Definite integral is undefined and not undefined

[dyn]

Hi
If i calculate the definite integral between the limits of L and 0 of sin(nπx/L)sin(kπx/L) using the trig formula 2sinAsinB = cos (A-B) - cos (A+B) it is undefined when n=k because (n-k) appears in the denominator. If i calculate the same integral with n=k using the formula
sin²(nπx/L) = ( 1 - cos(2nπx/L))/2 i get the value L/2. So my question is , how can the integral be undefined when calculated one way but defined with an exact value when calculated another way ?
Thanks

 

[Orodruin]

because (n-k) appears in the denominator

No it doesn’t. The integral of cos(0) = 1 is not sin(0)/0. You will have to treat that case separately.

 

[Orodruin]

Alternatively, consider $n$ and $k$ to be reals rather than integers and then take the limit $n \to k$ for $\sin((n-k)x)/(n-k) \to x$.

 

[dyn]

If i integrate cos (πx/L) ( n-k) , i get L/(π(n-k))sin (πx/L)(n-k) which is undefined at n=k

 

[Orodruin]

If i integrate cos (πx/L) ( n-k) , i get L/(π(n-k))sin (πx/L)(n-k) which is undefined at n=k

Please use LaTeX to make it clearer what you mean.

I assume the (n-k) is inside the cosine because that is what it should be - even if not clear from your post.

In this case my first comment applies, and second comment. You simply cannot integrate like that because the formula you are using for the primitive function assumes that $n \neq k$.

 

[dyn]

You simply cannot integrate like that because the formula you are using for the primitive function assumes that $n \neq k$.

The trig formula 2sinAsinB = cos (A-B) - cos (A+B) applies for A=B and A≠B so when i use that formula in the integral why does it not apply for A=B ? I see your point that it gives cos(A-B) which is 1 when A=B so integrating 1 ends up giving the correct answer. But what i don't understand is why if the trig formula is valid for any A and B is it not valid in the integral for A=B ? In other words , why can't the formula just be followed as a prescription ?

 

[Orodruin]

The trigonometric formula is fine. The primitive function is not.

 

[pasmith]

Hi
If i calculate the definite integral between the limits of L and 0 of sin(nπx/L)sin(kπx/L) using the trig formula 2sinAsinB = cos (A-B) - cos (A+B) it is undefined when n=k because (n-k) appears in the denominator. If i calculate the same integral with n=k using the formula
sin²(nπx/L) = ( 1 - cos(2nπx/L))/2 i get the value L/2. So my question is , how can the integral be undefined when calculated one way but defined with an exact value when calculated another way ?
Thanks

$\cos(0x) = \cos(0) = 1$ and $\sin(0x) = \sin 0 = 0$ are constants; accordingly the relevant antiderivative is $\int k\,dx = kx + C$. The same applies to $\exp(0x) = \exp(0) = 1$.

Another well-known exception which would produce a zero denominator according to the general rule is $\int x^{-1}\,dx = \ln x + C$.

 

[PeroK]

In general if
$$\int f(x) dx = F(x)$$Then$$\int f(ax)dx = \frac 1 a F(ax) \ (a \ne 0)$$It ought to be clear why that doesn't work for $a=0$.