$int_{[0,2pi]}f dlambda=int_{[alpha,alpha+2pi]}f dlambda$ for complex $f$ with $f(x)=f(x+2pi)$












0












$begingroup$


Let $f$ be a complex-valued, measurable function defined on $mathbb R$ with $f(x)=f(x+2pi)$



I want to show $$int_{[0,2pi]}f dlambda=int_{[alpha,alpha+2pi]}f dlambda $$



$(alpha in mathbb R)$



My original attempt was to write $int_{[alpha,alpha+2pi]}f dlambda -int_{[0,2pi]}f dlambda=int_{alpha}^{alpha+2pi}f(x)dx-int_{0}^{2pi}f(x)dx=F(alpha+2pi)-F(2pi)-(F(alpha)-F(0))=int_{2pi}^{alpha+2pi}f(x)dx-int_{0}^{alpha}f(x)dx=0$



But I think I can't simply write is a Riemann Integral.



How can I show $$int_{[0,2pi]}f dlambda=int_{[alpha,alpha+2pi]}f dlambda $$ instead?






complex-analysis analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Simply split $[alpha,alpha+2pi] = [alpha,2pi] cup [2pi , 2pi+alpha]$.
    $endgroup$
    – Yanko
    Jan 12 at 16:04










  • $begingroup$
    How does that help?
    $endgroup$
    – user626880
    Jan 12 at 16:12










  • $begingroup$
    I post a more detailed answer.
    $endgroup$
    – Yanko
    Jan 12 at 16:16








  • 1




    $begingroup$
    Possible duplicate of An integrable and periodic function $f(x)$ satisfies $int_{0}^{T}f(x)dx=int_{a}^{a+T}f(x)dx$.
    $endgroup$
    – user587192
    Jan 12 at 16:44
















0












$begingroup$


Let $f$ be a complex-valued, measurable function defined on $mathbb R$ with $f(x)=f(x+2pi)$



I want to show $$int_{[0,2pi]}f dlambda=int_{[alpha,alpha+2pi]}f dlambda $$



$(alpha in mathbb R)$



My original attempt was to write $int_{[alpha,alpha+2pi]}f dlambda -int_{[0,2pi]}f dlambda=int_{alpha}^{alpha+2pi}f(x)dx-int_{0}^{2pi}f(x)dx=F(alpha+2pi)-F(2pi)-(F(alpha)-F(0))=int_{2pi}^{alpha+2pi}f(x)dx-int_{0}^{alpha}f(x)dx=0$



But I think I can't simply write is a Riemann Integral.



How can I show $$int_{[0,2pi]}f dlambda=int_{[alpha,alpha+2pi]}f dlambda $$ instead?






complex-analysis analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Simply split $[alpha,alpha+2pi] = [alpha,2pi] cup [2pi , 2pi+alpha]$.
    $endgroup$
    – Yanko
    Jan 12 at 16:04










  • $begingroup$
    How does that help?
    $endgroup$
    – user626880
    Jan 12 at 16:12










  • $begingroup$
    I post a more detailed answer.
    $endgroup$
    – Yanko
    Jan 12 at 16:16








  • 1




    $begingroup$
    Possible duplicate of An integrable and periodic function $f(x)$ satisfies $int_{0}^{T}f(x)dx=int_{a}^{a+T}f(x)dx$.
    $endgroup$
    – user587192
    Jan 12 at 16:44














0












0








0





$begingroup$


Let $f$ be a complex-valued, measurable function defined on $mathbb R$ with $f(x)=f(x+2pi)$



I want to show $$int_{[0,2pi]}f dlambda=int_{[alpha,alpha+2pi]}f dlambda $$



$(alpha in mathbb R)$



My original attempt was to write $int_{[alpha,alpha+2pi]}f dlambda -int_{[0,2pi]}f dlambda=int_{alpha}^{alpha+2pi}f(x)dx-int_{0}^{2pi}f(x)dx=F(alpha+2pi)-F(2pi)-(F(alpha)-F(0))=int_{2pi}^{alpha+2pi}f(x)dx-int_{0}^{alpha}f(x)dx=0$



But I think I can't simply write is a Riemann Integral.



How can I show $$int_{[0,2pi]}f dlambda=int_{[alpha,alpha+2pi]}f dlambda $$ instead?






complex-analysis analysis







share|cite|improve this question









$endgroup$




Let $f$ be a complex-valued, measurable function defined on $mathbb R$ with $f(x)=f(x+2pi)$



I want to show $$int_{[0,2pi]}f dlambda=int_{[alpha,alpha+2pi]}f dlambda $$



$(alpha in mathbb R)$



My original attempt was to write $int_{[alpha,alpha+2pi]}f dlambda -int_{[0,2pi]}f dlambda=int_{alpha}^{alpha+2pi}f(x)dx-int_{0}^{2pi}f(x)dx=F(alpha+2pi)-F(2pi)-(F(alpha)-F(0))=int_{2pi}^{alpha+2pi}f(x)dx-int_{0}^{alpha}f(x)dx=0$



But I think I can't simply write is a Riemann Integral.



How can I show $$int_{[0,2pi]}f dlambda=int_{[alpha,alpha+2pi]}f dlambda $$ instead?






complex-analysis analysis




complex-analysis analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 16:03









user626880user626880

204




204












  • $begingroup$
    Simply split $[alpha,alpha+2pi] = [alpha,2pi] cup [2pi , 2pi+alpha]$.
    $endgroup$
    – Yanko
    Jan 12 at 16:04










  • $begingroup$
    How does that help?
    $endgroup$
    – user626880
    Jan 12 at 16:12










  • $begingroup$
    I post a more detailed answer.
    $endgroup$
    – Yanko
    Jan 12 at 16:16








  • 1




    $begingroup$
    Possible duplicate of An integrable and periodic function $f(x)$ satisfies $int_{0}^{T}f(x)dx=int_{a}^{a+T}f(x)dx$.
    $endgroup$
    – user587192
    Jan 12 at 16:44


















  • $begingroup$
    Simply split $[alpha,alpha+2pi] = [alpha,2pi] cup [2pi , 2pi+alpha]$.
    $endgroup$
    – Yanko
    Jan 12 at 16:04










  • $begingroup$
    How does that help?
    $endgroup$
    – user626880
    Jan 12 at 16:12










  • $begingroup$
    I post a more detailed answer.
    $endgroup$
    – Yanko
    Jan 12 at 16:16








  • 1




    $begingroup$
    Possible duplicate of An integrable and periodic function $f(x)$ satisfies $int_{0}^{T}f(x)dx=int_{a}^{a+T}f(x)dx$.
    $endgroup$
    – user587192
    Jan 12 at 16:44
















$begingroup$
Simply split $[alpha,alpha+2pi] = [alpha,2pi] cup [2pi , 2pi+alpha]$.
$endgroup$
– Yanko
Jan 12 at 16:04




$begingroup$
Simply split $[alpha,alpha+2pi] = [alpha,2pi] cup [2pi , 2pi+alpha]$.
$endgroup$
– Yanko
Jan 12 at 16:04












$begingroup$
How does that help?
$endgroup$
– user626880
Jan 12 at 16:12




$begingroup$
How does that help?
$endgroup$
– user626880
Jan 12 at 16:12












$begingroup$
I post a more detailed answer.
$endgroup$
– Yanko
Jan 12 at 16:16






$begingroup$
I post a more detailed answer.
$endgroup$
– Yanko
Jan 12 at 16:16






1




1




$begingroup$
Possible duplicate of An integrable and periodic function $f(x)$ satisfies $int_{0}^{T}f(x)dx=int_{a}^{a+T}f(x)dx$.
$endgroup$
– user587192
Jan 12 at 16:44




$begingroup$
Possible duplicate of An integrable and periodic function $f(x)$ satisfies $int_{0}^{T}f(x)dx=int_{a}^{a+T}f(x)dx$.
$endgroup$
– user587192
Jan 12 at 16:44










1 Answer
1






active

oldest

votes


















0












$begingroup$

Edit: This proof assumes that $alpha<2pi$. But since $f$ is $2pi$ periodic the function $f$ on $[alpha,2pi+alpha]$ behaves the same as $f$ on $[alpha-2pi,alpha]$ and we can keep removing $2pi$ until $alpha-2mpi<2pi$, then denote $beta=alpha-2mpi$ and argue as below with $beta$ instead of $alpha$:



We can write $[alpha,alpha+2pi] = [alpha,2pi]cup [2pi,2pi+alpha]$. Since the union is disjoint (up to one element) we have



$$int_alpha^{alpha+2pi} f(x)dx = int_alpha^{2pi} f(x)dx+int_{2pi}^{2pi+alpha} f(x)dx$$



By assumption $f(x)=f(x+2pi)$ and so by changing variables the second integral becomes



$$int_{2pi}^{2pi+alpha} f(x)dx = int_0^alpha f(x+2pi)dx = int_0^alpha f(x)dx$$



Insert to the first equality we have



$$int_alpha^{alpha+2pi} f(x)dx = int_alpha^{2pi} f(x)dx+int_{0}^{alpha} f(x)dx = int_0^{2pi}f(x)dx$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    What if $alpha>2pi$? How do you make sense of the interval $[alpha,2pi]$?
    $endgroup$
    – user587192
    Jan 12 at 16:37






  • 1




    $begingroup$
    This would work if, instead of your "$2pi$", you used the (unique) integer multiple of $2pi$ between $alpha$ and $alpha+2pi$.
    $endgroup$
    – paul garrett
    Jan 12 at 16:40










  • $begingroup$
    @user587192 If $alpha>2pi$ then move the entire interval by $2pi$ as many times as needed.
    $endgroup$
    – Yanko
    Jan 12 at 16:54











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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0












$begingroup$

Edit: This proof assumes that $alpha<2pi$. But since $f$ is $2pi$ periodic the function $f$ on $[alpha,2pi+alpha]$ behaves the same as $f$ on $[alpha-2pi,alpha]$ and we can keep removing $2pi$ until $alpha-2mpi<2pi$, then denote $beta=alpha-2mpi$ and argue as below with $beta$ instead of $alpha$:



We can write $[alpha,alpha+2pi] = [alpha,2pi]cup [2pi,2pi+alpha]$. Since the union is disjoint (up to one element) we have



$$int_alpha^{alpha+2pi} f(x)dx = int_alpha^{2pi} f(x)dx+int_{2pi}^{2pi+alpha} f(x)dx$$



By assumption $f(x)=f(x+2pi)$ and so by changing variables the second integral becomes



$$int_{2pi}^{2pi+alpha} f(x)dx = int_0^alpha f(x+2pi)dx = int_0^alpha f(x)dx$$



Insert to the first equality we have



$$int_alpha^{alpha+2pi} f(x)dx = int_alpha^{2pi} f(x)dx+int_{0}^{alpha} f(x)dx = int_0^{2pi}f(x)dx$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    What if $alpha>2pi$? How do you make sense of the interval $[alpha,2pi]$?
    $endgroup$
    – user587192
    Jan 12 at 16:37






  • 1




    $begingroup$
    This would work if, instead of your "$2pi$", you used the (unique) integer multiple of $2pi$ between $alpha$ and $alpha+2pi$.
    $endgroup$
    – paul garrett
    Jan 12 at 16:40










  • $begingroup$
    @user587192 If $alpha>2pi$ then move the entire interval by $2pi$ as many times as needed.
    $endgroup$
    – Yanko
    Jan 12 at 16:54
















0












$begingroup$

Edit: This proof assumes that $alpha<2pi$. But since $f$ is $2pi$ periodic the function $f$ on $[alpha,2pi+alpha]$ behaves the same as $f$ on $[alpha-2pi,alpha]$ and we can keep removing $2pi$ until $alpha-2mpi<2pi$, then denote $beta=alpha-2mpi$ and argue as below with $beta$ instead of $alpha$:



We can write $[alpha,alpha+2pi] = [alpha,2pi]cup [2pi,2pi+alpha]$. Since the union is disjoint (up to one element) we have



$$int_alpha^{alpha+2pi} f(x)dx = int_alpha^{2pi} f(x)dx+int_{2pi}^{2pi+alpha} f(x)dx$$



By assumption $f(x)=f(x+2pi)$ and so by changing variables the second integral becomes



$$int_{2pi}^{2pi+alpha} f(x)dx = int_0^alpha f(x+2pi)dx = int_0^alpha f(x)dx$$



Insert to the first equality we have



$$int_alpha^{alpha+2pi} f(x)dx = int_alpha^{2pi} f(x)dx+int_{0}^{alpha} f(x)dx = int_0^{2pi}f(x)dx$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    What if $alpha>2pi$? How do you make sense of the interval $[alpha,2pi]$?
    $endgroup$
    – user587192
    Jan 12 at 16:37






  • 1




    $begingroup$
    This would work if, instead of your "$2pi$", you used the (unique) integer multiple of $2pi$ between $alpha$ and $alpha+2pi$.
    $endgroup$
    – paul garrett
    Jan 12 at 16:40










  • $begingroup$
    @user587192 If $alpha>2pi$ then move the entire interval by $2pi$ as many times as needed.
    $endgroup$
    – Yanko
    Jan 12 at 16:54














0












0








0





$begingroup$

Edit: This proof assumes that $alpha<2pi$. But since $f$ is $2pi$ periodic the function $f$ on $[alpha,2pi+alpha]$ behaves the same as $f$ on $[alpha-2pi,alpha]$ and we can keep removing $2pi$ until $alpha-2mpi<2pi$, then denote $beta=alpha-2mpi$ and argue as below with $beta$ instead of $alpha$:



We can write $[alpha,alpha+2pi] = [alpha,2pi]cup [2pi,2pi+alpha]$. Since the union is disjoint (up to one element) we have



$$int_alpha^{alpha+2pi} f(x)dx = int_alpha^{2pi} f(x)dx+int_{2pi}^{2pi+alpha} f(x)dx$$



By assumption $f(x)=f(x+2pi)$ and so by changing variables the second integral becomes



$$int_{2pi}^{2pi+alpha} f(x)dx = int_0^alpha f(x+2pi)dx = int_0^alpha f(x)dx$$



Insert to the first equality we have



$$int_alpha^{alpha+2pi} f(x)dx = int_alpha^{2pi} f(x)dx+int_{0}^{alpha} f(x)dx = int_0^{2pi}f(x)dx$$






share|cite|improve this answer











$endgroup$



Edit: This proof assumes that $alpha<2pi$. But since $f$ is $2pi$ periodic the function $f$ on $[alpha,2pi+alpha]$ behaves the same as $f$ on $[alpha-2pi,alpha]$ and we can keep removing $2pi$ until $alpha-2mpi<2pi$, then denote $beta=alpha-2mpi$ and argue as below with $beta$ instead of $alpha$:



We can write $[alpha,alpha+2pi] = [alpha,2pi]cup [2pi,2pi+alpha]$. Since the union is disjoint (up to one element) we have



$$int_alpha^{alpha+2pi} f(x)dx = int_alpha^{2pi} f(x)dx+int_{2pi}^{2pi+alpha} f(x)dx$$



By assumption $f(x)=f(x+2pi)$ and so by changing variables the second integral becomes



$$int_{2pi}^{2pi+alpha} f(x)dx = int_0^alpha f(x+2pi)dx = int_0^alpha f(x)dx$$



Insert to the first equality we have



$$int_alpha^{alpha+2pi} f(x)dx = int_alpha^{2pi} f(x)dx+int_{0}^{alpha} f(x)dx = int_0^{2pi}f(x)dx$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 12 at 16:57

























answered Jan 12 at 16:16









YankoYanko

6,5471528




6,5471528








  • 1




    $begingroup$
    What if $alpha>2pi$? How do you make sense of the interval $[alpha,2pi]$?
    $endgroup$
    – user587192
    Jan 12 at 16:37






  • 1




    $begingroup$
    This would work if, instead of your "$2pi$", you used the (unique) integer multiple of $2pi$ between $alpha$ and $alpha+2pi$.
    $endgroup$
    – paul garrett
    Jan 12 at 16:40










  • $begingroup$
    @user587192 If $alpha>2pi$ then move the entire interval by $2pi$ as many times as needed.
    $endgroup$
    – Yanko
    Jan 12 at 16:54














  • 1




    $begingroup$
    What if $alpha>2pi$? How do you make sense of the interval $[alpha,2pi]$?
    $endgroup$
    – user587192
    Jan 12 at 16:37






  • 1




    $begingroup$
    This would work if, instead of your "$2pi$", you used the (unique) integer multiple of $2pi$ between $alpha$ and $alpha+2pi$.
    $endgroup$
    – paul garrett
    Jan 12 at 16:40










  • $begingroup$
    @user587192 If $alpha>2pi$ then move the entire interval by $2pi$ as many times as needed.
    $endgroup$
    – Yanko
    Jan 12 at 16:54








1




1




$begingroup$
What if $alpha>2pi$? How do you make sense of the interval $[alpha,2pi]$?
$endgroup$
– user587192
Jan 12 at 16:37




$begingroup$
What if $alpha>2pi$? How do you make sense of the interval $[alpha,2pi]$?
$endgroup$
– user587192
Jan 12 at 16:37




1




1




$begingroup$
This would work if, instead of your "$2pi$", you used the (unique) integer multiple of $2pi$ between $alpha$ and $alpha+2pi$.
$endgroup$
– paul garrett
Jan 12 at 16:40




$begingroup$
This would work if, instead of your "$2pi$", you used the (unique) integer multiple of $2pi$ between $alpha$ and $alpha+2pi$.
$endgroup$
– paul garrett
Jan 12 at 16:40












$begingroup$
@user587192 If $alpha>2pi$ then move the entire interval by $2pi$ as many times as needed.
$endgroup$
– Yanko
Jan 12 at 16:54




$begingroup$
@user587192 If $alpha>2pi$ then move the entire interval by $2pi$ as many times as needed.
$endgroup$
– Yanko
Jan 12 at 16:54


















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