Frullani integral with multiplication

Question

The integral is this one:

$$\int_{0}^{\infty} \frac{\sin(ax)\cos(bx)}{x}\, dx$$

I tried to convert them to $\sin(ax+bx)-\sin(ax-bx)$ but that doesn't help so much.Somehow I must get a sum but I need a small hint for that

Answer 1

Well, we can use the Laplace transform:

$$\text{f}\left(x\right):=\frac{\sin\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)}{x}\tag1$$

So, we get:

$$\text{F}_{\text{a},\text{b}}\left(\text{s}\right):=\mathscr{L}_x\left[\frac{\sin\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)}{x}\right]_{\left(\text{s}\right)}:=\int\limits_0^\infty\frac{\sin\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)}{x}\cdot e^{-\text{s}x}\space\text{d}x\tag2$$

Now, using the 'frequency-domain integration' property of the Laplace transform:

$$\text{F}_{\text{a},\text{b}}\left(\text{s}\right)=\int\limits_\text{s}^\infty\mathscr{L}_x\left[\sin\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag3$$

We need to use:

$$\sin\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)=\frac{\sin\left(x\cdot\left(\text{a}-\text{b}\right)\right)+\sin\left(x\cdot\left(\text{a}+\text{b}\right)\right)}{2}\tag4$$

So, for the Laplace transform of the $\sin$ function we get:

\begin{equation} \begin{split} \mathscr{L}_x\left[\sin\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)\right]_{\left(\sigma\right)}&=\mathscr{L}_x\left[\frac{\sin\left(x\cdot\left(\text{a}-\text{b}\right)\right)+\sin\left(x\cdot\left(\text{a}+\text{b}\right)\right)}{2}\right]_{\left(\sigma\right)}\\ \\ &=\frac{1}{2}\cdot\left\{\mathscr{L}_x\left[\sin\left(x\cdot\left(\text{a}-\text{b}\right)\right)\right]_{\left(\sigma\right)}+\mathscr{L}_x\left[\sin\left(x\cdot\left(\text{a}+\text{b}\right)\right)\right]_{\left(\sigma\right)}\right\}\\ \\ &=\frac{1}{2}\cdot\left\{\frac{\text{a}-\text{b}}{\sigma^2+\left(\text{a}-\text{b}\right)^2}+\frac{\text{a}+\text{b}}{\sigma^2+\left(\text{a}+\text{b}\right)^2}\right\} \end{split}\tag5 \end{equation}

So, for the integral we get:

\begin{equation} \begin{split} \text{F}_{\text{a},\text{b}}\left(\text{s}\right)&=\frac{1}{2}\int\limits_\text{s}^\infty\left\{\frac{\text{a}-\text{b}}{\sigma^2+\left(\text{a}-\text{b}\right)^2}+\frac{\text{a}+\text{b}}{\sigma^2+\left(\text{a}+\text{b}\right)^2}\right\}\space\text{d}\sigma\\ \\ &=\frac{1}{2}\lim_{\text{n}\to\infty}\left[\arctan\left(\frac{\sigma}{\text{a}-\text{b}}\right)+\arctan\left(\frac{\sigma}{\text{a}+\text{b}}\right)\right]_\text{s}^\text{n} \end{split}\tag6 \end{equation}

When $\displaystyle\text{a}-\text{b}>0$ we can simplify further:

$$\text{F}_{\text{a},\text{b}}\left(\text{s}\right)=\frac{\pi-\arctan\left(\frac{\text{s}}{\text{a}-\text{b}}\right)-\arctan\left(\frac{\text{s}}{\text{a}+\text{b}}\right)}{2}\tag7$$

Now, when $\text{s}=0$ and $\text{a}-\text{b}>0$:

\begin{equation} \begin{split} \text{F}_{\text{a},\text{b}}\left(0\right)&=\int\limits_0^\infty\frac{\sin\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)}{x}\cdot e^{-0\cdot x}\space\text{d}x\\ \\ &=\int\limits_0^\infty\frac{\sin\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)}{x}\space\text{d}x\\ \\ &=\frac{\pi-\arctan\left(\frac{0}{\text{a}-\text{b}}\right)-\arctan\left(\frac{0}{\text{a}+\text{b}}\right)}{2}=\frac{\pi}{2} \end{split}\tag8 \end{equation}

Answer 2

You may also use the Frullani theorem for complex parameters and get $$\int_{0}^{\infty}\frac{\sin\left(ax\right)\cos\left(bx\right)}{x}dx=\frac{i}{4}\int_{0}^{\infty}\frac{e^{-ix\left(a+b\right)}+e^{-ix\left(a-b\right)}-e^{ix\left(a-b\right)}-e^{ix\left(a+b\right)}}{x}dx$$ $$=\frac{i}{4}\int_{0}^{\infty}\frac{e^{-ix\left(a+b\right)}-e^{ix\left(a+b\right)}}{x}dx+\frac{i}{4}\int_{0}^{\infty}\frac{e^{-ix\left(a-b\right)}-e^{ix\left(a-b\right)}}{x}dx$$ $$=\lim_{\epsilon\rightarrow0^{+}}\left(\frac{i}{4}\int_{0}^{\infty}\frac{e^{\epsilon-ix\left(a+b\right)}-e^{\epsilon+ix\left(a+b\right)}}{x}dx+\frac{i}{4}\int_{0}^{\infty}\frac{e^{\epsilon-ix\left(a-b\right)}-e^{\epsilon+ix\left(a-b\right)}}{x}dx\right)$$ $$=\lim_{\epsilon\rightarrow0^{+}}\frac{i}{4}\left(\log\left(\frac{\epsilon-i\left(a+b\right)}{\epsilon+i\left(a+b\right)}\right)+\log\left(\frac{\epsilon-i\left(a-b\right)}{\epsilon+i\left(a-b\right)}\right)\right)$$ $$=\color{red}{\frac{\pi}{4}\left(\textrm{sign}\left(a+b\right)+\textrm{sign}\left(a-b\right)\right)}.$$

Answer 3

Hint:

$$\int _{ 0 }^{ \infty } \frac { \sin \left( ax+ab \right) }{ x } \, dx=\frac { \pi }{ 2 } { sgn }(a+b)\\ $$

Answer 4

$2\sin (ax)\cos (bx)=\sin ( (b+a)x)-\sin ((b-a)x) $ . You can set $b+a=A, b-a=B $ and use the theorem.

Answer 5

Without loss of generality, we can consider the case $a,b>0$. Since the integrand function is even, extend the integral to the real line and perform the substitution $ax=u$: $$I:=\int_0^{\infty}\frac{\sin (ax)\cos (bx)}{x}\mathrm dx=\frac 12\int_{-\infty}^{\infty}\frac{\sin u}u\cos\left(\frac ba u\right) \mathrm du =\frac 14\int_{-\infty}^{\infty}\frac{\sin u}u (e^{i\frac ba u}+e^{-i\frac ba u})\mathrm du$$ If $a=b$, then the integral evaluates to $\pi/4$. Suppose $a\neq b$, recalling that the Fourier Transform of $\frac {\sin x}x$ is $\pi\chi_{[-1,1]}(\xi)$ and that $\mathcal F[f(x)e^{ipx}]=\mathcal F[f](\xi-p)$ you have: $$I= \frac{\pi} 4 \left(\chi_{[-1,1]}\left(\xi-\frac ba\right)+\chi_{[-1,1]}\left(\xi+\frac ba\right)\right)\Bigg|_{\xi=0}=\begin{cases} 0 &\text{if} \,\, b>a\\ \frac{\pi}2 &\text{if} \,\,b<a\end{cases} $$