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

Hint:

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

Answer 2

$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 3

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:

$$\text{F}_{\text{a},\text{b}}\left(\text{s}\right):=\mathscr{L}_x\left[\frac{\sin\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)}{x}\right]_{\left(\text{s}\right)}:=\int_0^\infty\frac{\sin\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)}{x}\cdot e^{-\text{s}x}\space\text{d}x\tag2$$

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

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

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:

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

So, for the integral we get:

$$\text{F}_{\text{a},\text{b}}\left(\text{s}\right)=\frac{1}{2}\int_\text{s}^\infty\left\{\frac{\text{a}-\text{b}}{\sigma^2+\left(\text{a}-\text{b}\right)^2}+\frac{\text{a}+\text{b}}{\sigma^2+\left(\text{a}+\text{b}\right)^2}\right\}\space\text{d}\sigma=$$ $$\frac{1}{2}\lim_{\text{n}\to\infty}\left[\arctan\left(\frac{\sigma}{\text{a}-\text{b}}\right)+\arctan\left(\frac{\sigma}{\text{a}+\text{b}}\right)\right]_\text{s}^\text{n}\tag6$$

When $\text{a}-\text{b}>0$ 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 $\text{a}-\text{b}>0$:

$$\text{F}_{\text{a},\text{b}}\left(0\right)=\int_0^\infty\frac{\sin\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)}{x}\cdot e^{-0\cdot x}\space\text{d}x=$$ $$\int_0^\infty\frac{\sin\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)}{x}\space\text{d}x=\frac{\pi-\arctan\left(\frac{0}{\text{a}-\text{b}}\right)-\arctan\left(\frac{0}{\text{a}+\text{b}}\right)}{2}=\frac{\pi}{2}\tag8$$

Answer 4

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