2017.3.4 Question 4

1. Notice that $a=e^{\text{ln}<!--\; FUNCTION\; APPLICATION\; -->a}$ and hence $a^x=e^{x\text{ln}<!--\; FUNCTION\; APPLICATION\; -->a}$, $a^{\frac{x}{\text{ln}<!--\; FUNCTION\; APPLICATION\; -->a}}=e^x$we have

   $$\begin{array}{llll}\hfill F(y)& =\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{1}{y}{\int \; <!--\; nolimits\; -->}_0^y\text{ln}<!--\; FUNCTION\; APPLICATION\; -->f(x)<!--\; FUNCTION\; APPLICATION\; -->dx\right)& \hfill & \\ \hfill & =a^{\frac{1}{y\text{ln}<!--\; FUNCTION\; APPLICATION\; -->a}\cdot {\int \; <!--\; nolimits\; -->}_0^y\text{ln}<!--\; FUNCTION\; APPLICATION\; -->f(x)<!--\; FUNCTION\; APPLICATION\; -->dx}& \hfill & \\ \hfill & =a^{\frac{1}{y}\cdot {\int \; <!--\; nolimits\; -->}_0^y\frac{\text{ln}<!--\; FUNCTION\; APPLICATION\; -->f(x)}{\text{ln}<!--\; FUNCTION\; APPLICATION\; -->a}dx}& \hfill & \\ \hfill & =a^{\frac{1}{y}\cdot {\int \; <!--\; nolimits\; -->}_0^y{\text{log}<!--\; FUNCTION\; APPLICATION\; -->}_{a}f(x)<!--\; FUNCTION\; APPLICATION\; -->dx}& \hfill & \end{array}$$

   as desired.

2. We have

   $$\begin{array}{llll}\hfill H(y)& =\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{1}{y}{\int \; <!--\; nolimits\; -->}_0^y\text{ln}<!--\; FUNCTION\; APPLICATION\; -->f(x)g(x)<!--\; FUNCTION\; APPLICATION\; -->dx\right)& \hfill & \\ \hfill & =\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left[\frac{1}{y}{\int \; <!--\; nolimits\; -->}_0^y\left(\text{ln}<!--\; FUNCTION\; APPLICATION\; -->f(x)+\text{ln}<!--\; FUNCTION\; APPLICATION\; -->g(x)\right)dx\right]& \hfill & \\ \hfill & =\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left[\frac{1}{y}\left({\int \; <!--\; nolimits\; -->}_0^y\text{ln}<!--\; FUNCTION\; APPLICATION\; -->f(x)<!--\; FUNCTION\; APPLICATION\; -->dx+{\int \; <!--\; nolimits\; -->}_0^y\text{ln}<!--\; FUNCTION\; APPLICATION\; -->g(x)<!--\; FUNCTION\; APPLICATION\; -->dx\right)\right]& \hfill & \\ \hfill & =\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{1}{y}{\int \; <!--\; nolimits\; -->}_0^y\text{ln}<!--\; FUNCTION\; APPLICATION\; -->f(x)<!--\; FUNCTION\; APPLICATION\; -->dx\right)\cdot \text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{1}{y}{\int \; <!--\; nolimits\; -->}_0^y\text{ln}<!--\; FUNCTION\; APPLICATION\; -->g(x)<!--\; FUNCTION\; APPLICATION\; -->dx\right)& \hfill & \\ \hfill & =F(y)\cdot G(y).& \hfill & \end{array}$$

3. Let $f(x)=b^x$.

   $$\begin{array}{llll}\hfill F(y)& =\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{1}{y}{\int \; <!--\; nolimits\; -->}_0^y\text{ln}<!--\; FUNCTION\; APPLICATION\; -->f(x)<!--\; FUNCTION\; APPLICATION\; -->dx\right)& \hfill & \\ \hfill & =b^{\frac{1}{y}{\int \; <!--\; nolimits\; -->}_0^y{\text{log}<!--\; FUNCTION\; APPLICATION\; -->}_{b}f(x)<!--\; FUNCTION\; APPLICATION\; -->dx}& \hfill & \\ \hfill & =b^{\frac{1}{y}{\int \; <!--\; nolimits\; -->}_0^y{\text{log}<!--\; FUNCTION\; APPLICATION\; -->}_b{b}^x<!--\; FUNCTION\; APPLICATION\; -->dx}& \hfill & \\ \hfill & =b^{\frac{1}{y}{\int \; <!--\; nolimits\; -->}_0^{y}xdx}& \hfill & \\ \hfill & =b^{\frac{1}{y}\cdot \frac{y^2}{2}}& \hfill & \\ \hfill & =b^{\frac{y}{2}}& \hfill & \\ \hfill & =\sqrt{b^y}.& \hfill & \end{array}$$

4. Since $F(y)=\sqrt{f(y)}$, we notice that $f(y)=F{(y)}^2=\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{2}{y}{\text{∫ <!-- nolimits -->}<!--\; FUNCTION\; APPLICATION\; -->}_0^y\text{ln}<!--\; FUNCTION\; APPLICATION\; -->f(x)<!--\; FUNCTION\; APPLICATION\; -->dx\right)$, and therefore $\text{ln}<!--\; FUNCTION\; APPLICATION\; -->f(y)=\frac{2}{y}{\text{∫ <!-- nolimits -->}<!--\; FUNCTION\; APPLICATION\; -->}_0^y\text{ln}<!--\; FUNCTION\; APPLICATION\; -->f(x)<!--\; FUNCTION\; APPLICATION\; -->dx$.

   We substitute $g(y)=\text{ln}<!--\; FUNCTION\; APPLICATION\; -->f(y)$, and therefore

   $$\mathit{yg}(y)=2{\int \; <!--\; nolimits\; -->}_0^{y}g(y)dx.$$

   Therefore, differentiating both sides with respect to $y$ gives us

   $$y{g}^{\prime }(y)+g(y)=2g(y),$$

   and therefore

   $$-g(y)+y{g}^{\prime }(y)=0.$$

   Multiplying $y^{-2}$ on both sides gives us

   $$-y^{-2}g(y)+y^{-1}{g}^{\prime }(y)=0,$$

   and therefore

   $$\frac{d}{dy}\frac{g(y)}{y}=0,$$

   and therefore

   $$\frac{g(y)}{y}=C⟹g(y)=\mathit{Cy}.$$

   Therefore, we have

   $$\begin{array}{llll}\hfill f(y)& =\text{exp}<!--\; FUNCTION\; APPLICATION\; -->g(y)& \hfill & \\ \hfill & =\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{Cy})& \hfill & \\ \hfill & =b^y& \hfill & \end{array}$$

   if we substitute $b=\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(C)>0$, and therefore $f(x)=b^y$ as desired.