$\int_0^{+\infty}{ \sin{(ax)} \sin{(bx)}}dx=?$ 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??
$\displaystyle{\sin{(A)} \sin{(B)}=\frac{\cos{(A-B)}-\cos{(A+B)}}{2}}$
 A: 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.)
A: 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.
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
\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}
A: This integral will not converge as the function oscillates forever without any damping and the primitive has no limit. 
A: 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.
