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PostPosted: Tue Apr 19, 2016 12:58 pm 
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Evaluate for $n,m\in\mathbb{N}$ the following integrals:

\(\begin{aligned}
1.\quad &\displaystyle \int_{0}^{+\infty}\frac{\log^n{x}}{1+x^2}\,{\rm{d}}x\\\\
2.\quad &\displaystyle \int_{0}^{+\infty}\frac{\log{x}}{(1+x^2)^m}\,{\rm{d}}x\\\\
3.\quad &\displaystyle \int_{0}^{+\infty}\frac{\log^n{x}}{(1+x^2)^m}\,{\rm{d}}x
\end{aligned}\)

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PostPosted: Tue Apr 19, 2016 2:13 pm 
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Grigorios Kostakos wrote:
$1.\displaystyle \int_{0}^{+\infty}\frac{\log^n{x}}{1+x^2}\,{\rm{d}}x$


Well, for the first one for the moment... We have successively:

\begin{align*} \int_{0}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x &= \int_{0}^{1} \frac{\log^n x}{1+x^2}\, {\rm d}x + \int_{1}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x\\ &= \int_{0}^{1} \frac{\log^n x}{1+x^2}\, {\rm d}x - \int_{1}^{0} \frac{\log^n \left ( \frac{1}{x} \right )}{1+ \left ( \frac{1}{x} \right )^2} \frac{1}{x^2} \, {\rm d}x\\ &= \int_{0}^{1} \frac{\log^n x}{1+x^2} \, {\rm d}x + \int_{0}^{1} \frac{\log^n \left ( \frac{1}{x} \right )}{1+x^2}\, {\rm d}x\\ &= \int_{0}^{1} \frac{\log^n x}{1+x^2}\, {\rm d}x+ \int_{0}^{1} \frac{(-1)^n \log^n x}{1+x^2} \, {\rm d}x \\ &= \left ( 1+(-1)^n \right ) \int_{0}^{1} \frac{\log^n x}{1+x^2} \, {\rm d}x\\ &=\left ( 1+(-1)^n \right )\int_{0}^{1} \log^n x \sum_{k=0}^{\infty} (-1)^k x^{2k} \, {\rm d}x \\ &= \left ( 1+(-1)^n \right ) \sum_{k=0}^{\infty} (-1)^k \int_{0}^{1}x^{2k} \log^n x \, {\rm d}x\\ &= \left ( 1+(-1)^n \right )(-1)^n n! \sum_{k=0}^{\infty} \frac{(-1)^k}{\left ( 2k+1 \right )^{n+1}} \\ &= \left ( 1+(-1)^n \right )(-1)^n n! \sum_{k=0}^{\infty} \left [ \frac{1}{\left ( 4k+1 \right )^{n+1}} - \frac{1}{\left ( 4k+3 \right )^{n+1}} \right ] \\ &=\left ( 1+(-1)^n \right )(-1)^n n! \frac{1}{4^{n+1}} \sum_{k=0}^{\infty} \left [ \frac{1}{\left ( k+ \frac{1}{4} \right )^{n+1}} - \frac{1}{\left ( k+\frac{3}{4} \right )^{n+1}} \right ] \\ &= \frac{1}{4^{n+1}}\left ( 1+(-1)^n \right ) \left [ \psi^{(n)} \left ( \frac{3}{4} \right ) - \psi^{(n)} \left ( \frac{1}{4} \right )\right ] \end{align*}

Thus, distinguishing cases for $n$ (either it is odd or even) we get the following beautiful closed form:

$$\int_{0}^{\infty}\frac{\log^n x}{1+x^2}\, {\rm d}x = \left\{\begin{matrix} 0 &, &n \; {\rm odd} \\\\ \displaystyle \frac{2}{4^{n+1}}\left [ \psi^{(n)} \left ( \frac{3}{4} \right ) - \psi^{(n)} \left ( \frac{1}{4} \right ) \right ]& ,& n \; {\rm even} \end{matrix}\right.$$

Recalling also that , if $n$ is even:

$$\psi^{(n)}(1-z)-\psi^{(n)} (z)= \pi \; \frac{\mathrm{d}^{n} }{\mathrm{d} x^{n}} \cot \pi z \tag{ polygamma reflection formula}$$

holds, the above formula reduces down to - the not so beautiful -

$$\int_{0}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x = \left[ \frac{2\pi}{4^{n+1}} \frac{\mathrm{d}^n }{\mathrm{d} x^n} \cot \pi z \right]_{z=1/4} \;\; n \; \; {\rm even}$$

:clap2: :clap2: :clap2:

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PostPosted: Wed Apr 20, 2016 10:40 am 
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Actually there is no need for polygammas. In the step

Tolaso J Kos wrote:
$$\left ( 1+(-1)^n \right )(-1)^n n! \sum_{k=0}^{\infty} \frac{(-1)^k}{\left ( 2k+1 \right )^{n+1}}$$


we can easily recognize the Beta Dirichlet function. Thus:

$$\int_{0}^{\infty} \frac{\log^n x}{ 1+x^2 } \, {\rm d}x= \left\{\begin{matrix}
0& , &n \; {\rm odd} \\
2 n! \beta(n+1)&, & n \; {\rm even}
\end{matrix}\right.$$

Of course evaluating $\beta(n+1)$ returns us back to the polygamma values, thus the first closed form is much more preferable.

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PostPosted: Wed Apr 20, 2016 1:18 pm 
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Grigorios Kostakos wrote:
$2.\displaystyle \int_{0}^{+\infty}\frac{\log{x}}{(1+x^2)^m}\,{\rm{d}}x$


Let us begin with the following two known represantations of the Beta function:

$${\rm B}(x, y)= \int_{0}^{\infty} \frac{t^{x-1}}{\left ( 1+t \right )^{x+y}}\, {\rm d}t = \frac{\Gamma (x)\Gamma (y)}{\Gamma \left ( x+y \right )}$$

Now, setting $t \mapsto t^2$ as well as $y \mapsto m -x$ we have that:

\begin{equation} {\rm B} \left ( x, m-x \right )= m \int_{0}^{\infty} \frac{t^{2x-1}}{\left ( 1+t^2 \right )^m}\, {\rm d}t \end{equation}

Now, differentiating $(1)$ with respect to $x$ once we have that:

$$\int_{0}^{\infty} \frac{t^{2x-1} \log t }{\left ( 1+t^2 \right )^m}\, {\rm d}t = \frac{\Gamma (x)\Gamma(m-x) \bigg [ \psi^{(0)}(x)- \psi^{(0)}(m-x) \bigg ]}{m^2 \Gamma(m)}$$

Thus our required integral is equal to:

$$ \int_{0}^{\infty} \frac{\log x}{\left ( 1+x^2 \right )^m}\, {\rm d}x ={\rm B}^{(1)} \left ( \frac{1}{2}, m- \frac{1}{2} \right ) = \frac{\Gamma \left ( \frac{1}{2} \right ) \Gamma \left ( m- \frac{1}{2} \right ) \bigg[ \psi^{(0)}\left ( \frac{1}{2}\right) - \psi^{(0)} \left ( m- \frac{1}{2} \right ) \bigg]}{m^2 \Gamma(m)}$$

:clap2: :clap2: :clap2:

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PostPosted: Wed Apr 20, 2016 3:46 pm 
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Grigorios Kostakos wrote:
$3.\displaystyle \int_{0}^{+\infty}\frac{\log^n{x}}{(1+x^2)^m}\,{\rm{d}}x$


As can be seen from the above solution ,

$$\int_{0}^{\infty} \frac{\log^n x}{\left ( 1+x^2 \right )^m}\, {\rm d}x = {\rm B}^{(n)} \left ( \frac{1}{2}, m- \frac{1}{2} \right )$$

However, we hardly know anything about the high order derivatives of the Beta function. What is known however are the following recursive formulae:

  • $$ \Gamma^{(n+1)} (1)= -\gamma \Gamma^{(n)} (1) +n ! \sum_{k=1}^{n} \frac{(-1)^{k+1}}{\left ( n-k \right )!} \zeta(k+1)\Gamma^{(n-k)} (1)$$
  • $$\Gamma^{(n+1)}\left ( \frac{1}{2} \right )= -\left ( \gamma +2 \log 2 \right ) \Gamma^{(n)} \left ( \frac{1}{2} \right ) +n ! \sum_{k=1}^{n}\frac{(-1)^{k+1}}{\left ( n-k \right )!} \left ( 2^{k+1}-1 \right )\zeta(k+1) \Gamma^{(n-k)}\left ( \frac{1}{2} \right )$$

The interested reader will undoubtely find more information in this paper.

So, the best we can do for the moment for the third integral is to leave it that way.

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PostPosted: Wed Apr 20, 2016 3:55 pm 
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From the paper above we also get that if $n \in \mathbb{N}$ and

$$I(n)= \int_{0}^{\infty} \frac{\left ( \log t \right )^n}{(1+t)\sqrt{t}} \, {\rm d}t $$

then:

$$I(n)= \left\{\begin{matrix}
0 &, &n \; \text{odd} \\
\displaystyle2\sum_{k=0}^{n-1} (-1)^k \binom{2n}{k} \Gamma^{(2n-k)} \left ( \frac{1}{2} \right )\Gamma^{(k)} \left ( \frac{1}{2} \right ) + (-1)^n \binom{2n}{n}\left [ \Gamma^{(n)} \left ( \frac{1}{2} \right ) \right ]^2& , & n \; \text{even}
\end{matrix}\right. $$

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PostPosted: Wed Apr 20, 2016 4:37 pm 

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A relative post can be found here.

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PostPosted: Thu Apr 21, 2016 9:29 am 
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Grigorios Kostakos wrote:
3.$\displaystyle \int_{0}^{+\infty}\frac{\log^n{x}}{(1+x^2)^m}\,{\rm{d}}x$


It is nice to share ideas with people around the world just to realize that you are ignoring the obvious. All we need for this is Liebniz's General Rule thus the last integral boils down to:

$$\int_{0}^{\infty} \frac{\log^n x}{\left ( 1+x^2 \right )^m}\, {\rm d}x = {\rm B}^{(n)} \left ( \frac{1}{2}, m- \frac{1}{2} \right )= \frac{1}{m^2 \Gamma(m)}\sum_{k=0}^{n} \binom{n}{k} \Gamma^{(k)} \left ( \frac{1}{2} \right )\Gamma^{(n-k)} \left ( m- \frac{1}{2} \right )$$

Then we are using the reduction formula for $\Gamma^{(n)} \left( \frac{1}{2}\right)$ which I quoted above:

Tolaso J Kos wrote:
$$\Gamma^{(n+1)}\left ( \frac{1}{2} \right )= -\left ( \gamma +2 \log 2 \right ) \Gamma^{(n)} \left ( \frac{1}{2} \right ) +n ! \sum_{k=1}^{n}\frac{(-1)^{k+1}}{\left ( n-k \right )!} \left ( 2^{k+1}-1 \right )\zeta(k+1) \Gamma^{(n-k)}\left ( \frac{1}{2} \right )$$

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