Calculate $\int _0^{\infty }\:\frac{\cos\left(a\omega \right)-1}{\omega }\sin\left(b\omega \right)d\omega $ using Fourier transform

Question

I have been stuck for quite a while at trying to calculate this integral: $$\int _0^{\infty }\:\frac{\cos\left(a\omega \right)-1}{\omega }sin\left(b\omega \right)d\omega $$ Where $a,b > 0$ are constants. The book gave me a guideline which was to use the fourier transform of this function: $$ f(x) = \begin{cases} 1, \hspace{1cm} 0<x\leq a\\ -1, \hspace{0.6cm} -a\leq x \leq 0 \\0, \hspace{1cm} \mathrm{otherwise} \end{cases} $$ Then the Fourier transform is: $$ F(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-i\omega x}dx =\frac{i(\cos a\omega -1)}{\pi \omega} $$ I tried calculating the integral by plugging in the definition of $F(\omega)$: $$\int _0^{\infty }\:\frac{\cos\left(a\omega \right)-1}{\omega }\sin\left(b\omega \right)d\omega = \frac{\pi}{2i}\int_{-\infty}^{\infty}\sin (b\omega) F(\omega)d\omega =\\ \frac{1}{4i}\int_{-\infty}^{\infty}\int_{-a}^{a}\sin (b\omega) e^{-i\omega x}dxd\omega$$ I couldn't solve it. Was stuck in the process at: $$\int_{-\infty}^{\infty}\frac{\sin (a\omega)\sin (b\omega)}{\omega}d\omega$$ Was also stuck at trying to find the Fourier transform of $\sin(bx)$. Could you please help me? Even just a hint?

Answer 1

You can use the Fourier transform to get the answer.

$f(x) = \begin{cases} \,\,\,\,1, \hspace{1cm} 0<x\leq a\\ -1, \hspace{0.6cm} -a\leq x \leq 0 \\\,\,\,\,0, \hspace{1cm} \mathrm{otherwise} \end{cases}\,\,$ and $\,\, F(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-i\omega x}dx =\frac{i(\cos a\omega -1)}{\pi \omega}$

But applying the inverse Fourier transform $$f(x)=\int_{-\infty}^{\infty}F(\omega)e^{i\omega x}d\omega$$ On the other hand $$ \int_{-\infty}^{\infty}F(\omega)e^{i\omega x}d\omega=\int_{-\infty}^{\infty}\frac{i(\cos a\omega -1)}{\pi \omega}e^{i\omega x}d\omega$$ $$=\int_{-\infty}^{0}\frac{i(\cos a\omega -1)}{\pi \omega}e^{i\omega x}d\omega+\int_0^{\infty}\frac{i(\cos a\omega -1)}{\pi \omega}e^{i\omega x}d\omega$$ Making change $ \omega\to-\omega$ in the first integral you get $$f(x)=\frac{i}{\pi}\int_0^{\infty}\frac{\cos a\omega -1}{\pi \omega}(e^{i\omega x}-e^{-i\omega x})\,d\omega=-\frac{2}{\pi}\int_0^{\infty}\frac{\cos a\omega -1}{\pi \omega}\sin\omega x\,d\omega$$ Now, put $x=b$, and the desired integral $$I=\int _0^{\infty }\frac{\cos a\omega-1}{\omega }\sin b\omega \,d\omega=-\frac{\pi}{2}\begin{cases} \,\,\,\,1, \hspace{1cm} 0<b\leq a\\ -1, \hspace{0.6cm} -a\leq b \leq 0 \\\,\,\,\,0, \hspace{1cm} \mathrm{otherwise} \end{cases}$$

You can also evaluate the integral directly: $$I=\int _0^{\infty }\frac{\cos a\omega-1}{\omega }\sin b\omega \,d\omega=\int _0^{\infty }\Big(\frac{\cos a\omega\sin b\omega}{\omega }-\frac{\sin b\omega}{\omega}\Big)d\omega$$ Using $\sin x+\sin y=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})$ $$I=\int _0^{\infty }\Big(\frac{\sin(a+b)\omega}{2\omega }+\frac{\sin(b-a)\omega}{2\omega }-\frac{\sin b\omega}{\omega}\Big)d\omega$$ Taking into consideration that $\,\int _0^{\infty }\frac{\sin cx}{x }dx=\frac{\pi}{2}sgn \,c$ $$I=\frac{\pi}{4}\big(sgn(a+b)+sgn(b-a)-2sgn \,b\big)$$

Answer 2

Clearly $$\int _0^{\infty }\:\frac{\cos\left(a\omega \right)-1}{\omega }\sin\left(b\omega \right)d\omega=\frac12\int_{-\infty}^{\infty }\:\frac{\cos\left(a\omega \right)-1}{\omega }\sin\left(b\omega \right)d\omega=\frac12\int_{-\infty}^{\infty }f(\omega)\bar g(\omega)d\omega $$ where $$ f(\omega)=\cos\left(a\omega \right)-1,g(\omega)=\frac{\sin\left(b\omega \right)}\omega. $$ Now $$ \mathcal F(\xi)=\frac12\sqrt{2\pi}\bigg[\delta(\xi-a)+\delta(\xi+a)-2\delta(\xi)\bigg] $$ and $$ \mathcal G(\xi)=\frac14\sqrt{2\pi}\bigg[\text{sign}(\xi+b)-\text{sign}(\xi-b)\bigg]. $$ By Plancherel's theorem, one has \begin{eqnarray} &&\int _0^{\infty }\:\frac{\cos\left(a\omega \right)-1}{\omega }\sin\left(b\omega \right)d\omega\\ &=&\frac12\int_{-\infty}^{\infty }f(\omega)g(\omega)d\omega=\frac12\int_{-\infty}^{\infty }\mathcal{F}(\xi)\bar{ \mathcal{G}}(\xi)d\xi\\ &=&\frac12\int_{-\infty}^{\infty }\frac12\sqrt{2\pi}\bigg[\delta(\xi-a)+\delta(s+a)-2\delta(s)\bigg]\frac14\sqrt{2\pi}\bigg[\text{sign}(\xi+b)-\text{sign}(\xi-b)\bigg]d\xi\\ &=&\frac{\pi}{4}\big[\text{sign}(a+b)+\text{sign}(b-a)-2\text{sign}(b)\big] \end{eqnarray}

Answer 3

We can transform $\frac{(\cos(aw)-1)\sin(bw)}{w}$ to $\frac{-\frac{i}{2}(\frac{e^{iaw}}{2}+\frac{e^{-iaw}}{2}-1)(e^{ibw}-e^{-ibw})}{w}$.

The Integral is:

$$ \int_{0}^{\infty}\frac{(\cos(aw)-1) \sin(bw)}{w}dw =\int_{0}^{\infty}\frac{-\frac{i}{2} (\frac{e^{iaw}}{2}+\frac{{\mathrm e}^{-iaw}}{2}-1) (e^{ibw}-e^{-ibw})}{w}dw $$

\begin{array}{l} =\lim_{w\to\infty}\left(\frac{i \,\mathrm{Ei}_{1}(i(a-b)w)}{4}+\frac{i\,\mathrm{Ei}_{1}(-i(a+b)w)}{4}+\frac{i\,\mathrm{Ei}_{1}(bwi)}{2}-\frac{i \,\mathrm{Ei}_{1}(-ibw)}{2}-\frac{i\,\mathrm{Ei}_{1}(i(a+b)w)}{4}-\frac{i\,\mathrm{Ei}_{1}(-i(a-b) w)}{4}+\frac{i\ln(i(a-b))}{4}+\frac{i\ln(-i(a+b))}{4}+\frac{i\ln(ib)}{2}-\frac{i\ln(-ib)}{2}-\frac{i\ln(i(a+b))}{4}-\frac{i\ln(-i(a-b))}{4}\right)\\ %=\frac{1}{4}\lim_{w\to\infty}\left(\ln((a-b)i)i+\ln(-i(a+b)) i+2 \,i\ln(bi)-2 \,i\ln(-ib)-i\ln((a+b)i)-i \ln(-i(a-b))+\pi \,\mathrm{sgn}((a-b)w)-\pi \,\mathrm{sgn}((a+b) w)+2 \pi \,\mathrm{sgn}(bw)-2 \,\mathrm{Si}((a-b) w)-4 \,\mathrm{Si}(b w)+2 \,\mathrm{Si}((a+b) w)\right)\\ =-\frac{1}{4}\pi(\text{sgn}(a-b)-\text{sgn}(a+b)+2 \text{sgn}(b)) \end{array}

Note that Ei denotes the Exponential Integral function.