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January 28th, 2012, 02:15 PM   #1
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Integral with Difficult Conditions

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January 29th, 2012, 01:57 PM   #2
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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.
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January 29th, 2012, 02:16 PM   #3
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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.
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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.
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January 30th, 2012, 04:35 AM   #5
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Re: Integral with Difficult Conditions

Ok, sorry, give me a hint and I'll try to solve it.
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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.
Thanks from greg1313 and SDK

Last edited by skipjack; August 5th, 2017 at 05:16 PM.
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