7
$\begingroup$

For $n \in \mathbb N$, $n \geq 1$, consider the following integral expression:

\begin{equation} I(n) := \frac{2n}{\pi}\int_0^{2\sqrt{2n-1}} \frac{\ln(x)\sqrt{-x^2 + 8n-4}}{4n^2-x^2} dx \end{equation}

My attempts to use any of the obvious (elementary) few methods that I am accustomed with (i.e partial integration, substitution, series expansion of the integrand) in order to find a general antiderivative were all futile. There might still be a way to compute an antiderivative using complex analysis, but this is currently beyond my capabilites. However, although Wolfram Alpha does also not seem to be able to either to compute a general antiderivate or compute the above expression for general $n \in \mathbb N$, it will yield (after possibly some refreshing) an explicit value for any $n$ I've tested so far (all $n$ between $1$ and $15$). Namely, one gets the following result for $1 \leq n \leq 15:$

\begin{equation} I(n)= (n-\frac{1}{2}) \ln(2n-1) - (n-1) \ln(2n). \end{equation}

This suggest that there is indeed a way to compute the above expression explicitly. Any help is highly appreciated.

$\endgroup$
8
$\begingroup$

Hint. By making the change of variable, $$ x=2\sqrt{2n-1}\:u ,\quad dx=2\sqrt{2n-1}\:du,\quad u=\frac{x}{2\sqrt{2n-1}}, $$ one has $$ \frac{\pi}{2n}\cdot I(n)=\ln(2\sqrt{2n-1})\int_0^1\frac{\sqrt{1-u^2}}{a^2-u^2}\:du+\int_0^1\frac{\ln u \cdot\sqrt{1-u^2}}{a^2-u^2}\:du \tag1 $$ with $$ a:=\frac{n}{\sqrt{2n-1}}>1, \quad (n>1).\tag2 $$ The first integral on the right hand side of $(1)$ may be evaluated by two changes of variable, $u=\sin \theta$ and $t=\tan\theta$, obtaining, for $a>1$,

$$ \int_0^1\frac{\sqrt{1-u^2}}{a^2-u^2}\:du=\frac12\int_0^\infty\frac{1}{\left((a^2-1)t^2+a^2\right)\cdot\left(t^2+1\right)}\:dt=\frac{\pi \left(a-\sqrt{a^2-1}\right)}{2 a}.\tag3 $$

Similarly, the second integral on the right hand side of $(1)$ is such that $$ \int_0^1\frac{\ln u \cdot\sqrt{1-u^2}}{a^2-u^2}\:du= \frac{1}{2}\int_0^\infty\frac{2\ln t-\ln(1+t^2)}{\left((a^2-1)t^2+a^2\right)\cdot\left(t^2+1\right)}\:dt,\quad (a>1),\tag4 $$ transforming the integrand by a partial fraction decomposition, using the standard results $$ \begin{align} \int_0^\infty\frac{\ln t}{t^2+\alpha^2}\:dt&=\frac{\pi}{2}\cdot\frac{\ln\alpha}{\alpha},\quad \alpha>0, \qquad (\star) \\\int_0^\infty\frac{\ln (1+t^2)}{t^2+\alpha^2}\:dt&=\pi\cdot\frac{\ln(\alpha+1)}{\alpha},\quad \alpha>0,\qquad (\star \star) \end{align} $$ one gets

$$ \int_0^1\frac{\ln u \cdot\sqrt{1-u^2}}{a^2-u^2}\:du=\frac{\pi\sqrt{a^2-1}}{2 a} \cdot\ln\left(\frac{\sqrt{a^2-1}+a}{a}\right)-\frac{\pi}2 \ln 2,\quad (a>1).\tag5 $$

Combining $(3)$ and $(5)$ with $(2)$ gives

$$ I(n) := \frac{2n}{\pi}\int_0^{2\sqrt{2n-1}} \frac{\ln(x)\sqrt{-x^2 + 8n-4}}{4n^2-x^2} dx= \left(n-\frac{1}{2}\right) \ln(2n-1) - (n-1) \ln(2n) $$

for all real numbers $n$ such that $n>1$.


Edit. To prove $(\star)$, one may perform the change of variable $t=\alpha u$, $\alpha>0$, getting $$ \int_0^\infty\frac{\ln t}{t^2+\alpha^2}\:dt=\frac1\alpha \cdot\int_0^\infty\frac{\ln \alpha+ \ln u}{u^2+1}\:du=\frac{\pi}{2}\cdot\frac{\ln\alpha}{\alpha} $$ since $$ \int_0^\infty\frac{1}{u^2+1}\:du=\left[\frac{}{}\arctan u \frac{}{}\right]_0^\infty=\frac \pi2, \quad \int_0^\infty\frac{\ln u}{u^2+1}\:du=0 $$ (the latter is seen by making $u \to 1/u$).

To prove $(\star\star)$, one may perform the change of variable $t=\alpha u$, $\alpha>0$, getting $$ \int_0^\infty\frac{\ln (1+t^2)}{t^2+\alpha^2}\:dt=\frac1\alpha \cdot\int_0^\infty\frac{\ln (1+\alpha^2u^2)}{u^2+1}\:du $$ differentiating the latter integral with respect to $\alpha$ gives a classic evaluation $$ \frac{d}{d\alpha}\int_0^\infty\frac{\ln (1+\alpha^2u^2)}{u^2+1}\:du=\int_0^\infty\frac{2\alpha u^2}{(u^2+1)(1+\alpha^2u^2)}\:du=\frac{\pi}{\alpha+1} $$ then integrating one gets $(\star\star)$.

$\endgroup$
  • $\begingroup$ Thanks. Am I right to assume that your two standard results originally both come from complex analysis and the method of countour integration ? $\endgroup$ – Berni Waterman Apr 3 '17 at 14:35
  • 1
    $\begingroup$ @BerniWaterman These results are completely real and do not require complex tools. $\endgroup$ – Jack Tiger Lam Apr 3 '17 at 14:40
  • $\begingroup$ I managed to compute both of the equalities using contour integration, but I don't see how to compute them using only real analysis. Feel free to elaborate on this, if you like. $\endgroup$ – Berni Waterman Apr 3 '17 at 16:25
  • $\begingroup$ @BerniWaterman Yes, I will elaborate on this. $\endgroup$ – Olivier Oloa Apr 3 '17 at 18:57
  • 2
    $\begingroup$ Yes, thank you so much! $\endgroup$ – Berni Waterman Apr 4 '17 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.