Home > Analysis, Olympiad, Real Analysis, Uncategorized > SEEMOUS 2016 Problem 4 – Solution

## SEEMOUS 2016 Problem 4 – Solution

Problem 4. Let ${n \geq 1}$ be an integer and set

$\displaystyle I_n = \int_0^\infty \frac{\arctan x}{(1+x^2)^n}dx.$

Prove that

a) ${\displaystyle \sum_{i=1}^\infty \frac{I_n}{n} =\frac{\pi^2}{6}.}$

b) ${\displaystyle \int_0^\infty \arctan x \cdot \ln \left( 1+\frac{1}{x^2}\right) dx = \frac{\pi^2}{6}}$.

Proof: First, let’s note that each of the integrals ${I_n}$ is convergent since ${I_n \leq \int_0^\infty \arctan x dx = \pi/2}$. We can deduce a better upper bound using the fact that ${(1+x^2)^n \geq 1+nx^2}$. We have

$\displaystyle I_n \leq \int_0^\infty \frac{\arctan x}{1+nx^2} dx \leq \frac{\pi^2}{4\sqrt{n}}.$

Now, note that the series of functions

$\displaystyle \sum_{i=n}^\infty \frac{\arctan x}{n(1+x^2)^n}$

converges uniformly on each interval of the form ${[a,\infty)}$, since we have a geometric series multiplied by ${\arctan x}$. Indeed, if we denote

$\displaystyle F(z) = \sum_{n \geq 1} \frac{z^n}{n},$

then this series is absolutely convergent for ${|z|<1}$ and is derivable term by term. Thus, we have ${F'(z) = \sum_{n \geq 0} z^n = 1/(1-z)}$. Integrating and noting that ${F(0)=0}$ we obtain ${F(z) = \displaystyle \ln \left( \frac{1}{1-z}\right)}$ for ${|z|<1}$. Thus, we can invert the series with the integrals and obtain that

$\displaystyle \sum_{n=1}^\infty \int_a^\infty \frac{\arctan x}{n(1+x^2)^n}dx = \int_a^\infty \sum_{n=1}^\infty \frac{\arctan x}{n(1+x^2)^n}dx$

which gives

$\displaystyle \sum_{n=1}^\infty \int_a^\infty \frac{\arctan x}{n(1+x^2)^n}dx= \int_a^\infty \arctan x \ln \left(1+\frac{1}{x^2}\right) dx.$

Since ${I_n/n \leq \displaystyle \frac{\pi^2}{4n\sqrt{n}}}$ we conclude that the series ${I_n}$ is convergent and thus we can pass to the limit as ${a \rightarrow 0}$ in the above equality of integral to deduce that

$\displaystyle \sum_{i=1}^\infty \frac{I_n}{n} = \int_0^\infty \arctan x \ln \left(1+\frac{1}{x^2}\right) dx.$

This proves that b) follows once we have a).

To prove a) we make the change of variables ${x = \tan y}$ and we deduce that

$\displaystyle I_n = \int_0^{\pi/2} y\cos^{2n-2} y dy.$

We note that ${G(z) = \sum_{n \geq 1} \frac{z^{2n-2}}{n} = \frac{1}{z^2} \ln (1/(1-z^2)}$ for ${|z|<1}$ and that by applying a similar method as in the first part we can switch the series and the integral to conclude that

$\displaystyle \sum_{n \geq 1} \frac{I_n}{n} = -2\int_0^{\pi/2} \frac{ y \ln (\sin y)}{\cos^2 y} dy.$

Noting that ${1/cos^2 = (\tan)'}$ we perform an integration by parts in the above integral and we obtain

$\displaystyle -2\int_0^{\pi/2} \frac{ y \ln (\sin y)}{\cos^2 y} dy =-2 [x\ln(\sin x) \tan x]_0^{\pi/2}+$

$\displaystyle +2 \int_0^{\pi/2} \ln(\sin x)\tan x dx+2\int_0^{\pi/2} x dx =$

$\displaystyle \frac{\pi^2}{4} +2 \int_0^{\pi/2} \ln(\sin x)\tan x dx.$

To finish it is enough to prove that

$\displaystyle I = \int_0^{\pi/2} \ln(\sin x)\tan x dx = -\frac{\pi^2}{24}.$

We have

$\displaystyle \int_0^{\pi/2} \ln(\sin x)\tan x dx = \frac{1}{2} \int_0^{\pi/2} \ln(1-\cos^2 x)\frac{\sin x}{\cos x} dx$

which by the change of variables ${\cos x = u}$ gives

$\displaystyle I = \frac{1}{2}\int_0^1 \frac{\ln(1-u^2)}{u} du.$

Since ${\ln(1-u^2) = -\sum_{n=1}^\infty \frac{u^{2n}}{n}}$ and the series converges uniformly for ${|u|<1}$ we have

$\displaystyle I = -\frac{1}{2} \sum_{n=1}^\infty \int_0^1 \frac{u^{2n-1}}{n}=-\frac{1}{2} \sum_{n=1}^\infty \frac{1}{2n^2}=-\frac{1}{4} \sum_{n=1}^\infty \frac{1}{n^2}=-\frac{\pi^2}{24},$

since ${\sum_{n=1}^\infty 1/n^2 = \pi^2/6}$. This finishes the proof.

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1. April 19, 2016 at 10:28 pm

Alternative proof for a): $I_{n+1}=int_0^{pi/2} y cos^{2n}(y) dy=int_0^{pi/2} y cos^{2n-1}(y) (sin(y))’ dy$ and integrating by parts we get: $I_{n+1}= – 1/(2n)+(2n-1)I_n – (2n-1)I_{n+1}$ which is equivalent with: $I_n/n = 2(I_n – I_{n+1})-1/(2n^2)$. Now, the series is equal to 2I_1-pi^2/12=pi^2/6.

2. April 19, 2016 at 10:31 pm

Alternative proof for a): $I_{n+1}=int_0^{pi/2} y cos^{2n}(y) dy=int_0^{pi/2} y cos^{2n-1}(y) (sin(y))’ dy$ and integrating by parts we get: $I_{n+1}= – 1/(2n)+(2n-1)I_n – (2n-1)I_{n+1}$ which is equivalent with: $I_n/n = 2(I_n – I_{n+1})-1/(2n^2)$. Now, the series is equal to $2I_1-pi^2/12=pi^2/6.$

1. March 6, 2016 at 2:58 pm