Fourier series of $frac{1}{x}$












1












$begingroup$



What is the Fourier series expansion of $frac{1}{x}$ ?




The best method I could come up with was shifting the function by 'k' (shifting the function to $frac{1}{x - k}$), so that while calculating the coefficients you don't run into the discontinuity of 1/x.



Is there a different method to calculate the Fourier series of $frac{1}{x}$.






fourier-series







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    On what interval?
    $endgroup$
    – user587192
    Dec 16 '18 at 0:25










  • $begingroup$
    Interval (0,2$pi$)
    $endgroup$
    – Nimish
    Dec 16 '18 at 0:33






  • 1




    $begingroup$
    On $(0,2pi)$ it doesn't make sense but on $(-pi,pi)$ it does, search about "principal value"
    $endgroup$
    – reuns
    Dec 16 '18 at 0:35












  • $begingroup$
    Let $h(y) = pv.(int_{-pi}^pi frac{e^{i xy}}{x}dx)$ then $h'(y) = int_{-pi}^pi i e^{i xy}dx =2i frac{sin(pi y)}{y}$ and since $h(0) = 0$ then $h(y) =2i int_0^y frac{sin(pi u)}{u} du$
    $endgroup$
    – reuns
    Dec 16 '18 at 0:38








  • 1




    $begingroup$
    @mathworker21 for example because $pv.(int_{-pi}^pi frac{f(x)}{x}dx ) = int_{-pi}^pi frac{f(x)-f(0)}{x}dx$ which can be taken as the definition
    $endgroup$
    – reuns
    Dec 16 '18 at 0:45


















1












$begingroup$



What is the Fourier series expansion of $frac{1}{x}$ ?




The best method I could come up with was shifting the function by 'k' (shifting the function to $frac{1}{x - k}$), so that while calculating the coefficients you don't run into the discontinuity of 1/x.



Is there a different method to calculate the Fourier series of $frac{1}{x}$.






fourier-series







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    On what interval?
    $endgroup$
    – user587192
    Dec 16 '18 at 0:25










  • $begingroup$
    Interval (0,2$pi$)
    $endgroup$
    – Nimish
    Dec 16 '18 at 0:33






  • 1




    $begingroup$
    On $(0,2pi)$ it doesn't make sense but on $(-pi,pi)$ it does, search about "principal value"
    $endgroup$
    – reuns
    Dec 16 '18 at 0:35












  • $begingroup$
    Let $h(y) = pv.(int_{-pi}^pi frac{e^{i xy}}{x}dx)$ then $h'(y) = int_{-pi}^pi i e^{i xy}dx =2i frac{sin(pi y)}{y}$ and since $h(0) = 0$ then $h(y) =2i int_0^y frac{sin(pi u)}{u} du$
    $endgroup$
    – reuns
    Dec 16 '18 at 0:38








  • 1




    $begingroup$
    @mathworker21 for example because $pv.(int_{-pi}^pi frac{f(x)}{x}dx ) = int_{-pi}^pi frac{f(x)-f(0)}{x}dx$ which can be taken as the definition
    $endgroup$
    – reuns
    Dec 16 '18 at 0:45
















1












1








1





$begingroup$



What is the Fourier series expansion of $frac{1}{x}$ ?




The best method I could come up with was shifting the function by 'k' (shifting the function to $frac{1}{x - k}$), so that while calculating the coefficients you don't run into the discontinuity of 1/x.



Is there a different method to calculate the Fourier series of $frac{1}{x}$.






fourier-series







share|cite|improve this question









$endgroup$





What is the Fourier series expansion of $frac{1}{x}$ ?




The best method I could come up with was shifting the function by 'k' (shifting the function to $frac{1}{x - k}$), so that while calculating the coefficients you don't run into the discontinuity of 1/x.



Is there a different method to calculate the Fourier series of $frac{1}{x}$.






fourier-series




fourier-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 0:20









NimishNimish

898




898








  • 3




    $begingroup$
    On what interval?
    $endgroup$
    – user587192
    Dec 16 '18 at 0:25










  • $begingroup$
    Interval (0,2$pi$)
    $endgroup$
    – Nimish
    Dec 16 '18 at 0:33






  • 1




    $begingroup$
    On $(0,2pi)$ it doesn't make sense but on $(-pi,pi)$ it does, search about "principal value"
    $endgroup$
    – reuns
    Dec 16 '18 at 0:35












  • $begingroup$
    Let $h(y) = pv.(int_{-pi}^pi frac{e^{i xy}}{x}dx)$ then $h'(y) = int_{-pi}^pi i e^{i xy}dx =2i frac{sin(pi y)}{y}$ and since $h(0) = 0$ then $h(y) =2i int_0^y frac{sin(pi u)}{u} du$
    $endgroup$
    – reuns
    Dec 16 '18 at 0:38








  • 1




    $begingroup$
    @mathworker21 for example because $pv.(int_{-pi}^pi frac{f(x)}{x}dx ) = int_{-pi}^pi frac{f(x)-f(0)}{x}dx$ which can be taken as the definition
    $endgroup$
    – reuns
    Dec 16 '18 at 0:45
















  • 3




    $begingroup$
    On what interval?
    $endgroup$
    – user587192
    Dec 16 '18 at 0:25










  • $begingroup$
    Interval (0,2$pi$)
    $endgroup$
    – Nimish
    Dec 16 '18 at 0:33






  • 1




    $begingroup$
    On $(0,2pi)$ it doesn't make sense but on $(-pi,pi)$ it does, search about "principal value"
    $endgroup$
    – reuns
    Dec 16 '18 at 0:35












  • $begingroup$
    Let $h(y) = pv.(int_{-pi}^pi frac{e^{i xy}}{x}dx)$ then $h'(y) = int_{-pi}^pi i e^{i xy}dx =2i frac{sin(pi y)}{y}$ and since $h(0) = 0$ then $h(y) =2i int_0^y frac{sin(pi u)}{u} du$
    $endgroup$
    – reuns
    Dec 16 '18 at 0:38








  • 1




    $begingroup$
    @mathworker21 for example because $pv.(int_{-pi}^pi frac{f(x)}{x}dx ) = int_{-pi}^pi frac{f(x)-f(0)}{x}dx$ which can be taken as the definition
    $endgroup$
    – reuns
    Dec 16 '18 at 0:45










3




3




$begingroup$
On what interval?
$endgroup$
– user587192
Dec 16 '18 at 0:25




$begingroup$
On what interval?
$endgroup$
– user587192
Dec 16 '18 at 0:25












$begingroup$
Interval (0,2$pi$)
$endgroup$
– Nimish
Dec 16 '18 at 0:33




$begingroup$
Interval (0,2$pi$)
$endgroup$
– Nimish
Dec 16 '18 at 0:33




1




1




$begingroup$
On $(0,2pi)$ it doesn't make sense but on $(-pi,pi)$ it does, search about "principal value"
$endgroup$
– reuns
Dec 16 '18 at 0:35






$begingroup$
On $(0,2pi)$ it doesn't make sense but on $(-pi,pi)$ it does, search about "principal value"
$endgroup$
– reuns
Dec 16 '18 at 0:35














$begingroup$
Let $h(y) = pv.(int_{-pi}^pi frac{e^{i xy}}{x}dx)$ then $h'(y) = int_{-pi}^pi i e^{i xy}dx =2i frac{sin(pi y)}{y}$ and since $h(0) = 0$ then $h(y) =2i int_0^y frac{sin(pi u)}{u} du$
$endgroup$
– reuns
Dec 16 '18 at 0:38






$begingroup$
Let $h(y) = pv.(int_{-pi}^pi frac{e^{i xy}}{x}dx)$ then $h'(y) = int_{-pi}^pi i e^{i xy}dx =2i frac{sin(pi y)}{y}$ and since $h(0) = 0$ then $h(y) =2i int_0^y frac{sin(pi u)}{u} du$
$endgroup$
– reuns
Dec 16 '18 at 0:38






1




1




$begingroup$
@mathworker21 for example because $pv.(int_{-pi}^pi frac{f(x)}{x}dx ) = int_{-pi}^pi frac{f(x)-f(0)}{x}dx$ which can be taken as the definition
$endgroup$
– reuns
Dec 16 '18 at 0:45






$begingroup$
@mathworker21 for example because $pv.(int_{-pi}^pi frac{f(x)}{x}dx ) = int_{-pi}^pi frac{f(x)-f(0)}{x}dx$ which can be taken as the definition
$endgroup$
– reuns
Dec 16 '18 at 0:45












1 Answer
1






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oldest

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4












$begingroup$

The Fourier series only exists for periodic functions which are integrable over a period. You can choose an interval and consider the periodic extension of $frac{1}{x}$ over that interval, but if that interval contains $0$ (even as an endpoint), it will not be integrable.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I would not make such a statement. Absolute integrability of $f$ and further conditions of regularity such as local bounded variation guarantees convergence of the Fourier series at a point. But the Fourier sine series of $f(x) = 1/x$ on $[0,pi]$ exists. Note that $b_n = frac{2}{pi}int_0^pi frac{sin nx}{x} , dx = frac{2}{pi} Si(npi)$ exists.
    $endgroup$
    – RRL
    Dec 16 '18 at 6:50













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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









4












$begingroup$

The Fourier series only exists for periodic functions which are integrable over a period. You can choose an interval and consider the periodic extension of $frac{1}{x}$ over that interval, but if that interval contains $0$ (even as an endpoint), it will not be integrable.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I would not make such a statement. Absolute integrability of $f$ and further conditions of regularity such as local bounded variation guarantees convergence of the Fourier series at a point. But the Fourier sine series of $f(x) = 1/x$ on $[0,pi]$ exists. Note that $b_n = frac{2}{pi}int_0^pi frac{sin nx}{x} , dx = frac{2}{pi} Si(npi)$ exists.
    $endgroup$
    – RRL
    Dec 16 '18 at 6:50


















4












$begingroup$

The Fourier series only exists for periodic functions which are integrable over a period. You can choose an interval and consider the periodic extension of $frac{1}{x}$ over that interval, but if that interval contains $0$ (even as an endpoint), it will not be integrable.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I would not make such a statement. Absolute integrability of $f$ and further conditions of regularity such as local bounded variation guarantees convergence of the Fourier series at a point. But the Fourier sine series of $f(x) = 1/x$ on $[0,pi]$ exists. Note that $b_n = frac{2}{pi}int_0^pi frac{sin nx}{x} , dx = frac{2}{pi} Si(npi)$ exists.
    $endgroup$
    – RRL
    Dec 16 '18 at 6:50
















4












4








4





$begingroup$

The Fourier series only exists for periodic functions which are integrable over a period. You can choose an interval and consider the periodic extension of $frac{1}{x}$ over that interval, but if that interval contains $0$ (even as an endpoint), it will not be integrable.






share|cite|improve this answer









$endgroup$



The Fourier series only exists for periodic functions which are integrable over a period. You can choose an interval and consider the periodic extension of $frac{1}{x}$ over that interval, but if that interval contains $0$ (even as an endpoint), it will not be integrable.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 0:27









AlexanderJ93AlexanderJ93

6,137823




6,137823












  • $begingroup$
    I would not make such a statement. Absolute integrability of $f$ and further conditions of regularity such as local bounded variation guarantees convergence of the Fourier series at a point. But the Fourier sine series of $f(x) = 1/x$ on $[0,pi]$ exists. Note that $b_n = frac{2}{pi}int_0^pi frac{sin nx}{x} , dx = frac{2}{pi} Si(npi)$ exists.
    $endgroup$
    – RRL
    Dec 16 '18 at 6:50




















  • $begingroup$
    I would not make such a statement. Absolute integrability of $f$ and further conditions of regularity such as local bounded variation guarantees convergence of the Fourier series at a point. But the Fourier sine series of $f(x) = 1/x$ on $[0,pi]$ exists. Note that $b_n = frac{2}{pi}int_0^pi frac{sin nx}{x} , dx = frac{2}{pi} Si(npi)$ exists.
    $endgroup$
    – RRL
    Dec 16 '18 at 6:50


















$begingroup$
I would not make such a statement. Absolute integrability of $f$ and further conditions of regularity such as local bounded variation guarantees convergence of the Fourier series at a point. But the Fourier sine series of $f(x) = 1/x$ on $[0,pi]$ exists. Note that $b_n = frac{2}{pi}int_0^pi frac{sin nx}{x} , dx = frac{2}{pi} Si(npi)$ exists.
$endgroup$
– RRL
Dec 16 '18 at 6:50






$begingroup$
I would not make such a statement. Absolute integrability of $f$ and further conditions of regularity such as local bounded variation guarantees convergence of the Fourier series at a point. But the Fourier sine series of $f(x) = 1/x$ on $[0,pi]$ exists. Note that $b_n = frac{2}{pi}int_0^pi frac{sin nx}{x} , dx = frac{2}{pi} Si(npi)$ exists.
$endgroup$
– RRL
Dec 16 '18 at 6:50




















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