+\frac{493}{8}\pi$

*...


From this data we can see that:
$$\begin{align}
\pi&=3+I_2=\frac{22}{7}-\frac17 I_4=\frac{7744}{2465}+\frac{8}{493}I_6 \\[4pt]
\log(2)&=0+I_1=\frac{7}{10}-\frac15I_3=\frac{341}{492}+I_5
\end{align}$$
And because $I_k$ decreases very rapidly($I_{10}$ is in the order of $1e6$) we can set $I_k\approx 0$ for high $k$ and we get rational approximations of both, $\pi$ and $\log(2)$, for $I_{2k}$ and $I_{2k+1}$, respectively, that apparently go on forever.

I see some equations that somehow "encode" the information of a given number, but how is it that this integral has the information of both $\pi$ and $\log(2)$, apparently unrelated numbers?

Thanks.
 A: I'm not sure exactly what kind of an answer you're looking for---but the following might be a helpful heuristic:


*

*A standard substitution transforms the integrals $$I_k=\int_0^\infty \left(\operatorname{sech} x \tanh\frac{x}{2}\right)^k\,dx ,$$ (for $k$ a nonnegative integer, which we henceforth assume) into integrals of rational functions of a new variable.

*After applying the method of partial fractions, we can integrate term-by-term, and the only non-rational functions that occur in the antiderivatives have the form $\log q(u)$ and $\arctan r(u)$ for some affine functions $q, r$ with rational coefficients.

*At typical limits of integration the functions $q(u)$ usually take on values of rational number with small numerator and denominator, and so the contribution of these terms is a sum of logarithms of small numbers---of course, the smallest positive integer with nonzero logarithm is $2$, so it's no surprise that $\log 2$ occurs in the values of such integrals often.

*Likewise, if we can write $\arctan v$ without $\arctan$, typically the result is some rational multiple of $\pi$. In particular, we have $\arctan 1 = \frac{\pi}{4}$ and $\lim_{v \to \infty} \arctan v = \frac{\pi}{2}$, no it's no surprise that $\pi$ occurs often either.
More explicitly: Applying the hyperbolic analogue $$x = 2 \operatorname{artanh} t$$ of the Weierstrass substitution transforms $I_k$ into an integral with a rational integrand,
$$I_k = 2 \int_0^1 \frac{(1 - t^2)^{k - 1} t^k \,dt}{(1 + t^2)^k}.$$
The next step for evaluating this integral in terms of elementary functions depends on the parity of $k$.
For $k$ odd, substituting $$u = t^2, \qquad du = 2 t \,dt$$ gives
$$I_k = \int_0^1 \frac{(1 - u)^{k - 1} u^{(k - 1) / 2}}{(u + 1)^k} .$$
Expanding the integrand using partial fractions gives
$$I_k = \int_0^1 \left(P(u) + \frac{A_t}{(u + 1)^k} + \cdots + \frac{A_2}{(u + 1)^2} + \frac{A_1}{u + 1} \right) du $$
for some rational polynomial $P(u)$ and rational coefficients $A_1, \ldots, A_k$. But the antiderivative of every term but $\frac{A_1}{u + 1}$ is a rational function, so each contributes some rational number, altogether contributing a rational total, call it $\alpha$. The value of the last term is $A_1 \int_0^1 \frac{du}{u + 1} = A_1 \log u\vert_0^1 = A_1 \log 2$, so
$$\boxed{I_k = \alpha_k + \beta_k \log 2}$$ for some rational numbers $\alpha_k$ and $\beta_k := A_1$. A tedious but straightforward induction shows that $\beta_k \neq 0$. 
For $k$ even, that substitution is not available (or more precisely, it makes the integrand worse), but applying the method of partial fractions again gives
$$I_k = \int_0^1 \left(P(t) + \frac{A_k t + B_k}{(t^2 + 1)^k} + \cdots + \frac{A_2 t + B_2}{(t^2 + 1)^2} + \frac{A_1 t + B_1}{(t^2 + 1)} \right) dt $$ for some polynomial $P(t)$ and rational coefficients $A_1, \ldots, A_k, B_1, \ldots, B_k$. But our integrand in $u$ is even, and so $A_1 = \cdots = A_k = 0$. On the other hand, for $m > 1$, the integral of $\frac{1}{(t^2 + 1)^m}$ satisfies the reduction formula of the form
$$\int \frac{dt}{(t^2 + 1)^m} = f_m(t) + \gamma_m \int \frac{dt}{(t^2 + 1)^{m - 1}}$$ for some rational function $f_m$ and rational constant $\gamma_m$ (both depending on $m$). In particular, induction gives $$\int \frac{dt}{(t^2 + 1)^m} = g_m(x) + \delta_m \int \frac{dx}{t^2 + 1} = g_m(t) + \delta_m \arctan t + C$$ for some rational function $g_m$ and rational constant $\delta_m$ (again both depending on $m$). Substituting back into our previous formula for $I_k$, we have that
$$I_k = [h(t) + \zeta \arctan t]\vert_0^1 = h(1) - h(0) + \beta_k \pi$$ for some rational function $h$ and rational constant $\beta_k$, and so
$$\boxed{I_k = \alpha_k + \beta_k \pi}$$
for rational numbers $\alpha_k := h(1) - h(0)$ and $\beta_k$. Again, an induction shows that $\beta_k \neq 0$ for $k > 0$.
Remark 1 Essentially the same phenomenon occurs for the similar family
$$J_j := \int_0^1 \frac{x^{2 j} (1 - x)^{2 j}dx}{1 + x^2} ,$$
which generalizes the so-called Dalzell integral (the case $j = 2$, which gives $\frac{22}{7} - \pi$). If $j$ is odd, we get an expression of the form $\alpha + \beta \log 2$, and if $j$ is even we get $\alpha + \beta \pi$, with $\alpha, \beta \neq 0$ in both cases.
Remark 2 We can use the resulting integral expressions to derive rational approximations of $\pi$ and $\log 2$. On the interval $[0, 1]$, $\frac{1}{2^k} \leq \frac{1}{(1 + t^2)^k} \leq 1$, giving the bounds
$$ \frac{1}{2^k} E_k < I_k < E_k, \\ \textrm{where} \quad E_k = 2 \int_0^1 u^k (1 - u^2)^k du = \frac{\Gamma(k) \Gamma\left(\frac{1}{2} k + \frac{1}{2}\right)}{\Gamma\left(\frac{3}{2} k + \frac{1}{2}\right)} \sim \frac{\sqrt{2 \pi}}{\sqrt{k}} \left(\frac{2}{3 \sqrt{3}}\right)^k ,$$
so for any particular $k$, rearranging gives rational bounds for $\pi$ or $\log 2$. (For odd $k = 2 l + 1$, by the way, we have $E_{2 l + 1} = \frac{(2 l)! l!}{(3 l + 1)!}$.)
For example, taking $k = 2$ gives $I_2 = \pi - 3$ and $L_2 = \frac{16}{105}$,
and rearranging gives
$$\frac{319}{105} < \pi < \frac{331}{105} .$$
A: Too long for a comment.
We have the convenient relation 
$$\text{sech}(x)\tanh(x/2)=\frac{2}{e^x+1}-\frac{2}{e^{2x}+1},$$
so let 
$$L_n=I_{2n+1}=2^{2n+1}\sum_{k=0}^{2n+1}(-1)^k{2n+1\choose k}\int_0^{\infty}\frac{dx}{(e^x+1)^{2n-k+1}(e^{2x}+1)^k}$$
and $$P_n=I_{2n}=2^{2n}\sum_{k=0}^{2n}(-1)^k{2n\choose k}\int_0^{\infty}\frac{dx}{(e^x+1)^{2n-k}(e^{2x}+1)^k}.$$
These integrals may be easier to evaluate explicitly.
