$\int_0^{+\infty}{ \sin{(ax)} \sin{(bx)}}dx=?$

Question

How can I calculate the integral: $$\int_0^{+\infty}{ \sin{(ax)} \sin{(bx)}}dx$$ ??

I got stuck.. :/

Could you give me some hint??

Do I have to use the following formula??

$\displaystyle{\sin{(A)} \sin{(B)}=\frac{\cos{(A-B)}-\cos{(A+B)}}{2}}$

Answer 1

This integral does not converge but one can make some sense of it anyway as a generalized function of $a$ and $b$. Specifically, the result is a sum of Dirac delta functions. To show this, use the Euler identity to express the sine functions as exponential functions. After a little simplification, you will have several terms and each can be evaluated using the Dirac delta function's Fourier expansion: \begin{equation} \delta(p) = {1\over2\pi}\int_{-\infty}^\infty dx e^{i p x} \end{equation}

Edit: To be clear, this is a somewhat advanced analysis. If this was a homework problem for an elementary class and you've never heard of any of the techniques or formulas I've mentioned, the sought-after answer is probably simply "integral does not exist" which can be demonstrated via the arguments made by other posters.

Answer 2

This integral will not converge as the function oscillates forever without any damping and the primitive has no limit.

Answer 3

For the nontrivial cases $a \ne 0$, $b \ne 0$ the integrand $$ f(x) = \sin(ax)\sin(bx) $$ is oscilating somewhere between $1$ and $-1$ and that means the net area $I$ below the integral oscilates too and won't converge towards a finite value for the upper bound going towards $\infty$. $$ I = \lim_{\beta\to\infty}\int\limits_0^\beta \sin(ax)\sin(bx) \, dx \in \left\{ \mbox{undefined}, \pm \infty \right\} $$ The infinite cases happen for $a=b$, because then $f(x)=\sin^2(ax) \ge 0$ and for $a=-b$ because then $f(x) =-\sin^2(ax) \le 0$.

Example plot:

Too lazy to do the integration myself, I asked Wolfram's Google killer for an answer and got $$ \int \sin(ax)\sin(bx) \, dx = \frac{b\sin(ax)\cos(bx)-a\cos(ax)\sin(bx)}{a^2 - b^2} + \mbox{const} \implies \\ I = \lim_{x->\infty} \frac{b\sin(ax)\cos(bx)-a\cos(ax)\sin(bx)}{a^2 - b^2} $$

which asures me somewhat that I hit all three cases.

Answer 4

If $a,b\neq0$, then the set $A:=\{x\in [0,2\pi/b]: |\sin(ax)|>1/2\}$ is non-empty (otherwise exchange $a,b$). $A$ is open, hence there exists $\epsilon>0$ so that $B:=\{x\in A: |\sin(bx)|>\epsilon\}$ is non empty. Since $B$ is open it has positive Lebesgue measure. It follows that

$$\int_0^\infty |\sin(ax)\sin(bx)|=\infty$$ and the integral doesnt exist. (Similar ways show that both the positive and negative parts are infinite as well.)

Answer 5

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align}&\color{#66f}{\large\int_{0}^{\infty}\sin\pars{ax}\sin\pars{bx} \,\dd x} ={1 \over 4}\int_{-\infty}^{\infty}\braces{% \cos\pars{\bracks{a - b}x} -\cos\pars{\bracks{a + b}x}}\,\dd x \\[3mm]&={\pi \over 2}\int_{-\infty}^{\infty}\bracks{% \expo{\ic\pars{a - b}x} -\expo{\ic\pars{a + b}x}}\,{\dd x \over 2\pi} =\color{#66f}{\large% {\pi \over 2}\bracks{\delta\pars{a - b} - \delta\pars{a + b}}} \end{align}