You need to prove that question displaystyle{int_{{0}}^{{1}}} sin{{left(pi{m}{x}right)}} sin{{left(pi{n}{x}right)}}{left.{d}{x}right.}={leftlbrace{left(begin{matrix}{0}{m}ne{n}{1}text{/}{2}{m}={n}end{matrix}right)}right.}

Question
Integrals
asked 2021-01-19
You need to prove that question
\(\displaystyle{\int_{{0}}^{{1}}} \sin{{\left(\pi{m}{x}\right)}} \sin{{\left(\pi{n}{x}\right)}}{\left.{d}{x}\right.}={\left\lbrace{\left(\begin{matrix}{0}{m}\ne{n}\\{1}\text{/}{2}{m}={n}\end{matrix}\right)}\right.}\)

Answers (1)

2021-01-20
We have to show
\(\displaystyle{\int_{{0}}^{{1}}} \sin{{\left(\pi{m}{x}\right)}} \sin{{\left(\pi{n}{x}\right)}}{\left.{d}{x}\right.}={\left\lbrace{\left(\begin{matrix}{0}{m}\ne{n}\\{1}\text{/}{2}{m}={n}\end{matrix}\right)}\right.}\)
We will use the following trigonometric identities:
\(\displaystyle \sin{{A}} \sin{{B}}=\frac{1}{{2}}{\left[ \cos{{\left({A}-{B}\right)}}- \cos{{\left({A}+{B}\right)}}\right]}\)
Thus we have
\(\displaystyle \sin{{\left(\pi{m}{x}\right)}} \sin{{\left(\pi{n}{x}\right)}}=\frac{1}{{2}}{\left[ \cos{{\left({m}-{n}\right)}}\pi{x}- \cos{{\left({m}+{n}\right)}}\pi{x}\right]}\)
First consider the case \(\displaystyle{m}\ne{n}\). Then we have
\(\displaystyle{\int_{{0}}^{{1}}} \sin{{\left(\pi{m}{x}\right)}} \sin{{\left(\pi{n}{x}\right)}}{\left.{d}{x}\right.}={\int_{{0}}^{{1}}}\frac{1}{{2}}{\left[ \cos{{\left({m}-{n}\right)}}\pi{x}- \cos{{\left({m}+{n}\right)}}\pi{x}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\left(\frac{{ \sin{{\left[{\left({m}-{n}\right)}\pi{x}\right]}}}}{{{2}{\left({m}-{n}\right)}\pi}}-\frac{{ \sin{{\left[{\left({m}+{n}\right)}\pi{x}\right]}}}}{{{2}{\left({m}+{n}\right)}\pi}}\right)}{{|}_{{0}}^{{1}}}\)
\(\displaystyle={0},\)
because \(\displaystyle \sin{{\left({k}\pi\right)}}={0}\), for all k in ZZ. Considering \(\displaystyle{m}={n}\) than we have
\(\displaystyle{\int_{{0}}^{{1}}}{{\sin}^{2}{\left(\pi{m}{x}\right)}}{\left.{d}{x}\right.}={\int_{{0}}^{{1}}}\frac{1}{{2}}{\left[{1}- \cos{{2}}{m}\pi{x}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\left(\frac{x}{{2}}-\frac{{ \sin{{\left[{2}{m}\pi{x}\right]}}}}{{{4}{m}\pi}}\right)}{{|}_{{0}}^{{1}}}\)
\(\displaystyle=\frac{1}{{2}},\)
And we get the finally answer is
\(\displaystyle{\int_{{0}}^{{1}}} \sin{{\left(\pi{m}{x}\right)}} \sin{{\left(\pi{n}{x}\right)}}{\left.{d}{x}\right.}={\left\lbrace{\left(\begin{matrix}{0}{m}\ne{n}\\{1}\text{/}{2}{m}={n}\end{matrix}\right)}\right.}\)
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