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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
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Re: Integral with Difficult Conditions

Quote:
 Originally Posted by Weiler 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 $\displaystyle - \frac{\pi^2}{12}$
No need for this method in dealing...

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
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Quote:
 Originally Posted by uniquesailor There is Method to solve this integral . . .
Does the method use the fact that ln²(-1) = -$\pi$²?

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. Tags conditions, difficult, integral Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Blomster Complex Analysis 1 May 7th, 2012 03:20 AM RDaneel177 Real Analysis 1 January 27th, 2011 03:32 AM kurdishmath Calculus 1 November 27th, 2010 01:57 PM Danman Complex Analysis 2 December 8th, 2009 01:49 PM Danman Calculus 2 November 2nd, 2008 01:51 PM

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