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How to prove the equality $$\int^{na}_{ma}\frac{\ln(x-a)}{x^2+a^2} \, dx= \int^{\frac{a}{m}}_{\frac{a}{n}}\frac{\ln(x+a)}{x^2+a^2}\, dx,$$

where $a, m, n$ are strictly positive numbers such that $mn=m+n+1$?

Michael Hardy
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medicu
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  • I'm not sure that "As demonstrated the equality...?" is English. What do you want people to help you do? What have you tried? – not all wrong May 25 '13 at 15:32
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    @Sharkos: Most of the users here do not speak English. – Inceptio May 25 '13 at 15:38
  • That's not a problem, but I was not sure what was being asked, which is. For example, is the question "Does the following hold?" or "When does it hold?" or "Prove it holds."? – not all wrong May 25 '13 at 15:42

2 Answers2

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$x=a^2/u$ should help you. For example, $dx/(x^2+a^2)=-du/(a^2+u^2)$. I believe the constraint comes into play when consisting the extra terms in the logarithm.

However, I'm not quite sure about the $\pm$ signs in the logarithm; are these correct?

This substitution is motivated by the observation that you go from $am\to a/m$ via $x\to a^2/x$.

not all wrong
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My solution:

To note $I_{1}=\int^{na}_{ma}\frac{ln(x-a)}{x^{2}+a^{2}}dx$ and $I_{2}= \int^{\frac{a}{m}}_{\frac{a}{n}}\frac{ln(x+a)}{x^{2}+a^{2}}dx.$

Using the substitution $x=\frac{at+a^{2}}{t-a}$ we find $I_{1}=\frac{ln(2a^{2})}{2a}[\arctan (n)- \arctan (m)]$.

Using the substitution $x=\frac{-at+a^{2}}{t+a}$ we find $I_{2}=\frac{ln(2a^{2})}{2a}(\arctan \frac{1}{m}- \arctan \frac{1}{n})$.

Because for $m > 0$ and $n > 0$ is true equality

$\arctan (m) + \arctan \frac{1}{m} = \frac{\pi}{2}= \arctan (n) + \arctan \frac{1}{n}$

then

$\arctan (n)-\arctan (m)= \arctan \frac{1}{m}-\arctan \frac{1}{n}$

and consequently

$I_{1}=I_{2}$.

For another solution see: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=536296

medicu
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