Value of $\int ^\infty_0 \frac{b\sin{ax} - a\sin{bx}}{x}dx$?

Question

$$\int ^\infty_0 \frac{b\sin{ax} - a\sin{bx}}{x}dx$$

Hello guys! I'm having trouble solving this integral...looks an awful lot like an Frullani Integral, and I've tried to get it to an appropriate form to apply that method, but I've failed so far. Also, Wolfram gives a result lacking any logarithms, which may hint that it cannot be solved in that way. Can anyone help me here? (I have to solve this without using any complex math)

Answer 1

Hint

Split your integral in txo pieces and make the appropriate changes of variable to make the integrand looking like $$\frac{\sin(y)}{y}~dy$$ and you will then arrive to $$\int \frac{b\sin{ax} - a\sin{bx}}{x}dx=b \text{Si}(a x)-a \text{Si}(b x)$$ from where $$\int ^\infty_0 \frac{b\sin{ax} - a\sin{bx}}{x}dx=\frac{1}{2} \pi \Big(b~~ \text{sgn}(a)-a~~ \text{sgn}(b)\Big)$$

Answer 2

Another method which doesn't involves the use of special functions. (assuming $a,b>0$)

The original integral can be written as:

$$\int_0^{\infty} \int_0^{\infty} \left(be^{-xt}\sin(ax)-ae^{-xt}\sin(bx)\right)\,dx\,dt=\int_0^{\infty} \left(\frac{ab}{a^2+t^2}-\frac{ab}{b^2+t^2}\right)\,dt$$ $$=\boxed{\dfrac{\pi}{2}(b-a)}$$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{b\sin\pars{ax} - a\sin\pars{bx} \over x}\,\dd x:\ {\large ?}}$.

With the identity $\quad\ds{{\sin\pars{t} \over t} = \half\int_{-1}^{1}\expo{\ic kt}\,\dd k\,,\quad}$ we'll have:

\begin{align}&\color{#c00000}{\int_{0}^{\infty} {b\sin\pars{ax} - a\sin\pars{bx} \over x}\,\dd x} =\half\,ab\int_{-\infty}^{\infty} \half\int_{-1}^{1}\pars{\expo{\ic kax} - \expo{\ic kbx}}\,\dd k\,\dd x \\[3mm]&={1 \over 4}\,ab\int_{-1}^{1}\pars{% \int_{-\infty}^{\infty}\expo{\ic kax}\,\dd x -\int_{-\infty}^{\infty}\expo{\ic kax}\,\dd x}\,\dd k ={1 \over 4}\,ab\int_{-1}^{1}\bracks{% 2\pi\,\delta\pars{ka} - 2\pi\,\delta\pars{kb}}\,\dd k \\[3mm]&={\pi \over 2}\,ab\int_{-1}^{1}\bracks{% {\delta\pars{k} \over \verts{a}} - {\delta\pars{k} \over \verts{b}}}\,\dd k \\[3mm]&={\pi \over 2}\bracks{b\sgn\pars{a}\int_{-1}^{1}\delta\pars{k}\,\dd k -a\sgn\pars{b}\int_{-1}^{1}\delta\pars{k}\,\dd k} ={\pi \over 2}\,\bracks{b\sgn\pars{a} - a\sgn\pars{b}} \end{align}

\begin{align}&\color{#66f}{\large\int_{0}^{\infty} {b\sin\pars{ax} - a\sin\pars{bx} \over x}\,\dd x ={\pi \over 2}\,\sgn\pars{a}\sgn\pars{b}\pars{\verts{b} - \verts{a}}} \\[3mm]&=\left\lbrace\begin{array}{rclrclrcl} {\pi \over 2}\,\pars{a - b} & \mbox{if} & a < 0\,,\quad b < 0 \\[3mm] -\,{\pi \over 2}\,\pars{a + b} & \mbox{if} & a < 0\,,\quad b > 0 \\[3mm] {\pi \over 2}\,\pars{a + b} & \mbox{if} & a > 0\,,\quad b < 0 \\[3mm] {\pi \over 2}\,\pars{b - a} & \mbox{if} & a > 0\,,\quad b > 0 \\[3mm] 0 && \mbox{otherwise} \end{array}\right. \end{align}