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

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)

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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 $$\int ^\infty_0 \frac{b\sin{ax} - a\sin{bx}}{x}dx=\frac{1}{2} \pi \Big(b~~ \text{sgn}(a)-a~~ \text{sgn}(b)\Big)$$

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  • $\begingroup$ I apologize, it was a minus, not multiplication between the sines...the question still stands, and I'm sorry for wasting your time $\endgroup$ – user155921 Jun 8 '14 at 14:07
  • $\begingroup$ No problem. It was fun ! $\endgroup$ – Claude Leibovici Jun 8 '14 at 14:15
  • $\begingroup$ I used the change of variable $y=ax$ and this led to $\text {Si} (x)$ not $\text {Si} (ax)$, which in turn gave a result without the sgn function. Is there another change of variable? $\endgroup$ – user155921 Jun 8 '14 at 14:34
  • $\begingroup$ No. The antiderivative is fine but the problem is with the integral. $\endgroup$ – Claude Leibovici Jun 8 '14 at 14:37
  • $\begingroup$ I got it. Thank you very much for your help! $\endgroup$ – user155921 Jun 8 '14 at 14:44
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Another method which doesn't involves the use of special functions. (assuming $a,b>0$)

The original integral can be written as:

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

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

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}

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