| Fórum de Matemática | DÚVIDAS? Nós respondemos! https://forumdematematica.org/ |
|
| int 0 to 1 {1/x}^2 *{1/(1-x)} dx https://forumdematematica.org/viewtopic.php?f=6&t=187 |
Página 1 de 1 |
| Autor: | kinu [ 06 fev 2012, 17:57 ] |
| Título da Pergunta: | int 0 to 1 {1/x}^2 *{1/(1-x)} dx |
\(\displaystyle%20\int_{0}^{1}\left\{\frac{1}{x}\right\}^2.\left\{\frac{1}{1-x}\right\}dx\) where \(\left\{x\right\} =\) fractional part function |
|
| Autor: | emsbp [ 11 mai 2012, 14:19 ] |
| Título da Pergunta: | Re: int 0 to 1 {1/x}^2 *{1/(1-x)} dx |
Boa tarde. Aconselho que visite o seguinte site: http://mathworld.wolfram.com/FractionalPart.html Obrigado. |
|
| Autor: | João P. Ferreira [ 11 mai 2012, 16:34 ] |
| Título da Pergunta: | Re: int 0 to 1 {1/x}^2 *{1/(1-x)} dx |
Hi kinu just some tips... You have to split it in parts the \(x=1/2\) is an important point you split in parts the integral on the set \(x\in[\frac{1}{2},1]\) \(\displaystyle%20\int_{0}^{1}\left\{\frac{1}{x}\right\}^2.\left\{\frac{1}{1-x}\right\}dx\) if \(f(x)=\left\{\frac{1}{x}\right\}^2.\left\{\frac{1}{1-x}\right\}\) Then \(\int_{1/2}^{2/3}f(x)dx=\int_{1/2}^{2/3}\left(\frac{1}{x}-1\right)^2\left(\frac{1}{1-x}-2\right)dx\) \(\int_{2/3}^{3/4}f(x)dx=\int_{1/2}^{2/3}\left(\frac{1}{x}-1\right)^2\left(\frac{1}{1-x}-3\right)dx\) \(\int_{3/4}^{4/5}f(x)dx=\int_{1/2}^{2/3}\left(\frac{1}{x}-1\right)^2\left(\frac{1}{1-x}-4\right)dx\) . . . \(\int_{\frac{n}{n+1}}^{\frac{n+1}{n+2}}f(x)dx=\int_{\frac{n}{n+1}}^{\frac{n+1}{n+2}}\left(\frac{1}{x}-1\right)^2\left(\frac{1}{1-x}-(n+1)\right)dx\) Then you do it for the second set \(x\in[0,\frac{1}{2}]\) \(\int_{1/3}^{1/2}f(x)dx=\int_{1/3}^{1/2}\left(\frac{1}{x}-1\right)^2\left(\frac{1}{1-x}-1\right)dx\) \(\int_{1/4}^{1/3}f(x)dx=\int_{1/4}^{1/3}\left(\frac{1}{x}-2\right)^2\left(\frac{1}{1-x}-1\right)dx\) . . . \(\int_{\frac{1}{n}}^{\frac{1}{n-1}}f(x)dx=\int_{\frac{1}{n}}^{\frac{1}{n-1}}\left(\frac{1}{x}-(n-2)\right)^2\left(\frac{1}{1-x}-1\right)dx\) Then calculate the finite integrals and at the end sum everything, you should get two infinite series... Take care. PS: kinu, after 57 posts with very dificult exercises to solve, your contributions to others have been ZERO We have no commitments here, and our target is to help other people on a philanthropic basis, but we have no ads, we don't want donations, and we solve problems for free, our core are our great math contributors, and at this stage you have posted 57 problems very difficult to solve and you have solved ZERO exercises. Do not destroy the spirit!!!! |
|
| Autor: | kinu [ 18 mai 2012, 15:34 ] |
| Título da Pergunta: | Re: int 0 to 1 {1/x}^2 *{1/(1-x)} dx |
Thanks João P. Ferreira for your Hint and extremelly Soory from my side. These types of things not happen in future Thanks |
|
| Página 1 de 1 | Os Horários são TMG [ DST ] |
| Powered by phpBB® Forum Software © phpBB Group https://www.phpbb.com/ |
|