Why is $ frac12picdotint_-pi/2^pi/2cos(x)cdot e^-jnx neq frac1picdotint_0^pi/2cos(x)cdot e^-jnx $

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Let's assume we have $$f(x) = frac12picdotint_-pi/2^pi/2 cos(x)cdot e^-jnx.$$



Usually it's possible to move the border and multiply the integral with $2$. Geometrically you cut the half and add it again afterwards to make calculation easier.

Obviously this isn't legal here but I can't see why (beside the obvious reason that it's simply not equal).



beginalign
frac12picdotint_-pi/2^pi/2 cos(x)cdot e^-jnx &neq frac1picdotint_0^pi/2 cos(x)cdot e^-jnx \
frac12picdotint_-pi/2^pi/2 cos(x)cdot e^-jnx &= fracn pi cdot sin[n pi]1 - n^2 \
frac1picdotint_0^pi/2 cos(x)cdot e^-jnx &=
fracj (1 + e^-j n pi) n(1 - n^2) pi.
endalign






integration







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  • This is legal. Integration is linear and you can divide your main integral into $3$. Two of them vanish.
    – Niklas
    Sep 10 at 21:16










  • $e^-jnxne e^-jn(-x)$ so no.
    – Andrei
    Sep 10 at 21:21














up vote
0
down vote

favorite












Let's assume we have $$f(x) = frac12picdotint_-pi/2^pi/2 cos(x)cdot e^-jnx.$$



Usually it's possible to move the border and multiply the integral with $2$. Geometrically you cut the half and add it again afterwards to make calculation easier.

Obviously this isn't legal here but I can't see why (beside the obvious reason that it's simply not equal).



beginalign
frac12picdotint_-pi/2^pi/2 cos(x)cdot e^-jnx &neq frac1picdotint_0^pi/2 cos(x)cdot e^-jnx \
frac12picdotint_-pi/2^pi/2 cos(x)cdot e^-jnx &= fracn pi cdot sin[n pi]1 - n^2 \
frac1picdotint_0^pi/2 cos(x)cdot e^-jnx &=
fracj (1 + e^-j n pi) n(1 - n^2) pi.
endalign






integration







share|cite|improve this question























  • This is legal. Integration is linear and you can divide your main integral into $3$. Two of them vanish.
    – Niklas
    Sep 10 at 21:16










  • $e^-jnxne e^-jn(-x)$ so no.
    – Andrei
    Sep 10 at 21:21












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let's assume we have $$f(x) = frac12picdotint_-pi/2^pi/2 cos(x)cdot e^-jnx.$$



Usually it's possible to move the border and multiply the integral with $2$. Geometrically you cut the half and add it again afterwards to make calculation easier.

Obviously this isn't legal here but I can't see why (beside the obvious reason that it's simply not equal).



beginalign
frac12picdotint_-pi/2^pi/2 cos(x)cdot e^-jnx &neq frac1picdotint_0^pi/2 cos(x)cdot e^-jnx \
frac12picdotint_-pi/2^pi/2 cos(x)cdot e^-jnx &= fracn pi cdot sin[n pi]1 - n^2 \
frac1picdotint_0^pi/2 cos(x)cdot e^-jnx &=
fracj (1 + e^-j n pi) n(1 - n^2) pi.
endalign






integration







share|cite|improve this question















Let's assume we have $$f(x) = frac12picdotint_-pi/2^pi/2 cos(x)cdot e^-jnx.$$



Usually it's possible to move the border and multiply the integral with $2$. Geometrically you cut the half and add it again afterwards to make calculation easier.

Obviously this isn't legal here but I can't see why (beside the obvious reason that it's simply not equal).



beginalign
frac12picdotint_-pi/2^pi/2 cos(x)cdot e^-jnx &neq frac1picdotint_0^pi/2 cos(x)cdot e^-jnx \
frac12picdotint_-pi/2^pi/2 cos(x)cdot e^-jnx &= fracn pi cdot sin[n pi]1 - n^2 \
frac1picdotint_0^pi/2 cos(x)cdot e^-jnx &=
fracj (1 + e^-j n pi) n(1 - n^2) pi.
endalign






integration




integration






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edited 2 days ago









Leucippus

19k102769




19k102769










asked Sep 10 at 21:11









TimSch

816




816











  • This is legal. Integration is linear and you can divide your main integral into $3$. Two of them vanish.
    – Niklas
    Sep 10 at 21:16










  • $e^-jnxne e^-jn(-x)$ so no.
    – Andrei
    Sep 10 at 21:21
















  • This is legal. Integration is linear and you can divide your main integral into $3$. Two of them vanish.
    – Niklas
    Sep 10 at 21:16










  • $e^-jnxne e^-jn(-x)$ so no.
    – Andrei
    Sep 10 at 21:21















This is legal. Integration is linear and you can divide your main integral into $3$. Two of them vanish.
– Niklas
Sep 10 at 21:16




This is legal. Integration is linear and you can divide your main integral into $3$. Two of them vanish.
– Niklas
Sep 10 at 21:16












$e^-jnxne e^-jn(-x)$ so no.
– Andrei
Sep 10 at 21:21




$e^-jnxne e^-jn(-x)$ so no.
– Andrei
Sep 10 at 21:21










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You can only exploit symmetry when the integrand in question is even, i.e. when $f(-x)=f(x)$. In this case, $f(x)=cos(x)cdot e^-jnx$. It follows that $f(-x)=cos(-x)cdot e^-jn(-x)=cos(x) cdot e^jnx$. Thus, $f(-x)neq f(x)$.






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    You can only exploit symmetry when the integrand in question is even, i.e. when $f(-x)=f(x)$. In this case, $f(x)=cos(x)cdot e^-jnx$. It follows that $f(-x)=cos(-x)cdot e^-jn(-x)=cos(x) cdot e^jnx$. Thus, $f(-x)neq f(x)$.






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      up vote
      1
      down vote



      accepted










      You can only exploit symmetry when the integrand in question is even, i.e. when $f(-x)=f(x)$. In this case, $f(x)=cos(x)cdot e^-jnx$. It follows that $f(-x)=cos(-x)cdot e^-jn(-x)=cos(x) cdot e^jnx$. Thus, $f(-x)neq f(x)$.






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        You can only exploit symmetry when the integrand in question is even, i.e. when $f(-x)=f(x)$. In this case, $f(x)=cos(x)cdot e^-jnx$. It follows that $f(-x)=cos(-x)cdot e^-jn(-x)=cos(x) cdot e^jnx$. Thus, $f(-x)neq f(x)$.






        share|cite|improve this answer












        You can only exploit symmetry when the integrand in question is even, i.e. when $f(-x)=f(x)$. In this case, $f(x)=cos(x)cdot e^-jnx$. It follows that $f(-x)=cos(-x)cdot e^-jn(-x)=cos(x) cdot e^jnx$. Thus, $f(-x)neq f(x)$.







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        answered Sep 10 at 22:07









        Stijn Dietz

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