Question

∫sin^2(mx) from x=-π to π

Answer 1

cos 2x = cos^2 x - sin^2 x = 1 - 2 sin^2 x
hence
sin^2 x = 1/2 . (1 - cos 2x)
and therefore
\[ \sin^2 (mx) = \frac{1}{2} ( 1 - 2 \cos(2mx) ) \]

Answer 2

Now integrate that

Answer 3

\[\int\limits_{-π}^{π}Sin^2(mx)dx\]

Answer 4

still there?

Answer 5

\[={1 \over 2} \int\limits\limits_{−π}^{π}[\cos(2mx)+1 ]dx\]

Answer 6

you dropped a minus sign

Answer 7

\[ {1 \over 2} \int\limits\limits_{−π}^{π}[1-\cos(2mx) ]dx \]

Answer 8

\[={1 \over 2 } \left[ x-{\sin(2mx) \over 2m }\right]_{π}^{-π}\]

Answer 9

is this right so far?,
i am trying to sown that the integral is equal to π for m≠0

Answer 10

If m is an integer yes. Note that in your last expression you've reversed the limits of integration.

Answer 11

\[={1 \over 2 } \left[ x-{sin(2mx) \over 2m }\right]_{-π}^{π}\]
\[={1 \over 2 } \left[ (π-{sin(2πm) \over 2m })-(-π-{sin(-2πm) \over2m})\right]\]

Answer 12

\[={1 \over 2}[2π−2{\sin(2πm) \over 2m} ]\]

\[=π−{\sin(2πm) \over 2m}\]

and the sine function will be equal to zero at multiples of 2π (m is a positive integer)
hence = \[π\delta _{0m}\]

(did i miss anything or make mistakes?

Answer 13

I think this is ok; writing the Dirac delta function as you have here is perhaps bit over the top and not quite accurate. I would just say the integral is equal to

pi , if m is an integer
pi + sin(2mpi)/m , if m is not an integer

Check your problem though; I'm guessing m is an integer in which case you can just get rid of the second case.

Answer 14

yeah m is an interger

Answer 15

so the integral equals
2π ; m=n=0
{ π ; m=n≠0
0 ; m≠n

Answer 16

Where is this parameter n coming from? I only see n in the problem.

Answer 17

ops. ive been doing a few similar problems
that is the result for an integral of cos(nx)cos(mx) over the same limits

Answer 18

I see what you're doing now. Makes sense.

Answer 19

back to the question i asked:
If m = 0, im getting confused

Answer 20

If m = n = 0, then cos(nx)cos(mx) = 1 . 1 = 1.
Hence the integral is 2pi

Answer 21

sin (2πm)/m ~> 0/0

Answer 22

\[\lim_{x \rightarrow 0} \frac{ \sin ax}{x} = a\]

Answer 23

Hence the limit you wrote down is 2pi

Answer 24

so, sin (2πm) = m^2 in the limit of large m
interesting

Answer 25

*small m not large m

Answer 26

No, I don't know where you're getting that from.

Answer 27

The "straight" limit here is that as x --> 0 then the limit of (sin x)/x is 1.

Answer 28

Now when we consider sin(ax)/x, we can write this as

Answer 29

\[\lim_{x \rightarrow 0} \frac{ \sin(ax)}{x} = \lim_{x \rightarrow 0} \frac{ \sin(x)}{x/a} = \lim_{x \rightarrow 0} \ a \frac{ \sin(x)}{x} = a \ \lim_{x \rightarrow 0} \frac{ \sin(x)}{x} = a . 1 = a\]

Answer 30

In other words, when ax is small sin(ax) goes like ax.
Or sin(m.pi.x) goes like m.pi.x

Answer 31

from
\[=π-{\sin(2πm) \over 2m}\] the integral would become for m=0
\[=π-{(2πm) \over 2m}, \]
\[=π-π\]
\[=0\]

Answer 32

The direct way to see this is to put m = 0 into the original integral of
sin^2(mx) = sin^2(0) = 0.

The integral of 0 over -pi to pi is zero.

Answer 33

for m≠0\[=π-{\sin(2πm) \over 2m}\]
\[=π-{0 \over 2m}\]
\[=π\]

Answer 34

yes

Answer 35

Can i wright this as
\[\int\limits_{-π}^{π}\sin^2 (mx) .dx\]
\[=π-{sin(2πm) \over 2m}\]
\[ π;{ m≠0 } \]={\[0; {m=0}\]
\[=πδ_{m0}\]

Answer 36

Where δ_m0 is the Kronecker delta

Answer 37

Thankyou so much for your help JamesJ

Answer 38

sure