Problem with Improper Integration - Is Zero the Correct Answer?

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SUMMARY

The discussion centers on the improper integration of the function g(t) = cos(2πf0t) - jsin(2πf0t). The user initially calculated the integral of 1 over the interval [-T/2, T/2] and obtained zero, questioning the correctness of this result. The consensus indicates that the integral of a constant function over a symmetric interval yields zero, confirming the user's calculation as accurate. The integration process and boundary conditions were clarified, emphasizing the importance of understanding the function being integrated.

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killerfish
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Hi again,

I have a problem regarding improper integration.

Homework Statement

feafea.JPG


Refer to the image. I tried to solve and got zero for the answer. Is that correct? I refer to my actual problem it seem like it don't won't this way...

Thanks
 
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suppose that g(t)=1, what is the value of the integral?
 
benorin said:
suppose that g(t)=1, what is the value of the integral?

If 1 is integrated, we get t and then sub in the boundary value the answer still 0?

here the actual g(t) suppose to be g(t) = cos(2\pif0t) - jsin(2\pif0t).
 
Last edited:
killerfish said:
If 1 is integrated, we get t and then sub in the boundary value the answer still 0?
Are you saying that you are not capable of integrating
\int_{-T/2}^{T/2} dt
or you just haven't bothered to?

here the actual g(t) suppose to be g(t) = cos(2\pif0t) - jsin(2\pif0t).
 

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