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Integration Contest - Season 5

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Post Fri Jul 31, 2015 5:32 am
Random Variable Integration Guru
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Posts: 381
This one is inspired by something that sos440 posted.


Problem number I have no idea

Show that

$$ \int_{0}^{\infty} \frac{\tanh^{3}(x)}{x^{3}} \, dx = \frac{1}{\pi^{4}} \Big(186 \, \zeta(5) - 7 \pi^{2} \zeta(3)\Big) .$$

Post Sat Aug 01, 2015 9:19 am

Posts: 50
Denoting $\alpha_n=\frac{(2n+1)\pi i}{2}$, we have the Laurent series expansion
$$\tanh^3{z}=\frac{1}{(z-\alpha_n)^3}+\frac{1}{z-\alpha_n}+\mathcal{O}\left((z-\alpha_n)\right)$$
Thus
\begin{align}
\operatorname*{Res}_{z=\alpha_n}\frac{\tanh^3{z}}{z^3}
&=\operatorname*{Res}_{z=\alpha_n}\left[\frac{1}{z^3(z-\alpha_n)^3}+\frac{1}{z^3(z-\alpha_n)}\right]\\
&=\frac{6}{\alpha_n^5}+\frac{1}{\alpha_n^3}=\frac{192}{(2n+1)^5\pi^5i}-\frac{8}{(2n+1)^3\pi^3i}
\end{align}
Integrating $\dfrac{\tanh^3{z}}{z^3}$ along a semicircle in the UHP and noting that $0$ is just a removable singularity, as well as the fact that the integral over the arc vanishes, we get
\begin{align}
\int^\infty_0\frac{\tanh^{3}{x}}{x^3}\ {\rm d}x
&=\pi\sum^\infty_{n=0}\frac{192}{(2n+1)^5\pi^5}-\frac{8}{(2n+1)^3\pi^3}\\
&=\frac{1}{\pi^4}\times 192\times\left(1-\frac{1}{32}\right)\times \zeta(5)-\frac{1}{\pi^2}\times 8\times\left(1-\frac{1}{8}\right)\times\zeta(3)\\
&=\frac{186\zeta(5)}{\pi^4}-\frac{7\zeta(3)}{\pi^2}
\end{align}

Post Sat Aug 01, 2015 3:03 pm
Random Variable Integration Guru
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Posts: 381
The idea from sos440's post (which was about $\int_{0}^{\infty} \frac{\tanh^{2n}(x)}{x^{2}} \, dx$) is that summing the residues in the upper half-plane, we find

$$ \begin{align} \sum_{n=0}^{\infty} \text{Res} \left[ \frac{\tanh^{3}(z)}{z^{3}} , i \pi \left(n+\frac{1}{2} \right)\right] &= \sum_{n=0}^{\infty} \text{Res} \left[\frac{\tanh^{3} (z+ i \pi n)}{(z+i \pi n)^{3}} , \frac{i \pi}{2} \right] \\ &= \sum_{n=0}^{\infty} \text{Res} \left[ \frac{\tanh^{3}(z)}{(z+i \pi n)^{3}} , \frac{i \pi}{2} \right] \\ &= \text{Res} \left[\tanh^{3}(z) \sum_{n=0}^{\infty} \frac{1}{(z+i \pi n)^{3}}, \frac{i \pi}{2} \right] \\ &= \text{Res} \left[\tanh^{3}(z) \frac{1}{2i \pi^{3}} \, \psi_{2} \left(\frac{z}{i \pi} \right) , \frac{i \pi}{2}\right] \\ &= \frac{1}{2 i \pi^{3}} \text{Res} \left[\coth^{3}(z) \, \psi_{2} \left(\frac{1}{2} + \frac{z}{i \pi} \right) , 0\right]. \end{align}$$

And since $$\coth^{3}(z) \, \psi^{2} \left(\frac{1}{2} + \frac{z}{i \pi} \right) = \left(\frac{1}{z^{3}} + \frac{1}{z} + \ldots \right) \left( \psi_{2} \left(\frac{1}{2} \right) + \frac{1}{i \pi} \psi_{3} \left(\frac{1}{2} \right)z - \frac{1}{ \pi^{2}} \psi_{4} \left( \frac{1}{2}\right) \frac{z^{2}}{2!} + \ldots \right) ,$$

we get

$$ \begin{align} \int_{0}^{\infty} \frac{\tanh^{3}{x}}{x^{3}} \, dx &= i \pi \, \frac{1}{2 i \pi^{3}} \left[- \frac{1}{2 \pi^{2}} \psi_{4} \left( \frac{1}{2} \right) + \psi_{2} \left(\frac{1}{2} \right) \right] \\ &= \frac{1}{4 \pi^{4}} \left[ 2 \pi^{2} \psi_{2} \left( \frac{1}{2}\right) - \psi_{4} \left(\frac{1}{2} \right) \right]. \end{align}$$

Post Sat Aug 01, 2015 3:21 pm
galactus User avatar
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Hey G and M

I finally made it back :):)

Post Sun Oct 04, 2015 4:22 pm
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Posts: 381
The following formulas are not particularly difficult to derive, but perhaps you'll find them somewhat interesting.


Assume that the complex function $f(z)$ is analytic on and inside the unit circle, and assume its Maclaurin series expansion has real coefficients.

Also assume that $b \ge a \ge 0$ and $p>0$.

Show that

$$\text{Re} \int_{0}^{\infty} \cos (ax) \, \frac{f(e^{ibx})}{x^{2}+p^{2}} \, dx = \frac{\pi}{2p} \Big[f(e^{-bp}) \cosh(ap) - f(0)\sinh(ap) \Big]$$

and

$$\text{Im} \int_{0}^{\infty} \sin (ax) \, \frac{f(e^{ibx})}{x^{2}+p^{2}} \, dx = \frac{\pi}{2p}\sinh(ap) \Big[f(e^{-bp})- f(0) \Big] .$$

Post Thu Oct 08, 2015 10:25 am
Random Variable Integration Guru
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Posts: 381
Under the conditions stated above,

$$ \begin{align} &\text{Re} \int_{0}^{\infty} \cos (ax) \, \frac{f(e^{ibx})}{x^{2}+p^{2}} \, dx \\ &= \int_{0}^{\infty} \frac{\cos (ax)}{x^{2}+p^{2}} \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \cos(nbx) \, dx \\ &= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \int_{0}^{\infty} \frac{\cos(ax) \cos(nbx)}{x^{2}+p^{2}} \, dx \\ &= f(0) \int_{0}^{\infty} \frac{\cos (ax)}{x^{2}+p^{2}} \, dx + \sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!} \int_{0}^{\infty} \frac{\cos(ax) \cos(nbx)}{x^{2}+p^{2}} \, dx \\ &= f(0) \frac{\pi}{2p} e^{-ap} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!} \left(\int_{0}^{\infty} \frac{\cos \big((a+nb)x\big)}{x^{2}+p^{2}} \, dx + \int_{0}^{\infty} \frac{\cos\big((a-nb)x \big)}{x^{2}+p^{2}} \, dx \right) \\ &= \frac{\pi}{2p} f(0) e^{-ap} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!} \frac{\pi}{2p} \left(e^{-(a+nb)p} + e^{-(nb-a)p} \right) \\ &= \frac{\pi}{2p} f(0) e^{-ap} + \frac{\pi}{2p} \cosh(ap) \sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!} e^{-nbp} \\ &= \frac{\pi}{2p} f(0) e^{-ap} + \frac{\pi}{2p} \cosh(ap) \Big(f(e^{-bp}) -f(0) \Big) \\ &= \frac{\pi}{2p} \Big(f(e^{-bp})\cosh(ap) - f(0) \sinh(ap) \Big). \end{align}$$

So, for example, $$\int_{0}^{\infty} \frac{\cos(ax) e^{\cos(bx)} \cos(\sin bx)}{x^{2}+p^{2}} \, dx = \frac{\pi}{2p} \Big(e^{e^{-bp}}\cosh(ap) - \sinh(ap) \Big).$$

The other one is similar.

Post Tue Oct 13, 2015 3:12 am
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Posts: 381
Assume $a \ge 0$ and $p >0$.

By integrating the complex function $ \displaystyle e^{ia \tan (z)} \, \frac{\cot (z)}{p^{2}+z^{2}}$ around a contour in the upper half-plane, show that $$\int_{0}^{\infty} \sin (a \tan x) \, \frac{\cot (x)}{p^{2}+x^{2}} \, dx = \frac{\pi}{2p} \coth(p) \left[1-e^{-a \tanh (p)}\right].$$
And by integrating $\displaystyle e^{ia \cot (z)} \, \frac{\tan (z)}{p^{2}+z^{2}} $ around a contour in the lower half-plane, show that $$\int_{0}^{\infty} \sin (a \cot x) \, \frac{\tan (x)}{p^{2}+x^{2}} \, dx = \frac{\pi}{2p} \tanh(p) \left[1-e^{-a \coth (p)}\right].$$

Post Sun Oct 18, 2015 5:18 pm
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I'm going to use the following facts:


$1)$ Since $\left|e^{i a \tan(z) }\right| = \left|e^{ia \tan(x+iy)}\right|= \exp \left(\frac{-a \sinh(2y)}{\cos(2x)+\cosh(2y)} \right)$, $\left|e^{ia \tan (z)}\right|$ is less than or equal to $1$ in the upper half-plane.

$2)$ Since $\left|e^{i a \cot(z) }\right| = \left|e^{ia \cot(x+iy)}\right|= \exp \left(\frac{-a \sinh(2y)}{\cos(2x) - \cosh(2y)} \right)$, $\left|e^{ia \cot (z)} \right|$ is less than or equal to $1$ in the lower half-plane.

$3)$ On the vertical line $\text{Re}(z) = \left(N+\frac{1}{4} \right) \pi$ (where $N \in \mathbb{Z}$), $|\cot(z)| = |\tan(z)| = 1$.

$4)$ As $\text{Im}(z) \to \pm \infty$, $|\cot(z)| \to 1$ and $|\tan(z)| \to 1$.

$5)$ For $|z|$ large in magnitude, $\displaystyle \frac{1}{p^{2}+z^{2}} = \mathcal{O}(z^{-2})$.

$6)$ The partial fraction expansion of $\coth(z)$ is $\displaystyle \sum_{n=-\infty}^{\infty} \frac{z}{z^{2} + n^{2} \pi^{2}}$.

$7)$ The partial fraction expansion of $\tanh(z)$ is $\displaystyle \sum_{n=-\infty}^{\infty} \frac{z}{z^{2} + \left(\frac{2n+1}{2}\right)^{2} \pi^{2}}$.


Along the real axis, the function $f(z) = \displaystyle e^{i a \tan (z)} \, \frac{\cot (z)}{p^{2}+z^{2}}$ has simple poles at $z= n \pi$ and essential singularities at $z=\left(n+\frac{1}{2} \right) \pi$.

So consider a rectangular contour with vertices at $z= \pm \left(N+\frac{1}{4}\right) \pi$ and $z= \pm \left(N+\frac{1}{4}\right) \pi + iy$ (for some $N \in \mathbb{N}$ and some $y>p$) that is indented along the real axis.

Letting $N \to \infty$ through the positive integers and $y \to \infty$, $\int f(z) \, dz$ vanishes on the left and right sides of the contour because of $(1)$, $(3)$, and $(5)$, and it vanishes along the top of contour because of $(1)$, $(4)$, and $(5)$.

And because of $(1)$, the total contribution from the indentations around the essential singularities is vanishingly small.

So we have $$ \text{PV} \int_{-\infty}^{\infty} e^{ia \tan (x)} \, \frac{\cot (x)}{p^{2}+x^{2}} \, dx - i \pi \sum_{n=-\infty}^{\infty}\text{Res} [f(z), n \pi ] = 2 \pi i \, \text{Res}[f(z), ip] $$

where $$\sum_{n=-\infty}^{\infty}\text{Res} [f(z), n \pi ] = \sum_{n=-\infty}^{\infty} \frac{1}{p^{2} + n^{2}\pi^{2}} = \frac{\coth (p)}{p}$$

and $$\text{Res}[f(z), ip] = -\frac{e^{-a \tanh (p)} \coth(p)}{2ip}.$$

Therefore, $$\text{PV}\int_{-\infty}^{\infty} e^{i a \tan (x)} \, \frac{\cot (x)}{p^{2}+x^{2}} \, dx = \frac{i \pi}{p} \coth(p) \left[1-e^{-a \tanh(p)}\right] .$$

We can then equate the imaginary parts on both sides of the equation and drop the $PV$ sign since the integral is clearly convergent. (The zeroes of $\sin (a \tan z)$ cancel the poles of $\cot (z)$.)


EDIT:

The other integral is similar, but we have to close the contour in the lower half-plane since $\left|e^{i a \cot (z)}\right|$ can be large in the upper half-plane.

Along the real axis, the function $ \displaystyle g(z) = e^{i a \cot (z)} \, \frac{\tan(z)}{p^{2}+z^{2}}$ has simple poles at $z=\left(n + \frac{1}{2} \right) \pi$ and essential singularities at $z=n\pi$.

So consider a rectangular contour with vertices at $z=\pm \left(N+ \frac{1}{4}\right) \pi$ and $z=\pm \left(N+ \frac{1}{4} \right)\pi {\color{red}{-}} iy$ (for some $N \in \mathbb{N}$ and some $y>p$) that is indented along the real axis.

Letting $N \to \infty$ through the positive integers and $y \to \infty$, $\int g(z) \, dz$ vanishes on the left and right sides of the contour because of $(2)$, $(3)$, and $(5)$, and it vanishes along the bottom of the contour because of $(2)$, $(4)$, and $(5)$.

And because of $(2)$, the total contribution from the indentations around the essential singularities is vanishingly small.

So we have $$ \text{PV} \int_{-\infty}^{\infty} e^{ia \cot (x)} \, \frac{\tan (x)}{p^{2}+x^{2}} \, dx \, {\color{red}{+}} \, i \pi \sum_{n=-\infty}^{\infty}\text{Res} \left[f(z), \left(n+ \frac{1}{2}\right)\pi \right] = {\color{red}{-}} 2 \pi i \, \text{Res}[f(z), -ip] $$

where $$\sum_{n=-\infty}^{\infty}\text{Res} \left[f(z), \left(n+ \frac{1}{2}\right)\pi \right] = -\sum_{n=-\infty}^{\infty} \frac{1}{p^{2} + \left(\frac{2n+1}{2} \right)^{2} \pi^{2}} = -\frac{\tanh (p)}{p}$$

and $$\text{Res}[f(z), -ip] = \frac{e^{-a \coth (p)} \tanh(p)}{2ip}.$$

Therefore, $$\text{PV}\int_{-\infty}^{\infty} e^{i a \cot (x)} \, \frac{\tan (x)}{p^{2}+x^{2}} \, dx = \frac{i \pi}{p} \tanh(p) \left[1-e^{-a \coth(p)}\right] .$$

We can then equate the imaginary parts on both sides of the equation and again drop the $PV$ sign since the integral is convergent. (The zeroes of $\sin (a \cot z)$ cancel the poles of $\tan (z)$.)

Post Sun Oct 18, 2015 5:26 pm
galactus User avatar
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Wow, look at you G. Very cool indeed.


Maybe try this one on for size:

$$\int_{0}^{1}\frac{1-x^{n-1}}{(1-x)(1-x^{n})}(-\ln(x))^{m-1}dx=\left(1-\frac{1}{n^{m}}\right)\Gamma(m)\zeta(m)$$

Post Fri Oct 23, 2015 11:49 pm

Posts: 50
A slightly different (but uncreative and lengthy) approach to evaluating the integral Random Variable posted is as follows.

\begin{align}
\int^\infty_0\sin(a\tan{x})\frac{\cot{x}}{x^2+p^2}\ {\rm d}x
&=\frac{1}{2}\sum^\infty_{n=-\infty}\int^{\frac{\pi}{2}+n\pi}_{-\frac{\pi}{2}+n\pi}\sin(a\tan{x})\frac{\cot{x}}{x^2+p^2}\ {\rm d}x\\
&=\frac{1}{2}\sum^\infty_{n=-\infty}\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin(a\tan{x})\frac{\cot{x}}{(x+n\pi)^2+p^2}\ {\rm d}x\\
&=\frac{1}{2}\int^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin(a\tan{x})\cot{x}\left[-\sum_{\pm}\operatorname*{Res}_{z=\frac{-x\pm ip}{\pi}}\frac{\pi\cot(\pi z)}{(x+z\pi)^2+p^2}\right]\ {\rm d}x\\
&=\frac{1}{4ip}\int^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin(a\tan{x})\cot{x}\left(\cot(x-ip)-\cot(x+ip)\right)\ {\rm d}x\\
&=\frac{\tanh{p}}{2p}\int^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin(a\tan{x})\cot{x}\left(\frac{\tan^2{x}+1}{\tan^2{x}+\tanh^2{p}}\right)\ {\rm d}x\\
&=\frac{\tanh{p}}{2p}\int^\infty_{-\infty}\frac{\sin(ax)}{x(x^2+\tanh^2{p})}\ {\rm d}x\\
&=\frac{\coth{p}}{2p}\int^\infty_{-\infty}\frac{\sin(ax)}{x}-\frac{x\sin(ax)}{x^2+\tanh^2{p}}\ {\rm d}x\\
&=\frac{\pi\coth{p}}{2p}\left(1-e^{-a\tanh{p}}\right)
\end{align}
I believe this approach applies to the other integral as well.

Also,
\begin{align}
\int^1_0\frac{1-x^{n-1}}{(1-x)(1-x^n)}(-\ln{x})^{m-1}\ {\rm d}x
&=\int^1_0\frac{1+\cdots+x^{n-2}+x^{n-1}-x^{n-1}}{(1-x)(1+\cdots+x^{n-1})}(-\ln{x})^{m-1}\\
&=\int^1_0\frac{(-\ln{x})^{m-1}}{1-x}-\frac{x^{n-1}(-\ln{x})^{m-1}}{1-x^n}\ {\rm d}x\\
&=\left(1-\frac{1}{n^m}\right)\int^1_0\frac{(-\ln{x})^{m-1}}{1-x}\ {\rm d}x\\
&=\left(1-\frac{1}{n^m}\right)\sum^\infty_{n=0}\int^1_0x^n(-\ln{x})^{m-1}\ {\rm d}x\\
&=\left(1-\frac{1}{n^m}\right)\sum^\infty_{n=0}\int^\infty_0x^{m-1}e^{-(n+1)x}\ {\rm d}x\\
&=\left(1-\frac{1}{n^m}\right)\sum^\infty_{n=0}\frac{1}{(n+1)^m}\Gamma(m)\\
&=\left(1-\frac{1}{n^m}\right)\Gamma(m)\zeta(m)\\
\end{align}

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