calculusdefinite integralsimproper-integralsintegration
How do I evaluate this integral ? I am a student who just studied calculus 1 and recently definite integral.
$$
\int_{-\infty}^{+\infty} \frac{\ln(1+x^{4})}{1+x^{2}} dx
$$
I substituted both $x = \tan t$ and $x = \frac1{t}$ but could not proceed further.
Here is an approach.
We give some preliminary results.
The poly-Hurwitz zeta function may initially be defined by the series $$ \begin{align} \displaystyle \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, \quad \Re a >-1, \, \Re b >-1, \, \Re s>0. \tag1 \end{align} $$ This special function is a natural extension of the Hurwitz zeta function initially defined as $$ \zeta(s,a)=\sum_{n=0}^{\infty} \frac{1}{(n+a)^s}, \quad \Re a>0, \Re s>1, \tag2 $$ which is a natural extension itself of the Riemann zeta function initially defined as $$ \zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}, \quad \Re s>1. \tag3 $$
The poly-Hurwitz function appears in different places with different notations, one may find it here: [Masri, p. 2 and p. 15 (2004)], [Murty, p. 17 (2006)], [Sinha, p. 45 (2002)]. In this answer we are dealing with a simplified version of a general poly-Hurwitz function.
The series in $(1)$ converges absolutely for $\displaystyle \Re s>0$. Moreover, the convergence of the series is uniform on every half-plane $$\displaystyle H_{\delta}=\left\{s \in \mathbb{C}, \Re s \geq \delta \right\}, \, \delta \in \mathbb{R},\, \delta>0,$$ therefore the poly-Hurwitz zeta function $\displaystyle \zeta(\cdot \mid a,b)$ is analytic on the half-plane $\displaystyle \Re s>0$.
Let $a$, $b$ and $s$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re s >0$. One may observe that $$ \begin{align} \zeta(s\mid a,b) & = \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}\\ & = \sum_{n=1}^{+\infty} \frac{(n+b)+(a-b)}{(n+a)^{s+1}(n+b)}\\ & = \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s+1}}+(a-b)\sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s+1}(n+b)} \tag4 \end{align} $$ giving the functional identity $$ \begin{align} \zeta(s \mid a,b) = \zeta(s+1,a+1) +(a-b) \zeta(s+1 \mid a,b) \tag5 \end{align} $$
where $\displaystyle \zeta(\cdot,\cdot)$ is the standard Hurwitz zeta function.
From $(5)$, we obtain by induction, for $n=1,2,3,\ldots $,
$$ \begin{align} \zeta(s\mid a,b) = \sum_{k=1}^{n}(a-b)^{k-1}\zeta(s+k,a+1) +(a-b)^n\zeta(s+n \mid a,b). \tag6 \end{align} $$
We use $(6)$ to extend $\displaystyle \zeta(\cdot \mid a,b)$ to a meromorphic function on each open set $\Re s>-n $, $n\geq 1$. Since the Hurwitz zeta function is analytic on the whole complex plane except for a simple pole at $1$ with residue $1$, then from $(6)$ the poly-Hurwitz zeta function $\displaystyle \zeta(\cdot\mid a,b)$ is analytic on the whole complex plane except for a simple pole at $0$ with residue $1$.
In 1885 Stieltjes has found that the Laurent series expansion around $1$ of the Riemann zeta function $$ \zeta(1+s) = \frac{1}{s} + \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\gamma_k s^k, \quad s \neq 0,\tag7 $$ is such that the coefficients of the regular part of the expansion are given by $$ \begin{align} \gamma_k& = \lim_{N\to \infty}\left(\sum_{n=1}^N \frac{\log^k n}{n}-\frac{\log^{k+1} \!N}{k+1}\right). \end{align} \tag8 $$ Euler was the first to define a constant of this form (1734) $$ \begin{align} \gamma & = \lim_{N\to\infty}\left(1+\frac12+\frac13+\cdots+\frac1N-\log N\right)=0.577215\ldots. \end{align} $$ The constants $\displaystyle \gamma_k$ are called the Stieltjes constants and due to the fact that $\displaystyle \gamma_0=\gamma$ they are sometimes called the generalized Euler's constants.
Similarly, Wilton (1927) and Berndt (1972) established that the Laurent series expansion in the neighbourhood of $1$ of the Hurwitz zeta function $$ \begin{align} \zeta(1+s,a) = \frac1s+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a)\:s^{k}, \quad \Re a>0, \,s\neq 0, \tag{9} \end{align} $$ is such that the coefficients of the regular part of the expansion are given by $$ \begin{align} \gamma_k(a)& = \lim_{N\to \infty}\left(\sum_{n=0}^N \frac{\log^k (n+a)}{n+a}-\frac{\log^{k+1} (N+a)}{k+1}\right), \quad \Re a>0, \end{align} \tag{10} $$ with $\displaystyle \gamma_{0}(a)=-\psi(a)=-\Gamma'(a)/\Gamma(a)$. The coefficients $\gamma_k(a)$ are called the generalized Stieltjes constants.
We have seen from $(6)$ that the poly-Hurwitz zeta function admits a Laurent series expansion around $0$. Let's denote by $\displaystyle\gamma_k(a,b)$ the coefficients of the regular part of $\displaystyle \zeta(\cdot\mid a,b)$ around $0$. I will call these coefficients the poly-Stieltjes constants.
Do we have an analog of $(10)$ for $\displaystyle\gamma_k(a,b)$?
The following result is new.
Theorem 1. Let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$. Consider The poly-Hurwitz zeta function $$ \begin{align} \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, \quad \Re s>0. \tag{11} \end{align} $$ Then the meromorphic extension of $\displaystyle \zeta(\cdot\mid a,b)$ admits the following Laurent series expansion around $0$, $$ \zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{+\infty} \frac{(-1)^{k}}{k!}\gamma_k(a,b) s^k, \quad s \neq 0,\tag{12} $$ where the poly-Stieltjes constants $\displaystyle \gamma_k(a,b)$ are given by $$ \begin{align} \gamma_k(a,b)& = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log^k (n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) \end{align} \tag{13} $$ with $$ \gamma_{0}(a,b)=-\psi(b+1)=-\Gamma'(b+1)/\Gamma(b+1). \tag{14}$$
Proof. Let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$.
We first assume $\Re s>0$. Observing that, for each $n \geq 1$, $$ \left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right| \leq \sum_{k=0}^{\infty}\left|\frac{\log^k(n+a)}{n+b}\right|\frac{|s|^k }{k!}<\infty $$ and that $$ \sum_{n=1}^{\infty}\left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right|=\sum_{n=1}^{\infty}\left|\frac1{(n+a)^s(n+b)}\right| = \sum_{n=1}^{\infty}\frac1{|n+a|^{\Re s}|n+b|}<\infty,$$ we obtain $$ \begin{align} &\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac{\log^k(n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) s^k \\\\ &= \lim_{N\to+\infty}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\left(\sum_{n=1}^N\frac{\log^k(n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) s^k \\\\ &=\lim_{N\to+\infty}\sum_{k=0}^{\infty}\left(\sum_{n=1}^N\frac{(-1)^{k}}{k!}\frac{\log^k(n+a)}{n+b}s^k -\frac{(-1)^{k}}{k!}\frac{\log^{k+1} \!N}{k+1}s^k\right) \\\\ &=\lim_{N\to+\infty}\left(\sum_{n=1}^N\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\frac{\log^k(n+a)}{n+b}s^k -\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\frac{\log^{k+1} \!N}{k+1}s^k\right) \\\\ &=\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac1{(n+a)^s(n+b)} +\frac1{N^s}-\frac1s\right) \\\\ &=\zeta(s \mid a,b)-\frac1{s} \end{align} $$ as desired. Then, using $(6)$, we extend the preceding identity by analytic continuation to all $s \neq 0$. To prove $(14)$, we start from a standard series representation of the digamma function (see Abram. & Steg. p. 258 6.3.16): $$ \begin{align} -\psi(b+1) &= \gamma - \sum_{n=1}^{\infty} \left( \frac1n - \frac1{b+n} \right) \\ &=\lim_{N\to+\infty}\left(\gamma - \sum_{n=1}^N\left( \frac1n - \frac1{b+n} \right)\right)\\ &=\lim_{N\to+\infty}\left(\left(\sum_{n=1}^N\frac1{b+n} -\ln N\right)-\left(\sum_{n=1}^N\frac1n-\ln N-\gamma \right)\right)\\ &=\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac1{b+n} -\ln N\right)\\\\ &=\gamma_0(a,b) \end{align} $$ using $(13)$.
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One of the consequences of Theorem 1 is the new possibility to express some series in terms of the poly-Stieltjes constants.
Theorem 2. Let $a,b,c$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re c >-1$.
Then $$ \begin{align} (b-a)\sum_{n=1}^{+\infty} \frac{\log (n+c)}{(n+a)(n+b)}=\gamma_1(c,a)-\gamma_1(c,b), \tag{15} \end{align} $$ similarly $$ \begin{align} \sum_{n=1}^{+\infty} \frac1{n+b}\left({\log (n+a)-\log (n+c)}\right)=\gamma_1(a,b)-\gamma_1(c,b), \tag{16} \end{align} $$ with the poly-Stieltjes constant $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). $$
Proof. Let $a,b,c$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re c >-1$.
We have $$ (b-a)\frac{\log (n+c)}{(n+a)(n+b)}=\frac{\log (n+c)}{n+a}-\frac{\log (n+c)}{n+b} $$ giving, for $N\geq1$, $$ \begin{align} (b-a)&\sum_{n=1}^N \frac{\log (n+c)}{(n+a)(n+b)}=\\\\ & \left(\sum_{n=1}^N\frac{\log (n+c)}{n+a}-\frac{\log^2 \!N}2\right)-\left(\sum_{n=1}^N\frac{\log (n+c)}{n+b}-\frac{\log^2 \!N}2\right) \tag{17} \end{align} $$ letting $N \to \infty$ and using $(13)$ gives $(15)$.
We have, for $N\geq1$, $$ \begin{align} &\sum_{n=1}^N \frac1{n+b}\left({\log (n+a)-\log (n+c)}\right)\\\\ &= \left(\sum_{n=1}^N\frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right)-\left(\sum_{n=1}^N\frac{\log (n+c)}{n+b}-\frac{\log^2 \!N}2\right) \tag{18} \end{align} $$ letting $N \to \infty$ and using $(13)$ gives $(16)$.
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Let’s first give a numerical evaluation of Juantheron’s integral.
I would like to thank Jonathan Borwein and David H. Bailey who obtained the result below to $1000$ digits, in just 3.9 seconds run time, using David's new MPFUN-MPFR software, along with the tanh-sinh quadrature program included with the MPFUN-MPFR package.
They also tried the integral with David's MPFUN-Fort package, which has a completely different underlying multiprecision system, and they obtained the same result below.
Finally, they computed the integral with Mathematica $11.0$; it agreed with the result below, although it required about 10 times longer to run.
Proposition 1. We have $$ \begin{align} \int_{0}^{\Large\frac{\pi}2}\!&\frac1{(1+x^2)(1+\tan x)}\mathrm dx\\\\= 0.&59738180945180348461311323509087376430643859042555\\ &67307703207161550311033249824121789098990404474443\\ &73300942847961727020952797366230453350097928752529\\ &62099371263365268445580755896768905606293308536674\\ &89639352215352393870280616186538538722285601087082\\ &81730013060929540132583577240799018025603130403772\\ &83596189879605956759516344861849456740112012597646\\ &30195536341071109827787231788650530475635336662512\\ &50757672078320586388500276160658476344052492489409\\ &64026178233152015087197531148322444147655936720008 \tag{19}\\ &40650450631581050321100329502169853063154902765446\\ &58804861176982696627707544105655815406116180984371\\ &54148587721902800400109013880620460529382772599713\\ &06874977209651994186527207589425408866256042399213\\ &80515694164361264997143539392018681691584285790381\\ &65536517701019826846772718498479534803417547866296\\ &23842162877309354675086691711521468623807334908897\\ &71491673168051054009130049879837629516862688171756\\ &13790927986073268994254629238035029442300668334396\\ &901581838911515359223628586133156893962372426055\cdots \end{align} $$
David H. Bailey confirmed that Mathematica $11.0$, in spite of the great numerical precision, could not find a closed-form of the integral.
The next result proves that the OP integral admits a closed form in terms of the poly-Stieltjes constants.
Proposition 2. We have $$ \begin{align} \int_{0}^{\Large\frac{\pi}2}\!\!\frac1{(1+x^2)(1+\tan x)}\mathrm dx &=\frac{(e^2+1)^2}{2(e^4+1)}\arctan \! \frac{\pi}{2}-\frac{e^4-1}{4(e^4+1)}\log\left(1+\frac{\pi^2}{4}\right)\\\\ &+\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)} \\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac34 +\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac34 +\frac{i}{\pi}\!\right)\tag{20}\\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac14 -\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac14 -\frac{i}{\pi}\!\right) \end{align} $$ with the poly-Stieltjes constant $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). $$
Proof. We proceed in three steps.
Step 1. One may write $$ \require{cancel} \begin{align} &\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\mathrm dx \\ &=\int_{0}^{\Large\frac{\pi}{2}}\frac{\cos x}{(1+x^2)(\cos x+\sin x)}\mathrm dx\\ &=\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{(\cos x+\sin x)+(\cos x-\sin x)}{(1+x^2)(\cos x+\sin x)}\mathrm dx\\ &=\frac12\int_{0}^{\Large\frac{\pi}{2}}\!\frac{1}{1+x^2}\mathrm dx+\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)}\frac{(\cos x-\sin x)}{(\cos x+\sin x)}\mathrm dx\\ &=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{1+x^2}\tan (x-\pi/4)\:\mathrm dx\\ &=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_{-\Large\frac{\pi}{4}}^{\Large\frac{\pi}4}\frac{1}{1+(x+\pi/4)^2}\tan x \:\mathrm dx\\ &=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_0^{\Large\frac{\pi}4}\left(\frac1{1+(x+\pi/4)^2}-\frac1{1+(x-\pi/4)^2}\right)\tan x \:\mathrm dx\\ &=\frac12 \arctan\! \frac{\pi}{2}-\frac{\Im}2 \!\int_{0}^{\Large\frac{\pi}{4}}\!\!\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx \tag{21} \end{align} $$ Let’s evaluate the latter integral.
Step 2. One may recall that the tangent function, as a meromorphic function, can be expressed as an infinite sum of rational functions: $$ \tan x = \sum_{n=0}^{+\infty} \frac{2x}{\pi^2 (n+1/2)^2-x^2}, \quad x \neq \pm \pi/2, \pm 3\pi/2,\pm 5\pi/2,\ldots. \tag{22} $$ We have the inequality $$ \sup_{x \in [0,\pi/4]}\left|\frac{2x}{\pi^2 (n+1/2)^2-x^2}\right|\leq \frac1{(n+1/2)^2}, \quad n=0,1,2,\ldots, \tag{23} $$ the convergence in $(22)$ is then uniform on $[0,\pi/4]$. Thus, plugging $(22)$ into $(21)$, we are allowed to integrate $(21)$ termwise.
Each term, via a partial fraction decomposition, is evaluated to obtain $$ \begin{align} \int_{0}^{\Large\frac{\pi}4}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right)\frac{2x}{\pi^2 (n+1/2)^2-x^2}\:\mathrm dx\\ &=\frac{2\tau}{\pi^2 (n+1/2)^2-\tau^2}\log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)\\ &+\frac1{\pi}\frac1{(n+1/2+\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)\\ &+\frac1{\pi}\frac1{(n+1/2-\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right) \end{align} \tag{24} $$ where for the sake of convenience we have set $\tau:=\pi/4+i$.
Step 3. We sum $(24)$ from $n=0$ to $\infty$ obtaining
$$ \begin{align} \int_{0}^{\Large\frac{\pi}{4}}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx\\ &=\sum_{n=0}^{\infty}\frac{2\tau}{\pi^2 (n+1/2)^2-\tau^2}\log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)\\ &+\frac1{\pi}\sum_{n=0}^{\infty}\frac1{(n+1/2+\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)\\ &+\frac1{\pi}\sum_{n=0}^{\infty}\frac1{(n+1/2-\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right), \tag{25} \end{align} $$ then, singling out the first terms in the two last series and using Theorem $2$ $(16)$, we get $$ \begin{align} \int_{0}^{\Large\frac{\pi}{4}}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx\\ &=\tan \tau \log \left( \frac{4\tau-\pi}{4\tau+\pi}\right) +\frac{4\pi}{\pi^2 -4\tau^2}\log 3 \\&+\frac1{\pi}\gamma_1\!\left(\!\frac34,\frac12 +\frac{\tau}{\pi}\!\right) -\frac1{\pi}\gamma_1\!\left(\!\frac14,\frac12 +\frac{\tau}{\pi}\!\right)\tag{26}\\ &+\frac1{\pi}\gamma_1\!\left(\!\frac34,\frac12 -\frac{\tau}{\pi}\!\right) -\frac1{\pi}\gamma_1\!\left(\!\frac14,\frac12 -\frac{\tau}{\pi}\!\right) \end{align} $$ and the substitution $\tau=\pi/4+i$ gives the desired result.
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Achille Hui's conjecture is true.
Achille Hui has announced in the comments that the OP integral is equal to $$ \begin{align} \frac{\arctan(\frac{\pi}{2}) - t\log\sqrt{1+\frac{\pi^2}{4}}}{1+t^2}+\frac{\pi^2}4 \sum_{n=0}^{\infty}\frac{ (2n+1)\left(\log\left(n+\frac34\right)-\log\left(n+\frac14\right)\right) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) } \tag{27} \end{align} $$ with $\displaystyle t := \tanh(1)$.
The first term in $(27)$, with a little algebra is seen to be equal to the sum of the first two terms on the right hand side of $(20)$.
We establish the veracity of the conjecture using Proposition $2$ and using the next result.
Proposition 3. We have $$ \begin{align} &\frac{\pi^2}4 \sum_{n=0}^{\infty}\frac{ (2n+1)\left(\log\left(n+\frac34\right)-\log\left(n+\frac14\right)\right) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) }\\\\ &=\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)} \\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac34 +\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac34 +\frac{i}{\pi}\!\right)\tag{28}\\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac14 -\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac14 -\frac{i}{\pi}\!\right) \end{align} $$ with the poly-Stieltjes constant $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right).$$
Proof. Observe that the first term of the series on the left hand side of $(28)$, given by $n=0$, is just equal to $$ \frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)}. $$ By a partial fraction decomposition, one may check that $$ \begin{align} \frac{\pi^2}4 &\frac{ (2n+1) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) } =\frac{\Im}{2\pi}\left(\!\frac1{n+\frac34+\frac{i}{\pi}}-\frac1{n+\frac14+\frac{i}{\pi}}\!\right) \tag{29} \end{align} $$ then, multiplying $(29)$ by $\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right)\right)$ and summing from $n=1$ to $\infty$ we get, using Theorem $2$ $(16)$, the result $(28)$.
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Consider a related integral $\displaystyle\;J(a) = \int_0^\infty e^{-ax}\tan^{-1}(x) dx\;$. We can integrate it by part to get: $$J(a) = \frac{1}{a}\int_0^\infty e^{-ax}\frac{dx}{1+x^2}$$
Using this, we can rewrite our integral as $$I(a) = \int_0^\infty \left(\sum_{n=1}^\infty e^{-nax}\right)\tan^{-1}(x)dx = \sum_{n=1}^\infty J(na) = \sum_{n=1}^\infty \frac{1}{na}\int_0^\infty e^{-nax}\frac{dx}{1+x^2}$$ Replace $x$ by $x/n$ in each terms, this can be transformed as $$I(a) = \frac{1}{a}\int_0^\infty e^{-at}\left(\sum_{n=1}^\infty \frac{1}{t^2+n^2}\right) dt $$ Start with a infinite product expansion of $\displaystyle\;\frac{\sinh(\pi x)}{\pi x}\;$. By taking logarithm and differentiate, $$\frac{\sinh(\pi x)}{\pi x} = \prod_{n=1}^\infty \left(1 + \frac{x^2}{n^2}\right) \quad\implies\quad \sum_{n=1}^\infty \frac{1}{x^2 + n ^2 } = \frac{1}{2x}\left( \pi\coth(\pi x) - \frac{1}{x}\right) $$ we can rewrite the sum inside above integrand of $I(a)$ and obtain
$$I(a) = \frac{\pi}{2a}F\left(\frac{a}{\pi}\right) \quad\text{ where }\quad F(x) = \int_0^\infty e^{-xt} \left(\coth(t) - \frac{1}{t}\right) \frac{dt}{t} $$ Notice $$-\frac{dF(x)}{dx} = \int_0^\infty e^{-xt}\left(\coth(t) - \frac{1}{t}\right)dt$$ and compare this with a integral representation of digamma function $\displaystyle\;\psi(x) = \frac{d\log\Gamma(x)}{dx}\;$, $$\psi(x) = \int_0^\infty\left(\frac{e^{-t}}{t} - \frac{e^{-xt}}{1-e^{-t}}\right)dt \quad\implies\quad \psi\left(\frac{x}{2}\right) = \int_0^\infty\left(\frac{e^{-2t}}{t} - \frac{2e^{-xt}}{1-e^{-2t}}\right)dt $$ We get $$\begin{array}{rrl} &-\frac{dF(x)}{dx} &= \int_0^\infty \frac{e^{-2t} - e^{-xt}}{t}dt - \frac12 \left[\psi\left(\frac{x}{2}\right) + \psi\left(\frac{x}{2}+1\right)\right]\\ &&= \log\frac{x}{2} - \frac12\left[\psi\left(\frac{x}{2}\right) + \psi\left(\frac{x}{2}+1\right)\right]\\ \implies & F(x) &= \text{const.} + \log\Gamma\left(\frac{x}{2}\right) + \log\Gamma\left(\frac{x}{2}+1 \right) - x\log\left(\frac{x}{2e}\right) \end{array}$$ Since $\lim\limits_{x\to\infty}F(x) = 0$, we can use Stirling's approximation to fix the constant in $F(x)$ to $-\log(2\pi)$. As a result, We find
$$I(a) = \frac{\pi}{2a}\left[ 2\log\Gamma\left(\frac{a}{2\pi}+1\right) - \log(a)\right] -\frac12\log\left(\frac{a}{2\pi e}\right) $$
Best Answer
A polynomial of the form $1+x^{2n}$ can always be written as the product of $n$ quadratic factors of the form $1-2x\cos\theta + x^2$. So that suggests we consider the integral $$ I(\theta) = \int_{-\infty}^\infty \frac{\ln(1-2x\cos\theta+x^2)}{1+x^2}dx. $$ Now, if we differentiate this with respect to $\theta$, we'll get a rational function of $x$, and the integral can then be evaluated through partial fractions: \begin{eqnarray} I'(\theta) &=& \int_{-\infty}^\infty \frac{2x\sin\theta}{(1-2x\cos\theta+x^2)(1+x^2)}dx \\&=&\tan\theta\int_{-\infty}^\infty\left(\frac{1}{1-2x\cos\theta+x^2}- \frac{1}{1+x^2}\right)dx \\ &=& \pi\frac{1-\sin\theta}{\cos\theta}=\pi\frac{\cos\theta}{1+\sin\theta} = \pi\frac{d}{d\theta}\ln(1+\sin\theta). \end{eqnarray} This gives us the $\theta$ dependence of $I$, but we need the value of $I$ at a particular $\theta$ to set the constant of integtration. We can evaluate the case $\theta = \pi/2$ with clever use of trigonometric identities and variable substitutions: \begin{eqnarray} I\left(\frac{\pi}{2}\right)& = &\int_{-\infty}^\infty \frac{\ln(1+x^2)}{1+x^2} = -4\int_{0}^{\pi/2}\ln(\cos\phi)d\phi = -4\int_{0}^{\pi/2}\ln(\sin\phi)d\phi \\ &=& -2\int_0^{\pi}\ln(\sin\phi)d\phi =-4\int_0^{\pi/2}\ln(\sin 2\psi)d\psi = -4\int_0^{\pi/2}\ln(2\sin\psi\cos\psi)d\psi \\ &=& -2\pi\ln 2-4\int_{0}^{\pi/2}\ln(\cos\psi)d\psi-4\int_{0}^{\pi/2}\ln(\sin\psi)d\psi \\ &=& -2\pi\ln 2 + 2I\left(\frac{\pi}{2}\right) \end{eqnarray} Thus $I(\pi/2) = 2\pi \ln2$, and we have $$ I(\theta) = \pi\ln[2(1+\sin\theta)]. $$ From here it's easy to show $$ \int_{-\infty}^\infty \frac{\ln(1+x^4)}{1+x^2}dx = I\left(\frac{\pi}{4}\right) + I\left(\frac{3\pi}{4}\right) = 2\pi\ln(2+\sqrt{2}) $$