Prove the following formula: integrate of sin(mx)sin(nx)dx = 0 if n does not equal m integrate of sin(mx)sin(nx)dx = pi if n=m
2 Answers
sciencesolve | Teacher | (Level 3) Educator Emeritus
The problem is inconsistent since it does not provide the limits of integration because only evaluating a definite integral yields a value such 0 or `pi` .
You only may evaluate the indefinite integral converting the product `sin(mx)sin(nx)` into a sum such that:
`sin(mx)sin(nx) = (1/2)(cos (mx - nx) - cos(mx + nx))`
`int sin(mx)sin(nx)dx = (1/2)int cos(m-n)x dx - (1/2)int cos(m+n)x dx`
`int sin(mx)sin(nx)dx = (1/2)((sin(m-n)x)/(m-n) - (sin(m+n)x)/(m+n)) + c`
Considering `m!=n` yields:
`int sin(mx)sin(nx)dx = (1/2)((sin(m-n)x)/(m-n) - (sin(m+n)x)/(m+n)) + c`
Considering `m=n` yields:
`int sin(mx)sin(nx)dx = 0/0 ` invalid
Hence, evaluating the given indefinite integral if `m!=n` yields `int sin(mx)sin(nx)dx = (1/2)((sin(m-n)x)/(m-n) - (sin(m+n)x)/(m+n)) + c` , but if m=n, the result is invalid.
quantatanu | Student, Undergraduate | (Level 1) Valedictorian
I suppose the problem is to integrate from
-Pi to Pi
or equivalently from
0 to 2Pi.
and also m and n are positive integers.
I will use the formula:
Sin(A)Sin(B) = (1/2)[Cos(A-B) - Cos(A+B)]
Therefore,
Sin(mx)Sin(nx) =(1/2)[Cos((m-n)x) - Cos((m+n)x)]
So gievn,
I = Integral[ Sin(mx) Sin(nx) dx] 0 to 2Pi
= Integral[(1/2){Cos((m-n)x) - Cos((m+n)x)} dx] 0 to 2Pi
= Integral[(1/2){Cos((m-n)x)] 0 to 2Pi
- (1/2) Sin((m+n)x)/(m+n)] 0 to 2Pi
as m+n =/= 0 always. So the second term in the result will be zero both at 0 and 2Pi.
therefore,
I = Integral[(1/2){Cos((m-n)x)] 0 to 2Pi
= Integral[(1/4) { Exp[(m-n) x] + Exp[ - (m-n) x]}]
=(1/4) (2Pi) Integral[(1/2Pi){ Exp[(m-n) x] + Exp[ - (m-n) x]}]
= (Pi/2) Integral[(1/2Pi){ Exp[(m-n) x] + Exp[(n-m) x]}]
= (Pi/2) * [CroneckerDelta(m,n) + CroneckerDelta(n,m)]
= (Pi/2) * 2 * CroneckerDelta(m,n)
= Pi * CroneckerDelta(m,n)
Where I have used the definition of Cronecker Delta function:
CroneckerDelta(m,n)=Integral[(1/2Pi){ Exp[(m-n) x]
and
CroneckerDelta(m,n)
= CroneckerDelta(n,m) = 1 {if m=n}
= 0 {if m =/= n}
Hence The Result:
I = Integral[ Sin(mx) Sin(nx) dx] 0 to 2Pi
= Pi * CroneckerDelta(m,n)
= Pi {if m=n}
= Pi *0 = 0 for m =/= n