Ahmed Integral

Calculus (Integrals, Series)
Post Reply
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

Ahmed Integral

#1

Post by Tolaso J Kos »

Prove that: $$\int_{0}^{1}\frac{\tan^{-1}\sqrt{2+x^2}}{\left ( 1+x^2 \right )\sqrt{x^2+2}}\,dx=\frac{5\pi^2}{96}$$


HINT
\( \displaystyle \frac{\pi^2}{16}=\int_{0}^{1}\int_{0}^{1}\frac{dx\,dy}{\left ( 1+x^2 \right )\left ( 1+y^2 \right )} \)
Imagination is much more important than knowledge.
User avatar
Riemann
Posts: 176
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Re: Ahmed Integral

#2

Post by Riemann »

Let us begin by the identity

$$\arctan x + \arctan \frac{1}{x} = \frac{\pi}{2} \quad \text{forall} \; x>0$$

Thus

\begin{align*} \mathcal{J} &= \int_{0}^{1} \frac{\arctan \sqrt{2+x^2}}{\left ( 1+x^2 \right ) \sqrt{2+x^2}} \, {\rm d}x \\
&= \frac{\pi}{2} \int_{0}^{1} \frac{{\rm d}x}{\left ( 1+x^2 \right ) \sqrt{2+x^2}} - \bigintsss_{0}^{1} \frac{\arctan \left ( \frac{1}{\sqrt{2+x^2}} \right )}{\left ( 1+x^2 \right )\sqrt{2+x^2}} \, {\rm d}x \\
&=\frac{\pi^2}{12} - \bigintsss_{0}^{1} \frac{\arctan \left ( \frac{1}{\sqrt{2+x^2}} \right )}{\left ( 1+x^2 \right )\sqrt{2+x^2}} \, {\rm d}x
\end{align*}

Now one of the $\arctan$ ‘s definition is

$$\arctan \frac{1}{a} = \int_{0}^{1} \frac{a}{x^2+a^2} \, {\rm d}x \Leftrightarrow \frac{1}{a} \arctan \frac{1}{a} = \int_{0}^{1} \frac{{\rm d}x}{x^2+a^2}$$

Hence

\begin{align*} \bigintsss_{0}^{1} \frac{\arctan \left ( \frac{1}{\sqrt{2+x^2}} \right )}{\left ( 1+x^2 \right )\sqrt{2+x^2}} \, {\rm d}x &= \int_{0}^{1} \int_{0}^{1} \frac{{\rm d}(x, y)}{\left ( 1+x^2 \right )\left ( 2+x^2+y^2 \right )} \\
&=\int_{0}^{1} \int_{0}^{1} \frac{1}{y^2+1} \bigg( \frac{1}{1+x^2} - \frac{1}{2+x^2+y^2} \bigg )\, {\rm d}(x, y) \\\\
&=\int_{0}^{1} \int_{0}^{1} \frac{{\rm d}(x, y)}{\left ( 1+x^2 \right ) \left ( 1+y^2 \right )} -\int_{0}^{1} \int_{0}^{1} \frac{{\rm d}(x, y)}{2+x^2+y^2} \\\\
&= \frac{1}{2} \int_{0}^{1} \int_{0}^{1} \frac{{\rm d}(x, y)}{\left ( 1+x^2 \right )\left ( 1+y^2 \right )} \\
&= \frac{1}{2}\left ( \int_{0}^{1} \frac{{\rm d}x}{1+x^2} \right )^2 \\
& = \frac{\pi^2}{32}
\end{align*}

and the result follows.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 11 guests