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| Интегралы http://mathhelpplanet.com/viewtopic.php?f=19&t=20459 |
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| Автор: | jagdish [ 14 дек 2012, 22:41 ] |
| Заголовок сообщения: | Интегралы |
If [math]I(k) = \int_{0}^{\infty}\frac{\ln(x)}{x^2+kx+k^2}.[/math] Then [math]kI(k)-I(1) =[/math] |
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| Автор: | Avgust [ 14 дек 2012, 23:36 ] |
| Заголовок сообщения: | Re: Интегралы |
[math]I(k) = \int_{0}^{\infty}\frac{\ln(x)}{x^2+kx+k^2}=\frac{2\pi \ln(k)}{3 \sqrt{3}\, k}[/math] [math]k \cdot I(k)-I(1)=\frac{2\pi }{3 \sqrt{3}}\ln(k)[/math]
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| Автор: | jagdish [ 15 дек 2012, 06:23 ] |
| Заголовок сообщения: | Re: Интегралы |
Thanks Avgust Got it. [math]\bf{\mathbb{I(\bold{k})}=\int_{0}^{\infty}\frac{\ln(x)}{x^2+kx+k^2}dx.................(1)}[/math] Put [math]\bf{x=kt\Leftrightarrow dx = kdt}[/math] and Changing Limits, We Get [math]\bf{\mathbb{I(\bold{k})}=\int_{0}^{\infty}\frac{\ln(kt).kdt}{k^2(t^2+t+1)}dt}[/math] [math]\bf{\mathbb{\bold{k}.I(\bold{k})}=\int_{0}^{\infty}\frac{\ln(k)}{t^2+t+1}dt+\int_{0}^{\infty}\frac{\ln(t)}{t^2+t+1}dt}[/math] [math]\bf{\mathbb{\bold{k}.I(\bold{k})}=\ln(k).\int_{0}^{\infty}\frac{1}{t^2+t+1}dt+\mathbb{I(\bold{1})}}[/math] Using eqation [math]\bf{(1)\;\;,}[/math] Put [math]\bf{k=1}[/math] in eqn...[math]\bf{(1)}[/math] We Get [math]\bf{I(\bold{1})=\int_{0}^{\infty}\frac{\ln(x)}{x^2+x+1}=\int_{0}^{\infty}\frac{\ln(t)}{t^2+t+1}\right)}[/math] So [math]\bf{\mathbb{\bold{k}.I(\bold{k})}-I(1)=\bf{\ln(k)\int_{0}^{\infy}\frac{1}{t^2+t+1}dt}}[/math] [math]\bf{=\ln(k)\int_{0}^{\infty}\frac{1}{\left(t+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}dt=\ln(k).\frac{2}{\sqrt{3}}.\tan^{-1}\left(\frac{2t+1}{\sqrt{3}}\right)\bigg|_{0}^{\frac{\infty}}}[/math] [math]\bf{=\ln(k).\frac{2}{\sqrt{3}}\left(\frac{\pi}{2}-\frac{\pi}{6}\right)=\ln(k).\frac{2\pi}{3\sqrt{3}}}[/math] So [math]\boxed{\boxed{\bf{\mathbb{\bold{k}.I(\bold{k})}-I(\bold{1})=\frac{2\pi}{3\sqrt{3}}.ln(k)}}}[/math] |
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