- #1
zenterix
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- Homework Statement
- The following is problem 7 of chapter 9.10 in Apostol's *Calculus*, Volume I.
(a) If ##m## and ##n## are integers prove that
$$\int_0^{2\pi} e^{inx}e^{-imx}dx=\begin{cases}0\ \ \ \ \ \text{if}\
m\neq n\\ 2\pi\ \ \ \ \text{if}\ m=n\end{cases}$$
(b) Use part (a) to deduce the orthogonality relations for the sine and
cosine (##m## and ##n## are integers, ##m^2\neq n^2##):
$$\int_0^{2\pi} \sin{nx}\cos{mx}dx=\int_0^{2\pi}
\sin{nx}\sin{mx}dx=\int_0^{2\pi} \cos{nx}\cos{mx}dx$$
- Relevant Equations
- $$\int_0^{2\pi} \sin^2{nx}dx=\int_0^{2\pi} \cos^2{nx}dx=\pi$$
To solve part (a), we write ##e^{inx}e^{-imx}=e^{ix(n-m)}##.
If ##m=n## then this expression is 1, and so the integral of 1 from 0 to ##2\pi## is ##2\pi##.
If ##m\neq n## then we use Euler's formula and integrate. The result is zero.
My question is how do we solve part (b) using part (a)?
I can solve part (b) by using the trigonometric identities for ##\sin{(a\pm b)}## and ##\cos{(a\pm b)}## with ##a=mx## and ##b=nx##.
But how do we solve (b) using part (a)?
What I tried to do was
$$e^{nxi}e^{-mxi}=(\cos{nx}+i\sin{mx})(\cos{mx}-i\sin{nx})\tag{1}$$
$$e^{nxi}e^{-mxi}=\cos{nx}\cos{mx}+\sin{nx}\sin{mx}+i(-\cos{nx}\sin{mx}+\sin{nx}\cos{mx})\tag{2}$$
If ##m\neq n## then when we integrate this expression the left-hand side is zero (by part (a)).
However, we are left with
$$0=\int_0^{2\pi} (\cos{nx}\cos{mx}+\sin{nx}\sin{mx}) dx + i\int_0^{2\pi} (-\cos{nx}\sin{mx}+\sin{nx}\cos{mx})dx$$
Thus
$$\int_0^{2\pi} (\cos{nx}\cos{mx}+\sin{nx}\sin{mx}) dx=0$$
$$\int_0^{2\pi} (-\cos{nx}\sin{mx}+\sin{nx}\cos{mx})dx=0$$
If ##m=n## then this expression is 1, and so the integral of 1 from 0 to ##2\pi## is ##2\pi##.
If ##m\neq n## then we use Euler's formula and integrate. The result is zero.
My question is how do we solve part (b) using part (a)?
I can solve part (b) by using the trigonometric identities for ##\sin{(a\pm b)}## and ##\cos{(a\pm b)}## with ##a=mx## and ##b=nx##.
But how do we solve (b) using part (a)?
What I tried to do was
$$e^{nxi}e^{-mxi}=(\cos{nx}+i\sin{mx})(\cos{mx}-i\sin{nx})\tag{1}$$
$$e^{nxi}e^{-mxi}=\cos{nx}\cos{mx}+\sin{nx}\sin{mx}+i(-\cos{nx}\sin{mx}+\sin{nx}\cos{mx})\tag{2}$$
If ##m\neq n## then when we integrate this expression the left-hand side is zero (by part (a)).
However, we are left with
$$0=\int_0^{2\pi} (\cos{nx}\cos{mx}+\sin{nx}\sin{mx}) dx + i\int_0^{2\pi} (-\cos{nx}\sin{mx}+\sin{nx}\cos{mx})dx$$
Thus
$$\int_0^{2\pi} (\cos{nx}\cos{mx}+\sin{nx}\sin{mx}) dx=0$$
$$\int_0^{2\pi} (-\cos{nx}\sin{mx}+\sin{nx}\cos{mx})dx=0$$