Λύση

(α)     $\int ^\infty _0 \frac{\sin x \cos x}{x} dx = \int ^\infty _0 \frac{\sin 2x}{2x... ...int ^\infty _0 \frac{\sin t}{t} dt = \frac{1}{2} \frac{\pi }{2} =\frac{\pi }{4}$ αφού όταν θέτω $t=2x \Rightarrow dt=2dx$.
(β)     Από την (α) έχω

$$\begin{eqnarray*} \frac{\pi }{4} = \int ^\infty _0 \frac{\sin x \cos x}{x} dx &=... ...\\ &=& 0 + \frac{1}{2} \int ^\infty _0 \frac{\sin ^2 x}{x^2} dx \end{eqnarray*}$$

Άρα $\int ^\infty _0 \frac{\sin ^2 x}{x^2} dx=\frac{\pi }{2}$. (Θυμηθείτε ότι $\lim _{x \rightarrow 0} \frac{\sin ^2 x}{2x} =\lim _{x\rightarrow 0} \frac{\sin x}{2} \ \cdot \ \frac{\sin x}{x}$ και $\lim _{x\rightarrow 0} \frac{\sin x}{x} =1$)

(γ)    

$$\begin{eqnarray*} \int ^\infty _0 \frac{\sin ^4 x}{x^2} dx &=& \int ^\infty _0 \... ...{\pi }{2} -\frac{1}{4} \int ^\infty _0\frac{\sin ^2 2x}{x^2} dx \end{eqnarray*}$$

Θέτω $2x=t \Rightarrow 2dx= dt$ άρα

$$\begin{eqnarray*} \int ^\infty _0 \frac{\sin ^4 x}{x^2} dx&=& \frac{\pi }{2} - \... ...rac{\pi }{2}\ \frac{1}{2}\ \frac{\pi }{2} \\ &=& \frac{\pi }{4} \end{eqnarray*}$$

(δ)    

$$\begin{eqnarray*} \int_0^\infty \frac{\sin^4 x}{x^4}\,dx &=& \int_0^\infty \left... ...2\int_0^\infty \frac{\sin^2 2x}{4x^2} \,dx -\frac23\frac{\pi}4 . \end{eqnarray*}$$

Τώρα θέτω $t=2x$, $dt=2dx$ και έχω:

$$\begin{eqnarray*} \int_0^\infty \frac{\sin^4 x}{x^4} &=& 2\int_0^\infty \frac{\s... ...ackrel{\hbox{(β)}}{=}& \frac{\pi}2\frac{\pi}6\\ &=& \frac{\pi}3 \end{eqnarray*}$$

Άσκηση 8 Υπόδειξη