STEP III 2007 question 7 solution

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(i)

$\displaystyle t(x) = \int_0^x \frac{1}{1+u^2} \text{d} u.$

$u = v^{-1} \Rightarrow \text{d} u = -v^{-2} \text{d} v$

$\displaystyle \therefore t(x) = \int_{u=0}^{u=x} \frac{1}{1+v^{-2}} \cdot -v^{-2} \text{d} v = -\int_\infty^{1/x} \frac{v^{-2}}{1 + v^{-2}} \text{d} v$

$\displaystyle = \int_{1/x}^\infty \frac{1}{1+v^2} \text{d} v$

$\displaystyle \therefore t(1/x) + t(x) = \int_{x}^\infty \frac{1}{1+v^2} \text{d} v + \int_0^x \frac{1}{1+u^2} \text{d} u = \int_0^\infty \frac{1}{1+u^2} \text{d} u = p/2.$

Choose x = 1 and result follows.

(ii)

$\displaystyle y = \frac{u}{\sqrt{1+u^2}} \Rightarrow y^2(1+u^2) = u^2 \Rightarrow u = \frac{y}{\sqrt{1-y^2}}$

$\displaystyle \frac{\text{d} u}{\text{d} y} = (1-y^2)^{-1/2} + y^2(1-y^2)^{-3/2} = (1-y^2+y^2)(1-y^2)^{-3/2} = \frac{1}{\sqrt{(1-y^2)^3}}$

Performing this substitution:

$\displaystyle t(x) = \int_{u=0}^{u=x} \frac{1}{1 + \frac{y^2}{1-y^2}} \cdot \frac{1}{\sqrt{(1-y^2)^3}} \text{d} y$

$\displaystyle = \int_{u=0}^{u=x} \frac{1-y^2}{\sqrt{(1-y^2)^3}} \text{d} y$

$\displaystyle = \int_0^{x/\sqrt{1+x^2}} \frac{1}{\sqrt{1-y^2}} \text{d} y = s\left( \frac{x}{\sqrt{1+x^2}} \right) .$

Again, choose x = 1 and result follows.

(iii)

$\displaystyle z = \frac{u + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}} u}$

$\displaystyle \Rightarrow u = \frac{z - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}} z}$ (detail omitted!)

$\displaystyle \Rightarrow \frac{\text{d} u}{\text{d} z} = \frac{(1 + \frac{1}{\sqrt{3}} z) - (z - \frac{1}{\sqrt{3}})(\frac{1}{ \sqrt{3}})}{(1 + \frac{1}{\sqrt{3}} z)^2} = \frac{4/3}{{(1 + \frac{1}{\sqrt{3}} z)^2}}$

$\displaystyle t(x) = \int_{u=0}^{u=x} \frac{1}{1 + \left( \frac{z - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}} z} \right)^2} \cdot \frac{4/3}{{(1 + \frac{1}{\sqrt{3}} z)^2}} \text{d} z$

$\displaystyle = \int_{u=0}^{u=x} \frac{4/3}{(1 + \frac{1}{\sqrt{3}} z)^2 + (z - \frac{1}{\sqrt{3}} )^2} \text{d} z$

$\displaystyle = \int_{u=0}^{u=x} \frac{4/3}{\frac{4}{3} (1 + z^2)} \text{d} z$

$\displaystyle = \int_{1/\sqrt{3}}^{\alpha (x)} \frac{1}{1+z^2} \text{d} z$

where

$\displaystyle \alpha (x) = \frac{x + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}} x} .$

Then

$\displaystyle \alpha \left( \frac{1}{\sqrt{3}} \right) = \sqrt{3}$

and so

$\displaystyle t\left( \frac{1}{\sqrt{3}} \right) = \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1+z^2} \text{d} z.$

$\displaystyle 3t\left( \frac{1}{\sqrt{3}} \right) = \int_0^{1/\sqrt{3}} \frac{1}{1+u^2} \text{d} u + \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1+z^2} \text{d} z + \int_{\sqrt{3}}^\infty \frac{1}{1+v^2} \text{d} v$

$\displaystyle = \int_0^\infty \frac{1}{1+u^2} \text{d} u = \frac{1}{2} p.$

Solution by generalebriety.

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