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January 28th, 2012, 02:15 PM  #1 
Member Joined: Aug 2011 Posts: 44 Thanks: 0  Integral with Difficult Conditions 
January 29th, 2012, 01:57 PM  #2 
Senior Member Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0  Re: Integral with Difficult Conditions
I think it is simply nonsensical to ask to solve such an integral without any of those tools. I mean, it's as if I said, "Without a hammer or hard object, nail down a spike on the wood." I know there are other tools, but it's just pushing the problem too hard. I have computed the value of that integral some time ago and it is $\displaystyle  \frac{\pi^2}{12}$
Last edited by skipjack; August 3rd, 2017 at 05:24 AM. 
January 29th, 2012, 02:16 PM  #3 
Senior Member Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0  Re: Integral with Difficult Conditions
You can use double integrals if you want. $\displaystyle \int\limits_0^1 {\frac{{\log u}}{{u + 1}}du} = \lambda$ $\displaystyle \int\limits_0^1 {\int\limits_0^{u  1} {\frac{{dx}}{{x + 1}}} \frac{{du}}{{u + 1}}} = \lambda$ or $\displaystyle \int\limits_0^1 {\int\limits_1^{u } {\frac{{dx}}{{x }}} \frac{{du}}{{u + 1}}} = \lambda$ But then again you will get to a step you'll need to use series or whatever useful tool you need for these type of integrals. Last edited by skipjack; August 3rd, 2017 at 05:21 AM. 
January 30th, 2012, 04:30 AM  #4  
Member Joined: Aug 2011 Posts: 44 Thanks: 0  Re: Integral with Difficult Conditions Quote:
There is Method to solve this integral according my conditions doesn't make that difficult. If you cannot solve it, let a chance for other... Last edited by skipjack; August 3rd, 2017 at 05:20 AM.  
January 30th, 2012, 04:35 AM  #5 
Senior Member Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0  Re: Integral with Difficult Conditions
Ok, sorry, give me a hint and I'll try to solve it.

February 24th, 2012, 11:13 AM  #6  
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038  Quote:
It's easy to show that $\displaystyle \int_0^1\frac{\ln(y)}{1\,+\,y}dy = \int_0^1\,\frac{\ln(1\,+\,y)}{y}\,dy = \frac12\int_0^1\,\frac{\ln(1\,\,y)}{y}\,dy = \frac12\zeta(2).$ Ignoring the specified conditions, one needs to show that $\zeta$(2) = $\pi$²/6, which is known as the Basel problem. Let L = $\displaystyle \int_0^1\,\frac{\ln(1\,+\,y)}{y}\,dy =\int_{{\small}1}^0\,\frac{\ln(1  y)}{y}\,dy,$ and M =$\displaystyle \int_0^1\,\frac{\ln(1  y)}{y}\,dy$ = 2L. 3L = L + M = $\displaystyle \int_{{\small}1}^1\,\frac{\ln(1\,\,y)}{y}\,dy.$ Substituting y = 1/x, dy = 1/x² dx, 3L = $\displaystyle \int_{{\small}1}^1\,\frac{\ln(1\,\,1/x)}{x}\,dx\,=\int_{{\small}1}^1\,\frac{\ln(1\,\,x)\,\,\ln(x)}{x}\,dx.$ Hence 6L = $\displaystyle \int_{{\small}1}^1\,\frac{\ln(x)}{x}\,dx = {\bigg[}\frac{\small1}{\small2}\ln^{\small2}(x){\bigg]}_{{\small}1}^1 = \frac{\pi^{\small2}}{\small2},$ and so L = $\pi$²/12. The above substitution step effectively introduces the use of complex numbers (and needs a little amplification to justify it), but I'm hoping this doesn't count as "Complex Analysis". See also this article. Last edited by skipjack; August 5th, 2017 at 05:16 PM.  

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