Definite integral is undefined and not undefined

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In summary, the concept of a definite integral can be perceived as both undefined and not undefined depending on the context. It is defined as the limit of a Riemann sum, representing the area under a curve between two points. However, if the function is discontinuous or if the interval of integration includes points where the function is not defined, the definite integral may be considered undefined in those scenarios. Thus, its status can vary based on the properties of the function and the limits of integration.
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dyn
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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
 
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  • #2
dyn said:
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.
 
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  • #3
Alternatively, consider and to be reals rather than integers and then take the limit for .
 
  • #4
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
 
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  • #5
dyn said:
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 .
 
  • #6
Orodruin said:
You simply cannot integrate like that because the formula you are using for the primitive function assumes that .


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 ?
 
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  • #7
The trigonometric formula is fine. The primitive function is not.
 
  • #8
dyn said:
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

and are constants; accordingly the relevant antiderivative is . The same applies to .

Another well-known exception which would produce a zero denominator according to the general rule is .
 
  • #9
In general if
ThenIt ought to be clear why that doesn't work for .
 

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