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

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: $$f(x) = \begin{cases} 1, \hspace{1cm} 0 Then the Fourier transform is: $$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?

• It might be helpful to note that the last integral you mention is $0$ if $a\neq b$. Also, it might be helpful to note that $\int_{\Bbb R}\frac{\sin(ax)}x\,\mathrm dx=\pi\operatorname{sgn}(a)$ Commented Aug 24, 2021 at 9:25

You can use the Fourier transform to get the answer.

$$f(x) = \begin{cases} \,\,\,\,1, \hspace{1cm} 0 and $$\,\, 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

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 $$\,\int _0^{\infty }\frac{\sin cx}{x }dx=\frac{\pi}{2}sgn \,c$$ $$I=\frac{\pi}{4}\big(sgn(a+b)+sgn(b-a)-2sgn \,b\big)$$

• corrected typo: $\sin x+\sin y=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})$ Commented Aug 25, 2021 at 6:58

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

• This is a nice answer. But it would be better to use the same convention as OP's. Commented Mar 22 at 19:20
• Yes, of course. Commented Mar 22 at 19:52

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.

• Could you elaborate more? If your transformed a function dependant on $\omega$, shouldn't you end up with a function dependant on $x$?
– Algo
Commented Aug 24, 2021 at 9:28
• I first transformed your term into a form using exponentials of $e$ (Fourier Transform) and integrating this function leads to a term that is not dependant from $w$ anymore.
– user736865
Commented Aug 24, 2021 at 9:35
• I still don't understand why $\int_{0}^{\infty}\frac{-\frac{\mathrm{I}}{2} (\frac{{\mathrm e}^{\mathrm{I} a w}}{2}+\frac{{\mathrm e}^{\mathrm{-I} a w}}{2}-1) ({\mathrm e}^{\mathrm{I} b w}-{\mathrm e}^{\mathrm{-I} b w})}{w}dw\\=-\frac{1}{4} \pi (\text{sgn}(a-b)-\text{sgn}(a+b)+2 \text{sgn}(b)$
– Algo
Commented Aug 24, 2021 at 9:37
• I'll add some more intermediate steps.
– user736865
Commented Aug 24, 2021 at 10:06