부분적분과 치환적분

조건에서 $f ( a ) = 0$이고 $f ( 2 x ) = 2 f ( x ) f' ( x )$이므로 $f ( 2 a ) = 2 f ( a ) f' ( a ) = 0$ 또한 $f ( 4 a ) = 2 f ( 2 a ) f' ( 2 a ) = 0$이다. $\displaystyle \int _{a} ^{2 a} {\frac{\left\{ f ( x ) \right\} ^{2}}{x ^{2}}} dx$ $\displaystyle = \int _{a} ^{2 a} {x ^{-2}} {\left\{ f ( x ) \right\} ^{2} dx}$ $\displaystyle = [ - x ^{- 1} \left\{ f ( x ) \right\} ^{2} ] _{a} ^{2 a} - \int _{a} ^{2 a} {- x ^{- 1} \bullet 2 f ( x ) f' ( x ) dx}$ $( \because$부분적분법) $\displaystyle = 0 + \int _{a} ^{2 a} {x ^{- 1} \bullet 2 f ( x ) f' ( x ) dx}$ $\displaystyle = \int _{a} ^{2 a} {\frac{2 f ( x ) f' ( x )}{x}} dx = \int _{a} ^{2 a} {\frac{f ( 2 x )}{x}} dx$ 여기서 $2 x = t$로 치환하면 $\displaystyle 2 dx = dt \leftrightarrow dx = \frac{1}{2} dt$ $\displaystyle \begin{cases} {x = a \rightarrow t = 2 a} \\ {x = 20 \rightarrow t = 4 a} \end{cases}$로 변환되므로 $\displaystyle = \int _{2 a} ^{4 a} {\frac{1}{2} \frac{f ( t )}{\frac{1}{2} t}} dt = \int _{2 a} ^{4 a} {\frac{f ( t )}{t}} dt = k$