Solving integrals looks like Fourier integral. $$\int_0^\infty \frac{\sin w}w \, \cos xw \, dw$$
How can I solve this integral,I was thinking it may be solved by using Fourier transform, but It seems that it wouldn't work.besides I've tried many exchanging variable it wasn't helpful too.Do you have any idea?
 A: You can use the sine addition formula and get
$$\sin w \cos xw = {1 \over 2}\sin(w + xw) + {1 \over 2}\sin(w - xw)$$
Hence your integral is equal to
$${1 \over 2} \int_0^{\infty} {\sin (1 + x)w \over w} \, dw+ {1 \over 2} \int_0^{\infty} {\sin (1 - x)w \over w}\, dw $$
Substituting $u= (1 + x)w$ in the first and $u = (1  - x) w$ in the second gives (when $x \neq 1$ or $-1$)
$${1 \over 2} sgn(1 + x)\int_0^{\infty} {\sin u \over u} \, du +  {1 \over 2}sgn(1 - x) \int_0^{\infty} {\sin u \over u} \,du$$
$$= sgn(1 + x){\pi \over 4} + sgn(1 - x){\pi \over 4}$$
So when $|x| > 1$, the integral is zero, and when $|x| < 1$, the integral is ${\displaystyle {\pi \over 2}}$. The cases $x = \pm 1$ can be done directly: Here the integral is
$${1 \over 2} \int_0^{\infty} {\sin 2w \over w} \, dw$$
Letting $u = 2w$ gives ${\displaystyle {\pi \over 4}}$ for these two cases.
Those familar with Fourier analysis might recognize these formulas, but the above shows them directly.
A: I believe it is an equivalent statement if you take the derivative and integrate with respect to x.
$$\int_0^\infty \frac{sin(w)}{w} cos(wx)dw=\int \frac{d}{d x}[\int_0^\infty \frac{sin(w)}{w} cos(wx)dw] dx$$
$$\rightarrow \int [\int_0^\infty \frac{sin(w)}{w} \frac{d}{d x}cos(wx)dw] dx$$
$$\rightarrow \int [\int_0^\infty \frac{sin(w)}{w} \frac{d}{d x}cos(wx)dw] dx$$
$$\rightarrow \int [\int_0^\infty \frac{sin(w)}{w} -wsin(wx)dw] dx$$
$$\rightarrow -\int [\int_0^\infty sin(w) sin(wx) dw] dx$$
Then expanding sin(a) as $\frac{e^{ia}-e^{-ia}}{2i}$:
$$\rightarrow -\int [\int_0^\infty \frac{e^{iw}-e^{-iw}}{2i}\frac{e^{iw}-e^{-iw}}{2i} dw] dx$$
$$\rightarrow \frac{-1}{2i}\int [\int_0^\infty \frac{e^{iw(1+x)}-e^{-iw(1+x)}+e^{iw(1-x)}-e^{-iw(1-x)}}{2i} dw] dx$$
$$\rightarrow \frac{-1}{2i}\int [\int_0^\infty sin(w(1+x))+sin(w(1-x)) dw] dx$$
$$\rightarrow \frac{1}{2i}\int [ \frac{cos(w(1+x))}{(1+x)}+\frac{cos(w(1-x))}{1-x} |_0^\infty] dx$$
$$\rightarrow \frac{-1}{2i}\int  \frac{1}{(1+x)}+\frac{1}{1-x}  dx$$
$$\rightarrow \frac{-1}{2i}(ln(1+x)-ln(1-x)$$
