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## Three similar improper integrals

Calculus (Integrals, Series)
Grigorios Kostakos
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### Three similar improper integrals

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}
Grigorios Kostakos
Tolaso J Kos
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### Re: Three similar improper integrals

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 -

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

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Tolaso J Kos
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### Re: Three similar improper integrals

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.
Imagination is much more important than knowledge.
Tolaso J Kos
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### Re: Three similar improper integrals

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:

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

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.
Tolaso J Kos
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### Re: Three similar improper integrals

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.
Imagination is much more important than knowledge.
Tolaso J Kos
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### Re: Three similar improper integrals

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:

$$I(n)= \left\{\begin{matrix} 0 &, &n \; \text{odd} \\ \displaystyle2\sum_{k=0}^{n-1} (-1)^k \binom{2n}{k} \Gamma^{(2n-k)} \left ( \frac{1}{2} \right )\Gamma^{(k)} \left ( \frac{1}{2} \right ) + (-1)^n \binom{2n}{n}\left [ \Gamma^{(n)} \left ( \frac{1}{2} \right ) \right ]^2& , & n \; \text{even} \end{matrix}\right.$$
Imagination is much more important than knowledge.
Riemann
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### Re: Three similar improper integrals

A relative post can be found here.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Tolaso J Kos
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### Re: Three similar improper integrals

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: $$\Gamma^{(n+1)}\left ( \frac{1}{2} \right )= -\left ( \gamma +2 \log 2 \right ) \Gamma^{(n)} \left ( \frac{1}{2} \right ) +n ! \sum_{k=1}^{n}\frac{(-1)^{k+1}}{\left ( n-k \right )!} \left ( 2^{k+1}-1 \right )\zeta(k+1) \Gamma^{(n-k)}\left ( \frac{1}{2} \right )$$
Imagination is much more important than knowledge.

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