Spicy Limits and Integrals!

Consider the following three facts :

$\begin{cases} \Omega_1 + \Omega_2 = \frac{\pi}{2} \;\;\;\;\;\;\;\; (1) \\ \Omega_3 = \ln 2 \;\;\;\;\;\;\;\;\;\;\;\;\; (2) \\ \Omega_2 - \Omega_1 \geq 0 \;\;\;\;\;\;\;\; (3) \end{cases}$

$\bullet \mathbf{\;Proof \; of \; (1) \; :}$

$\displaystyle \Omega_1 + \Omega_2 = \int_0^1 \sin^{-1} \left( \dfrac{x}{\sqrt{1+x^2}} \right) \mathrm{d}x + \int_0^1 \cos^{-1} \left( \dfrac{x}{\sqrt{1+x^2}} \right) \mathrm{d}x$

$\displaystyle = \int_0^1 \sin^{-1} \left( \dfrac{x}{\sqrt{1+x^2}} \right) + \cos^{-1} \left( \dfrac{x}{\sqrt{1+x^2}} \right) \mathrm{d}x = \int_0^1 \frac{\pi}{2} \mathrm{d}x = \frac{\pi}{2}$

$\bullet \mathbf{\;Proof \; of \; (2) \; :}$

$\displaystyle \Omega_3 = \lim_{n \to \infty} \left( \dfrac{1}{n+1} + \dfrac{1}{n+2} + \ldots + \dfrac{1}{2n} \right) = \lim_{n \to \infty} \sum_{k=1}^{n} \dfrac{1}{n+k} = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \dfrac{1}{1+\frac{k}{n}}$

$\displaystyle = \int_0^1 \frac{dx}{1+x} = \ln 2$

$\bullet \mathbf{\;Proof \; of \; (3) \; :}$

Using the facts that $\sin^{-1} \left( \dfrac{x}{\sqrt{1+x^2}} \right)$ is increasing and $\cos^{-1} \left( \dfrac{x}{\sqrt{1+x^2}} \right)$ is decreasing, and considering the integration limits of $\Omega_1 \; \text{and} \; \Omega_2$ , we can bound the integrand of $\Omega_1$ in $[0,\frac{\pi}{4}]$ and the integrand of $\Omega_2$ in $[\frac{\pi}{4},\frac{\pi}{2}]$ . Therefore:

$\displaystyle \Omega_2 - \Omega_1 \geq 0$

Hence our facts are proved. We know that only one of the given $8$ choices should be correct.

Using facts $(1)$ and $(2)$ , we can conclude that $\Omega_1 +\Omega_2+\Omega_3 = \frac{\pi}{2} + \ln 2 \; \text{and} \; \Omega_1 +\Omega_2 \neq \Omega_3$ ,

which eliminates $6$ choices .

The remaining $2$ choices say that either :

$\displaystyle \Omega_3 = \Omega_2 - \Omega_1 \; \text{or} \; \Omega_3 =\Omega_1 - \Omega_2$

But since $\Omega_3 >0$ and using fact $(3)$ , we are left with $\Omega_3 = \Omega_2 - \Omega_1$ .

So the correct answer is:

$\displaystyle \boxed{\Omega_1 + \Omega_3 = \Omega_2 }$

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For Clarity, I will solve one of the integrals; Just basic integration techniques :

$\displaystyle \Omega_1 = \int_0^1 \sin^{-1} \left( \dfrac{x}{\sqrt{1+x^2}} \right) \mathrm{d}x$

$\displaystyle \overset{{\color{#D61F06}{ x \to \tan u }}}{=} \int_0^{\frac{\pi}{4}} u\sec^2u du$

$\displaystyle \overset{{\color{#D61F06}{ \text{Integration By Parts} }}}{=} u\tan u |_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}} \tan u du$

$\displaystyle = (u\tan u - \ln \sec u ) |_0^{\frac{\pi}{4}} = \frac{\pi}{4} - \frac{1}{2} \ln 2$

Using this result and fact $(1)$ , we can find $\Omega_2 - \Omega_1$ and assure the correct answer.