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 Post subject: Three similar improper integrals Posted: Tue Apr 19, 2016 12:58 pm
 Team Member Joined: Mon Nov 09, 2015 1:36 am
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Location: Ioannina, Greece
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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Grigorios Kostakos

Top   Post subject: Re: Three similar improper integrals Posted: Tue Apr 19, 2016 2:13 pm
 Administrator  Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
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 -

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

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Imagination is much more important than knowledge. Top   Post subject: Re: Three similar improper integrals Posted: Wed Apr 20, 2016 10:40 am
 Administrator  Joined: Sat Nov 07, 2015 6:12 pm
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Location: Larisa
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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Imagination is much more important than knowledge. Top   Post subject: Re: Three similar improper integrals Posted: Wed Apr 20, 2016 1:18 pm
 Administrator  Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
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:

$$\int_{0}^{\infty} \frac{\log x}{\left ( 1+x^2 \right )^m}\, {\rm d}x ={\rm B}^{(1)} \left ( \frac{1}{2}, m- \frac{1}{2} \right ) = \frac{\Gamma \left ( \frac{1}{2} \right ) \Gamma \left ( m- \frac{1}{2} \right ) \bigg[ \psi^{(0)}\left ( \frac{1}{2}\right) - \psi^{(0)} \left ( m- \frac{1}{2} \right ) \bigg]}{m^2 \Gamma(m)}$$   _________________
Imagination is much more important than knowledge. Top   Post subject: Re: Three similar improper integrals Posted: Wed Apr 20, 2016 3:46 pm
 Administrator  Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
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:

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

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

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Imagination is much more important than knowledge. Top   Post subject: Re: Three similar improper integrals Posted: Wed Apr 20, 2016 3:55 pm
 Administrator  Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
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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Imagination is much more important than knowledge. Top   Post subject: Re: Three similar improper integrals Posted: Wed Apr 20, 2016 4:37 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
A relative post can be found here.

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

Top   Post subject: Re: Three similar improper integrals Posted: Thu Apr 21, 2016 9:29 am
 Administrator  Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
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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