Discovering $I_n=\int_0^{\frac{\pi}{2}} \frac{dx}{(1+\tan x)^{4n+2}}$ is rational.

Question

Latest Edit

Great thanks to Quanto who gave a complete proof to the question by setting up a simple reduction formula on $I_n$ below:

\begin{align} \boxed{I_{n} =\ \frac1{2(4n+1)}+ \frac1{2(4n)}+ \frac1{4(4n-1)}-\frac14I_{n-1}} \end{align}

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Let’s start with the easy one. $$ \begin{aligned}I_2& =\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{\left[1+\tan \left(\frac{\pi}{4}-x\right)\right]^2} d x \\ & =\frac{1}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(1+\tan x)^2 d x \\ & =\frac{1}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(1+2 \tan x+\tan ^2 x\right) d x \\ & =\frac{1}{2} \int_0^{\frac{\pi}{4}}\left(1+\tan ^2 x\right) d x \quad \textrm{ (Properties of odd and even functions.)} \\ & =\frac{1}{2} \int_0^{\frac{\pi}{4}} \sec ^2 x d x \\ & =\frac{1}{2}[\tan x]_0^{\frac{\pi}{4}} \\ & =\frac{1}{2} \end{aligned} $$

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To proceed, I tried further with the same method of evaluating $I_2$ by putting $x\mapsto \frac{\pi}{4} -x$ which transforms

$$ \begin{aligned} I_n & =\frac{1}{2^n} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(1+\tan x)^n d x \\ & =\frac{1}{2^n} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) \tan ^k x d x \\ & =\frac{1}{2^{n-1}} \sum_{k=0}^{\left[\frac{n}{2}\right]}\left(\begin{array}{l} n \\ 2 k \end{array}\right) \int_0^{\frac{\pi}{4}} \tan ^{2 k} x d x \quad \textrm{ (Properties of odd and even functions.)} \end{aligned} $$ Using the reduction formula for $\int_0^{\frac{\pi}{4}} \tan ^{2 k} x d x,$ we can continue our evaluation on $I_n$ as below: $$ \begin{aligned} I_3=-\frac{\pi}{8}+\frac{3}{4},\quad I_4=\frac{2}{3} -\frac{\pi}{8}, \quad I_5=\frac{5}{12}-\frac{\pi}{16}, \end{aligned} $$

For $n=6$, letting $x\mapsto \frac{\pi}{4}-x $ transforms the integral into $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{d x}{(1+\tan x)^6} & =\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left[1+\tan \left(\frac{\pi}{4}-x\right)\right]^6} \\ & =\frac{1}{64} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(1+\tan x)^6 d x\\\\ \end{aligned} $$ Expanding the integrand into $3$ odd and $4$ even functions simplifies it into $$ \begin{aligned}I_6&=\frac{1}{32} \int_0^{\frac{\pi}{4}}\left(1 +\tan ^6 x +15 \tan ^2 x+15 \tan ^4 x\right) d x\\&= \frac{1}{32} \int_0^{\frac{\pi}{4}}\left(1-\tan ^2 x+\tan ^4 x+15 \tan ^2 x\right) d(\tan x)\\& =\frac{1}{32}\left[\tan x+\frac{14 \tan ^3 x}{3}+\frac{\tan ^5 x}{5}\right]_0^{\frac{\pi}{4}}\\&=\frac{11}{60} \end{aligned} $$

When evaluating $I_6$, I found the symmetry of the binomial coefficients in expansion, I can group the corresponding terms together and found that $I_n$ has no $\pi$ when $n\equiv 2 \quad \pmod 4$, i.e. $$I_{4n+2}\in Q.$$ Proof: $$ \begin{aligned} I_{n} & =\frac{1}{2^{4 n+2}} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(1+\tan x)^{4 n+2} d x \\ & = \frac{1}{2^{4 n+2}} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sum_{k=0}^{2n+1} {4n+2\choose 2k} \tan ^k x d x\\ \end{aligned} $$ Using the properties of odd and even function simplifies the integral $$ I_{n}=\frac{1}{2^{4 n+1}} \sum_{k=0}^{2 n+1}\int_0^{\frac{\pi}{4}}\left(\begin{array}{c} 4 n+2 \\ 2 k \end{array}\right) \tan ^{2 k} x d x $$ Grouping the symmetric terms gives $$ I_{n}=\frac{1}{2^{4 n+1}} \sum_{k=0}^{n+1}\left(\begin{array}{c} 4 n+2 \\ 2 k \end{array}\right) \int_0^{\frac{\pi}{4}} \left(\tan ^{2 k} x+\tan ^{4 n+2-2 k} x\right) d x $$ $(-1)^k+(-1)^{2 n+1-k} =(-1)^k\left[1+(-1)^{2(n-k)+1}\right] =0 \Rightarrow 1+y| y^k+y^{2 n+1-k}. $

Therefore $\tan ^{2 k} x+\tan ^{4 n+2-2 k} x$ is divisible by $1+\tan^2x$ and hence

$$\tan ^{2 k} x+\tan ^{4 n+2-2 k} x=(1+\tan^2x)P_k(\tan x)$$

for some polynomial $P_k(y)$ with integer coefficients. $$ \begin{aligned} I_{n}= & \frac{1}{2^{4 n+1}} \sum_{k=0}^{n+1}\left(\begin{array}{c} 4 n+2 \\ 2 k \end{array}\right) \int_0^{\frac{\pi}{4}} P_k(\tan x) d(\tan x) \\ = & \frac{1}{2^{4 n+1}} \sum_{k=0}^{n+1}\left(\begin{array}{c} 4 n+2 \\ 2 k \end{array}\right) Q_k(1) \in \mathbb{Q} \\ \end{aligned} $$ where $ Q_k^{\prime}(y)=P_k(y).$

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Is there any other simpler proof?

Comments and alternative methods are highly appreciated.

Answer 1

Here is a simpler proof. Utilize the reduction formula $$\int_0^{\frac\pi2}\frac1{(1+\tan x)^m}dx=K_m= \frac1{2(m-1)}+K_{m-1}-\frac12K_{m-2} $$ to reduce \begin{align} I_n=&\int_0^{\frac\pi2}\frac1{(1+\tan x)^{4n+2}}dx\\ =& \ \frac1{2(4n+1)}+ \frac1{2(4n)}+ \frac1{4(4n-1)}-\frac14I_{n-1} \end{align} Thus, the starting rational value $I_0= \int_0^{\frac\pi2}\frac1{(1+\tan x)^{2}}dx=\frac12 $ yields successively the rational values below $$I_1=\frac{11}{60},\>\>\> I_2=\frac{34}{315},\>\>\> I_3=\frac{27341}{360360},\>\>\> \cdots $$

Answer 2

Write $a\sim b$ if $a-b\in\Bbb Q$. Taking $t=\tan x$ and using a keyhole contour,$$\begin{align}-I_{n}&=\frac{1}{2i\pi}\oint\frac{\ln zdz}{\left(1+z\right)^{4n+2}\left(1+z^{2}\right)}\\&=\sum_{w\in\{-1,\,i,\,-i\}}\operatorname{Res}_{z=w}\frac{\ln z}{\left(1+z\right)^{4n+2}\left(1+z^{2}\right)}\\&=\frac{1}{\left(4n+1\right)!}\lim_{z\to-1}\frac{d^{4n+1}}{dz^{4n+1}}\frac{\ln z}{1+z^{2}}+\sum_{\pm}\lim_{z\to\pm i}\frac{\ln z}{\left(1+z\right)^{4n+2}\left(z\pm i\right)}\\&\sim\frac{1}{\left(4n+1\right)!}\lim_{z\to-1}\ln z\frac{d^{4n+1}}{dz^{4n+1}}\frac{1}{1+z^{2}}+\frac{\ln i+\ln\left(-i\right)}{\left(-4\right)^{n+1}}.\end{align}$$Since $\ln(-i)=\frac{3i\pi}{2}$,$$I_{n}\sim-i\pi\left(\frac{1}{\left(4n+1\right)!}\left.\frac{d^{4n+1}}{dz^{4n+1}}\frac{1}{1+z^{2}}\right|_{z=-1}+\frac{2}{\left(-4\right)^{n+1}}\right).$$Since $I_n\in\Bbb R$, this must simplify to $0$, completing the proof.

Answer 3

Applying the substitution $x = \arctan t$, $dx = \frac{dt}{1 + x^2}$ rationalizes the integral, giving $$I_n :=\int_0^\infty \frac{dt}{(t^2 + 1)(t + 1)^{4 n + 2}} .$$ We show by induction that for all $n \geq 0$, the integrand, $f_n(t)$, can be written as $$f_n(t) = a_n \left(\frac{1}{t + 1} - \frac{t}{t^2 + 1}\right) + R_n(t) \qquad (\ast),$$ where $a_n \in \Bbb Q$ and $R_n(t) \in \operatorname{span}_\Bbb Q\{(t + 1)^{-2}, \ldots, (t + 1)^{-(4 n + 2)}\}$. With that result in hand, we can establish the claim:

Proposition. $I_n$ is rational for all integers $n \geq 0$.

Proof. We analyze $$I_n = \int_0^{\infty} \left[a_n \left(\frac{1}{t + 1} - \frac{t}{t^2 + 1}\right) + R_n(t)\right] \,dt.$$ Direct computation shows that $\int_0^\infty \left(\frac{1}{t + 1} - \frac{t}{t^2 + 1}\right) = 0$, so $I_n = \int_0^\infty R_n(t)\,dt$, but $R_n(t)$ has a rational antiderivative with rational coefficients, so $I_n$ is rational. $\square$

We now carry out the induction. We take $n = 0$ for the base case; indeed: $f_0(t)$ has the form $(\ast)$, with $a_0 = \frac{1}{2}$ and $R_0(t) = \frac{1}{2} \frac{1}{(t + 1)^2}$. Now, $f_{n + 1}(t) = \frac{1}{(t + 1)^4} f_n(t)$, so by the inductive hypothesis \begin{align*} f_{n + 1}(t) &= \frac{1}{(t + 1)^4} \left[a_n \left(\frac{1}{t + 1} - \frac{t}{t^2 + 1}\right) + R_n(t)\right] \\ &= \underbrace{\frac{a_n}{4}}_{a_{n + 1}} \left(\frac{t}{t^2 + 1} - \frac{1}{t + 1}\right) \underbrace{- \frac{a_n}{4} \frac{1}{(t + 1)^2} + \frac{a_n}{2} \frac{1}{(t + 1)^4} + \frac{a_n}{(t + 1)^5} + \frac{R_n(t)}{(t + 1)^4}}_{R_{n + 1}(t)} , \\ \end{align*} which in particular also has the form $(\ast)$, completing the induction step.

Remark We can give an explicit formula for $I_n$ in terms of the Lerch transcedent function, $$\Phi(x, s, \alpha) := \sum_{k = 0}^\infty \frac{x^k}{(k + \alpha)^s} : $$ \begin{multline}I_n = \frac1{8n} - \frac18\Phi(-4, 1, n) + \frac14\Phi\left(-4, 1, \frac{3}{4} + n\right) + \frac12\Phi\left(-4, 1, \frac54 + n\right) \\ + \frac{1}{(-4)^n} \left[\frac{\log 5}{8} + \frac12 - \frac14\Phi\left(-4, 1, \frac{3}{4}\right) - \frac12\Phi\left(-4,1,\frac54\right)\right]\end{multline} The appearance of $\log 5$ here is a consequence of the special case $\Phi(x, 1, 1) = \operatorname{Li}_1(x) = -\frac{\log(1 - x)}{x}$.

We could also express $I_n$ in terms of ordinary hypergeometric functions, ${}_2 F_1$, since $$\Phi(x, 1, \alpha) = \frac{1}{\alpha} \cdot {}_2 F_1\left(\left.\matrix{1,\alpha\\1 + \alpha}\right\vert x\right) .$$

Answer 4

Nothing more than Quanto’s idea $$\begin{align} 4 I_n+I_{n-1}&= \int_0^{\frac{\pi}{2}} \frac{4+(1+\tan x)^4}{(1+\tan x)^{4 n+2}} d x\\&=\int_0^{\frac{\pi}{2}} \frac{4 \tan x\left(1+\tan ^2 x\right)+\left(5+\tan ^2 x\right)\left(1+\tan ^2 x\right)}{(1+\tan x)^{4 n+2}}dx \\&= \int_0^{\infty} \frac{(1+t)^2+2(1+t)+2}{(1+t)^{4 n+2}}dt , \quad \textrm{ where }t=\tan x \\&= \frac{1}{4 n-1}+\frac{2}{4 n}+\frac{2}{4 n+1}\end{align} $$ Using $I_0=\frac 12$ and Mathematical Induction completed the proof.