$u=x^{-4} \Rightarrow du=-4x^{-5}dx \Rightarrow x^{-1}dx= \frac{-x^4}{4} du= \frac{-u^{-1}}{4} du$
$, \lim_{x\to 0^-}u=+ \infty , \lim_{x\to + \infty } u =0$
$ \Rightarrow A=\int _0^ \infty \frac{Sin^3(x^{-4})}{x} dx=- \frac{1}{4} \int _ \infty ^0 \frac{Sin^3(u)}{u} du=\frac{1}{4} \int _0 ^ \infty \frac{Sin^3(u)}{u} du=\frac{1}{4} \int _0 ^ \infty \frac{Sin^3(u)}{u} du$
$=\frac{1}{4} \int _0 ^ \infty \frac{ \frac{1}{4} (3sin(u)-sin(3u)}{u} du= \frac{1}{16} \int _0 ^ \infty \frac{ 3sin(u)-sin(3u)}{u} du$
حالا از انتگرالهای وابسته به پارامتر کمک بگیرید.قرار دهید:
$I(a):= \int _0^ \infty \frac{sin(x)e^{-ax})}{x} dx \Rightarrow I'(a)=\int _0^ \infty \frac{-xsin(x)e^{-ax})}{x} dx = -\int _0^ \infty sin(x)e^{-ax}dx$
$=- \frac{-e^{-ax}(cos(x)+asin(x)}{a^2+1} ]_0^ \infty =- \frac{1}{a^2+1} \Rightarrow I(a)= \frac{ \pi }{2} -tan^{-1}(a)$(چرا؟)
$ \Rightarrow \int _0^ \infty \frac{sin(x)}{x} dx=I(0)= \frac{ \pi }{2} -tan^{-1}(0)=\frac{ \pi }{2}-0=\frac{ \pi }{2}$
$, \int _0^ \infty \frac{sin(3x)}{x} dx=\int _0^ \infty \frac{sin(x)}{x} dx= \frac{\pi }{2} $
$ \Rightarrow A=3\int _0^ \infty \frac{sin(x)}{x} dx-\int _0^ \infty \frac{sin(3x)}{x} dx= \frac{1}{16} (\frac{3\pi }{2}-\frac{ \pi }{2})= \frac{ \pi }{16} $
$ \Box $
