#2 03. 04. 2018 17:47 — Editoval laszky (03. 04. 2018 18:02)
Re: Nevlastní integrál
Ahoj. Pravdepodobne je kratsi postup nejakou chytrou substituci, ale...
a vsiml si, ze je 
. Potom muzeme integral upravit na (subs.t z=1/x)
![$\int_0^1\frac{x^2+1}{x^2}\arccos^2g(x)\mathrm dx = \left[\left(x-\frac{1}{x}\right)\arccos^2g(x)\right]_0^1 + \int_0^1\left(x-\frac{1}{x}\right)\frac{2\arccos g(x)}{\sqrt{1-g^2(x)}}g'(x)\mathrm dx$](/mathtex/7a/7a04ca254a7b7460dab2a502068ae584.gif "kopírovat do textarea")
.
lze upravit na 
, takze (subst g(x)=cos(y))
, je![$ 4\int_{0}^{\pi} \frac{y\sin y}{1-\cos y}\mathrm{d} y = 8\mathrm{Im} \sum_{k=1}^{\infty} \int_{0}^{\pi} y\mathrm{e}^{iky}\mathrm{d}y = 8\mathrm{Im} \sum_{k=1}^{\infty} \left[\left(\frac{1}{k^2}-\frac{iy}{k}\right)\mathrm{e}^{iky}\right]_0^{\pi} = $](/mathtex/a2/a299320448d73afd374eaf77287cd2ec.gif "kopírovat do textarea")
![$= 8\mathrm{Im} \sum_{k=1}^{\infty} \left[\left(\frac{1}{k^2}-\frac{iy}{k}\right)(\cos(ky)+i\sin(ky))\right]_0^{\pi} = -8\pi\sum_{k=1}^{\infty}\frac{\cos(k\pi)}{k} = -8\pi\sum_{k=1}^{\infty}\frac{(-1)^k}{k}$](/mathtex/b8/b83396956625ee680b43f7848de0198c.gif "kopírovat do textarea")
, pak 
, takze 
, a tedy 
.
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#4 03. 04. 2018 23:25 — Editoval Pavel (04. 04. 2018 01:19)
Re: Nevlastní integrál
↑ laszky:
Elegantní řešení. Já jsem použil substituce, reálnou analýzou a několik limit. Řešení mám ale delší.
HINT:
.ŘEŠENÍ:
![$
I=\int_0^{\infty}\arccos^2\left(\frac{1-6x^2+x^4}{(1+x^2)^2}\right)\,\mathrm dx
&=\int_0^{\infty}\arccos^2\left(\frac{1-2x^2+x^4}{(1+x^2)^2}-\frac{4x^2}{(1+x^2)^2}\right)\,\mathrm dx\\[.5\baselineskip]
&=\int_0^{\infty}\arccos^2\left(\left(\frac{1-x^2}{1+x^2}\right)^2-\left(\frac{2x}{1+x^2}\right)^2\right)\,\mathrm dx
$](/mathtex/77/7704bcfc7959ca6c3fa7f50ade6af97b.gif "kopírovat do textarea")
. Pak zřejmě
![$
\int_0^{\infty}\arccos^2\left(\left(\frac{1-x^2}{1+x^2}\right)^2-\left(\frac{2x}{1+x^2}\right)^2\right)\,\mathrm dx
&=\int_0^{\frac{\pi}2}\arccos^2\left(\cos^2(2y)-\sin^2(2y)\right)\cdot\frac 1{\cos^2y}\,\mathrm dy\\[.5\baselineskip]
&=\int_0^{\frac{\pi}2}\arccos^2\left(\cos(4y)\right)\cdot\frac 1{\cos^2y}\,\mathrm dy
$](/mathtex/e6/e634784fa74ba59d0e02b417aa6d8cb4.gif "kopírovat do textarea")
![$
\int_0^{\frac{\pi}2}\arccos^2\left(\cos(4y)\right)\cdot\frac 1{\cos^2y}\,\mathrm dy
&=\int_0^{\frac{\pi}4}16y^2\cdot\frac 1{\cos^2y}\,\mathrm dy+\int_{\frac{\pi}4}^{\frac{\pi}2}(2\pi-4y)^2\cdot\frac 1{\cos^2y}\,\mathrm dy\\[.5\baselineskip]
&=\int_0^{\frac{\pi}4}16y^2\cdot\frac 1{\cos^2y}\,\mathrm dy+\int_{\frac{\pi}4}^{\frac{\pi}2}(4\pi^2-16\pi y+16y^2)\cdot\frac 1{\cos^2y}\,\mathrm dy\\[.5\baselineskip]
&=\lim_{t\to\frac{\pi}2^-}\biggl(4\pi^2\underbrace{\int_{\frac{\pi}4}^t\frac 1{\cos^2}\,\mathrm dy}_{I_1}-16\pi\underbrace{\int_{\frac{\pi}4}^t \frac y{\cos^2 y}\,\mathrm dy}_{I_2}+16\underbrace{\int_0^t \frac{y^2}{\cos^2y}\,\mathrm dy}_{I_3}\biggr).
$](/mathtex/fa/fa287b7207f900b79f41936988a3cfd7.gif "kopírovat do textarea")
, 
, 
řešíme přímo nebo metodou per partes:![$
I_1&=\int_{\frac{\pi}4}^t\frac 1{\cos^2}\,\mathrm dy=\left[\tan y\right]_{\frac{\pi}4}^t=\tan t-1,\\[.5\baselineskip]
I_2&=\int_{\frac{\pi}4}^t\frac y{\cos^2}\,\mathrm dy=\left[y\tan y+\log\cos y\right]_{\frac{\pi}4}^t=t\tan t+\log\cos t-\frac{\pi}{4}+\frac 12\,\log 2,\\[.5\baselineskip]
I_3&=\int_{\frac{\pi}4}^t\frac {y^2}{\cos^2}\,\mathrm dy=\left[y^2\tan y+2y\log\cos y\right]_0^t-2\int_0^t\log\cos y\,\mathrm dy=t^2\tan t+2t\log\cos t-2\int_0^t\log\cos y\,\mathrm dy.
$](/mathtex/42/429270cf083556e0a06cbf453b121ddb.gif "kopírovat do textarea")
![$
I&=\lim_{t\to\frac{\pi}2^-}(4\pi^2I_1-16\pi I_2+16I_3)\\[.5\baselineskip]
&=\lim_{t\to\frac{\pi}2^-}\left(4\pi^2(\tan t-1)-16\pi\left(t\tan t+\log\cos t-\frac{\pi}{4}+\frac 12\,\log 2\right)+16\left(t^2\tan t+2t\log\cos t-2\int_0^t\log\cos y\,\mathrm dy\right)\right)\\[.5\baselineskip]
&=\lim_{t\to\frac{\pi}2^-}\left((4\pi^2-16\pi t+16t^2)\tan t+(32t-16\pi)\log\cos t\right)-8\pi\log 2-32\int_0^{\frac{\pi}2}\log\cos y\,\mathrm dy\\[.5\baselineskip]
&=\lim_{t\to\frac{\pi}2^-}\left((4\pi^2-16\pi t+16t^2)\tan t+(32t-16\pi)\log\cos t\right)+8\pi\log 2
$](/mathtex/b6/b6971657e4ea608bd0a8168497ae7dba.gif "kopírovat do textarea")
↑ Marian:
Myslím, že některé postupy se používají i zde.
Backslash je v TeXu tak důležitý jako nekonečno při dělení nulou v tělesech charakteristiky 0.
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