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Seeking Methods to solve the following two definite integrals:

\begin{equation} I_(n) = \int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt \qquad J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt \end{equation}

For $n \in \mathbb{R},\:n \gt 1$

The method I took was to take the following integral:

\begin{equation} \int_0^\infty \frac{t^k}{\left(t^n + 1\right)^m}\:dt = \frac{1}{n}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right) \end{equation}

Where: $0 \leq k \lt n$. Here let $m = 1$.

Differentiate under the curve with respect to $k$ and taking the limit as $k \rightarrow 0^+$ (via the Dominated Convergence Theorem), i.e.

\begin{align} \lim_{k \rightarrow 0+} \frac{d}{dk} \left[ \int_0^\infty \frac{t^k}{t^n + 1}\:dt \right] &= \lim_{k \rightarrow 0+} \frac{d}{dk} \left[\frac{1}{n}B\left(1 - \frac{k + 1}{n}, \frac{k + 1}{n} \right) \right] \\ \lim_{k \rightarrow 0+} \int_0^\infty \frac{t^k \ln(t)}{t^n + 1}\:dt &= \lim_{k \rightarrow 0+} \left[\frac{1}{n^2} B\left(1 - \frac{k + 1}{n}, \frac{k + 1}{n}\right)\left[\psi^{(0)}\left(\frac{k + 1}{n} \right) - \psi^{(0)}\left(1 - \frac{k + 1}{n} \right)\right] \right] \\ \int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt&= \frac{1}{n^2} B\left(1 - \frac{1}{n}, \frac{1}{n}\right)\left[\psi^{(0)}\left(\frac{1}{n} \right) - \psi^{(0)}\left(1 - \frac{1}{n} \right)\right] \\ &=- \frac{\pi^2}{n^2}\operatorname{cosec}\left(\frac{\pi}{n}\right)\cot\left(\frac{\pi}{n} \right) \end{align}

Which is our expression for $I_n$. Taking the same approach but differentiating twice with respect to $k$ we arrive at our expression for $J_n$:

\begin{equation} J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt = \frac{\pi^3}{n^3}\operatorname{cosec}\left(\frac{\pi}{n} \right)\left[\operatorname{cosec}^2\left(\frac{\pi}{n}\right) + \cot^2\left(\frac{\pi}{n}\right) \right] \end{equation}

And in fact we may generalise:

\begin{equation} \int_0^\infty \frac{\ln^p(t)}{\left(t^n + 1\right)^m}\:dt = \lim_{k\rightarrow 0}\frac{d^p}{dk^p}\left[\frac{1}{n} B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right)\right] \end{equation}

Where $p \in \mathbb{N}$

This method however was just an extension of another integral. I'm curious, if I had just started with $I_n, J_n$ what alternative methods could be used?

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  • $\begingroup$ Solution to the first integral here: math.stackexchange.com/a/1055066/42969. $\endgroup$ – Martin R Jan 14 at 10:31
  • $\begingroup$ @MartinR - Yes, that uses the exact method I've prescribed here. I was hoping to find alternative approaches to solving these integrals. $\endgroup$ – user150203 Jan 14 at 10:33
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    $\begingroup$ See here: math.stackexchange.com/q/269081/515527 $\endgroup$ – Zacky Jan 14 at 11:41
  • $\begingroup$ Cheers @Zacky!! much appreciated. If you know of any others, please post up. $\endgroup$ – user150203 Jan 14 at 11:44
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    $\begingroup$ This one is the core: math.stackexchange.com/q/110457/515527 $\endgroup$ – Zacky Jan 14 at 11:51
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You can use the pole expansions \begin{align} \csc^2 (z) &= \sum_{m \in \mathbb{Z}} \frac{1}{[m \pi + z]^2} \, , \, z \in \mathbb{C}\setminus\pi\mathbb{Z} \, , \\ \sec^2 (z) &= \sum_{m \in \mathbb{Z}} \frac{1}{\left[\left(m+\frac{1}{2}\right) \pi + z\right]^2} \, , \, z \in \mathbb{C}\setminus\pi \left(\mathbb{Z} + \frac{1}{2}\right) \, , \end{align} and their derivatives to find the integrals. We have \begin{align} -I_n &= \int \limits_0^\infty \frac{-\ln(t)}{1+t^n} \, \mathrm{d}t = \int \limits_0^1 [-\ln(t)] \frac{1-t^{n-2}}{1+t^n} \, \mathrm{d} t = \sum \limits_{k=0}^\infty (-1)^k \int\limits_0^1 [-\ln(t)] (1-t^{n-2})t^{n k} \, \mathrm{d} t \\ &= \sum \limits_{k=0}^\infty (-1)^k \left[\frac{1}{(nk+1)^2} - \frac{1}{(n k + n - 2 + 1)^2}\right] = \sum \limits_{k=0}^\infty (-1)^k \left[\frac{1}{(nk+1)^2} - \frac{1}{(n(k+1)-1)^2}\right] \\ &= \frac{\pi^2}{4 n^2} \sum \limits_{l=0}^\infty \left[\frac{1}{\left(l \pi + \frac{\pi}{2n}\right)^2} - \frac{1}{\left(\left(l+\frac{1}{2}\right) \pi + \frac{\pi}{2n}\right)^2} + \frac{1}{\left((l+1) \pi - \frac{\pi}{2n}\right)^2} - \frac{1}{\left(\left(l+\frac{1}{2}\right) \pi - \frac{\pi}{2n}\right)^2}\right] \\ &= \frac{\pi^2}{4 n^2} \sum_{m \in \mathbb{Z}} \left[\frac{1}{\left(m \pi + \frac{\pi}{2n}\right)^2} - \frac{1}{\left(\left(m+\frac{1}{2}\right) \pi + \frac{\pi}{2n}\right)^2} \right] = \frac{\pi^2}{4n^2} \left[\csc^2 \left(\frac{\pi}{2n}\right) - \sec^2 \left(\frac{\pi}{2n}\right)\right] \\ &= \frac{\pi^2}{n^2} \frac{\cos\left(\frac{\pi}{n}\right)}{\sin^2\left(\frac{\pi}{n}\right)} \end{align} and, similarly, \begin{align} J_n &= \int \limits_0^1 \ln^2(t) \frac{1+t^{n-2}}{1+t^n} \, \mathrm{d} t = 2 \sum \limits_{k=0}^\infty (-1)^k \left[\frac{1}{(nk+1)^3} + \frac{1}{(n(k+1)-1)^3}\right] \\ &= \frac{\pi^3}{4 n^3} \sum_{m \in \mathbb{Z}} \left[\frac{1}{\left(m \pi + \frac{\pi}{2n}\right)^3} - \frac{1}{\left(\left(m+\frac{1}{2}\right) \pi + \frac{\pi}{2n}\right)^3}\right] \\ &= \frac{\pi^3}{4 n^3} \left[\csc^2 \left(\frac{\pi}{2n}\right) \cot \left(\frac{\pi}{2n}\right) + \sec^2 \left(\frac{\pi}{2n}\right) \tan \left(\frac{\pi}{2n}\right)\right] \\ &= \frac{\pi^3}{n^3} \frac{1+\cos^2 \left(\frac{\pi}{n}\right)}{\sin^3 \left(\frac{\pi}{n}\right)} \end{align} for $n > 1$ .

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    $\begingroup$ Fantastic solution. Thanks for posting. If you know of any other approaches, please post up. $\endgroup$ – user150203 Jan 14 at 11:39
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Here is an alternative strategy that can be used to find $I^{(1)}_n$ (your $I_n$), $I^{(2)}_n$ (your $J_n$), and $I^{(p)}_n$ (the general case where $p \in \mathbb{N}$).

Writing $$I^{(1)}_n = \int_0^\infty \frac{\ln t}{1 + t^n} \, dt = \int_0^1 \frac{\ln t}{1 + t^n} \, dt + \int_1^\infty \frac{\ln t}{1 + t^n} \, dt.$$ Enforcing a substitution of $t \mapsto 1/t$ in the second of the integrals appearing to the right yields $$I^{(1)}_n = \int_0^1 \frac{\ln t}{1 + t^n} \, dt - \int_0^1 \frac{\ln t}{1 + t^n} \, t^{n - 2} dt.$$ Exploiting the geometric sum for the term $1/(1 + t^n)$ leads to $$I^{(1)}_n = \sum_{k = 0}^\infty (-1)^k \int_0^1 t^{kn} \ln t \, dt - \sum_{k = 0}^\infty (-1)^k \int_0^1 t^{kn + n - 2} \ln t \, dt.$$ After integrating by parts we have \begin{align} I^{(1)}_n &= -\frac{1}{n^2} \sum_{k = 0}^\infty \frac{(-1)^k}{(k + 1/n)^2} + \frac{1}{n^2} \sum_{k = 0}^\infty \frac{(-1)^k}{(k + 1 - 1/n)^2}\\ &= \frac{1}{n^2} \sum_{\substack{k = 0\\k \in \text{odd}}} \frac{1}{(k + 1/n)^2} - \frac{1}{n^2} \sum_{\substack{k = 0\\k \in \text{even}}} \frac{1}{(k + 1/n)^2}\\ & \qquad -\frac{1}{n^2} \sum_{\substack{k = 0\\k \in \text{odd}}} \frac{1}{(k + 1 - 1/n)^2} + \frac{1}{n^2} \sum_{\substack{k = 0\\k \in \text{even}}} \frac{1}{(k + 1 - 1/n)^2}. \end{align} Shifting the odd indices by: $k \mapsto 2k + 1$, and the even indices by: $k \mapsto 2k$, gives \begin{align} I^{(1)}_n &= \frac{1}{4n^2} \sum_{k = 0}^\infty \frac{1}{(k + 1/2 + 1/2n)^2} - \frac{1}{4n^2} \sum_{k = 0}^\infty \frac{1}{(k + 1/2n)^2}\\ & \qquad - \frac{1}{4n^2} \sum_{k = 0}^\infty \frac{1}{(k + 1 - 1/2n)^2} + \frac{1}{4n^2} \sum_{k = 0}^\infty \frac{1}{(k + 1/2 - 1/2n)^2}\\ &= -\frac{1}{4n^2} \left [ \psi^{(1)} \left (1 - \frac{1}{2n} \right ) + \psi^{(1)} \left (\frac{1}{2n} \right ) \right ]\\ & \qquad + \frac{1}{4n^2} \left [\psi^{(1)} \left (1 - \left (\frac{1}{2} - \frac{1}{2n} \right ) \right ) + \psi^{(1)} \left (\frac{1}{2} - \frac{1}{2n} \right ) \right ]\\ &= -\frac{\pi^2}{4n^2} \left [\operatorname{cosec}^2 \left (\frac{\pi}{2n} \right ) - \operatorname{cosec}^2 \left (\frac{\pi}{2} + \frac{\pi}{2n} \right ) \right ]\\ &= -\frac{\pi^2}{4n^2} \left [\operatorname{cosec}^2 \left (\frac{\pi}{2n} \right ) - \sec^2 \left (\frac{\pi}{2n} \right ) \right ]\\ &= -\frac{\pi^2}{n^2} \left [\frac{\cos^2(\pi/2n) - \sin^2 (\pi/2n)}{\{2 \sin (\pi/2n) \cos (\pi/2n)\}^2} \right ]\\ &= -\frac{\pi^2}{n^2} \operatorname{cosec} \left (\frac{\pi}{n} \right ) \cot \left (\frac{\pi}{n} \right ), \end{align} as expected. Note we have made use of the reflexion formula for the trigamma function $\psi^{(1)}(z)$.

In a similar way to how $I^{(1)}_n$ was found above, $I^{(2)}_n$ and $I^{(p)}_n$ can also be found.

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  • $\begingroup$ I always enjoy your solutions/methods and this one is fantastic. That initial 'trick' of splitting the integral and then make the change of variable is something I will certainly be keeping in my toolbox. Much appreciated. $\endgroup$ – user150203 Jan 15 at 6:45

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