$I:= \int _0^ \infty \frac{sin(ln^2(x))}{ln^2(x)(1+x)} dx,u:= \frac{1}{x} \Rightarrow U(0)= \infty .u( \infty )=0,dx=- \frac{1}{u^2} du$
$I=\int _0^ \infty \frac{sin(ln^2(u))}{ln^2(u)(u(1+u))} du=\int _0^ \infty \frac{sin(ln^2(u))}{ln^2(u)u} du-\int _0^ \infty \frac{sin(ln^2(u))}{ln^2(u)(1+u)} du$
$=\int _0^ \infty \frac{sin(ln^2(u))}{ln^2(u)u} du-I \Rightarrow 2I=\int _0^ \infty \frac{sin(ln^2(u))}{ln^2(u)u} du$
حالا تغییر متغیر $x:=ln(u) $ را بکار ببرید:
$ \Rightarrow 2I= \int _{- \infty }^ \infty \frac{sin(x^2)}{x^2} dx =2\int _0^ \infty \frac{sin(x^2)}{x^2} dx \Rightarrow I=\int _0^ \infty \frac{sin(x^2)}{x^2} dx$
حالا تابه $J:(0,+ \infty ) \rightarrow R$ را به صورت زیر تعریف کنید:
$J(a)=\int _0^ \infty \frac{sin(ax^2)}{x^2} dx \Rightarrow J'(a)=\int _0^ \infty \frac{x^2cos(ax^2)}{x^2} dx= \int _0^ \infty cos(ax^2)dx= \frac{ \sqrt{2 \pi } }{4 \sqrt{a} } $
$ \Rightarrow J(a)= \sqrt{\frac{ \pi a}{2}} +C,J(0)=0+C \Rightarrow C=0 \Rightarrow J(a)= \sqrt{\frac{ \pi a}{2}}$
$ \Rightarrow I=J(1)=\sqrt{\frac{ \pi}{2}}$
$ \Box $
