i know application but want proof of following definite integration property .












5














Question: prove



(a) If $f(x)=f(-x)$ and $f(x+pi)=-f(x)$ then,



$$int_{0_{+}}^{infty} f(x)dfrac{sin x}{x}dx=int_{0_{+}}^{dfrac{pi}{2}}f(x) cos x dx$$



Then, use this identity in proving result (b):



$$int_{0_{+}}^{infty}dfrac{tan x}{x}dx=dfrac{pi}{2}$$



My attempt: I easily solved part (b) as follows



$$int_{0_{+}}^{infty}dfrac{tan x}{x}dx=int_{0_{+}}^{infty}sec x dfrac{sin x}{x}dx$$



Now using identity (a) because $sec(-x)= sec x$ and $sec(x+pi)=-sec x$ we get:
$$int_{0_{+}}^{infty}sec x dfrac{sin x}{x}dx=int_{0_{+}}^{pi/2}sec xcos x dx=dfrac{pi}{2}$$



But I don't know how to prove identity (a) and how can I use identity to evaluate some famous definite integrals.










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  • 2




    How do you even define $int_0^infty frac{tan x}{x},dx$?
    – Zachary
    2 days ago












  • @Zachary What's wrong with that?
    – Andrei
    2 days ago










  • yes $dfrac{tanx}{x} $is undefined at $x=0$
    – deleteprofile
    2 days ago






  • 2




    $f(x)=frac{tan(x)}{x}$ is undefined at $x=frac{pi}{2}+kpi, kinmathbb{Z}$ and $x=0$.
    – Ixion
    2 days ago






  • 4




    Well, it has limit $1$ there, but I think @Zachary was more worried about the infinite areas of alternating sign either side of all the other asymptotes.
    – J.G.
    2 days ago
















5














Question: prove



(a) If $f(x)=f(-x)$ and $f(x+pi)=-f(x)$ then,



$$int_{0_{+}}^{infty} f(x)dfrac{sin x}{x}dx=int_{0_{+}}^{dfrac{pi}{2}}f(x) cos x dx$$



Then, use this identity in proving result (b):



$$int_{0_{+}}^{infty}dfrac{tan x}{x}dx=dfrac{pi}{2}$$



My attempt: I easily solved part (b) as follows



$$int_{0_{+}}^{infty}dfrac{tan x}{x}dx=int_{0_{+}}^{infty}sec x dfrac{sin x}{x}dx$$



Now using identity (a) because $sec(-x)= sec x$ and $sec(x+pi)=-sec x$ we get:
$$int_{0_{+}}^{infty}sec x dfrac{sin x}{x}dx=int_{0_{+}}^{pi/2}sec xcos x dx=dfrac{pi}{2}$$



But I don't know how to prove identity (a) and how can I use identity to evaluate some famous definite integrals.










share|cite|improve this question




















  • 2




    How do you even define $int_0^infty frac{tan x}{x},dx$?
    – Zachary
    2 days ago












  • @Zachary What's wrong with that?
    – Andrei
    2 days ago










  • yes $dfrac{tanx}{x} $is undefined at $x=0$
    – deleteprofile
    2 days ago






  • 2




    $f(x)=frac{tan(x)}{x}$ is undefined at $x=frac{pi}{2}+kpi, kinmathbb{Z}$ and $x=0$.
    – Ixion
    2 days ago






  • 4




    Well, it has limit $1$ there, but I think @Zachary was more worried about the infinite areas of alternating sign either side of all the other asymptotes.
    – J.G.
    2 days ago














5












5








5


5





Question: prove



(a) If $f(x)=f(-x)$ and $f(x+pi)=-f(x)$ then,



$$int_{0_{+}}^{infty} f(x)dfrac{sin x}{x}dx=int_{0_{+}}^{dfrac{pi}{2}}f(x) cos x dx$$



Then, use this identity in proving result (b):



$$int_{0_{+}}^{infty}dfrac{tan x}{x}dx=dfrac{pi}{2}$$



My attempt: I easily solved part (b) as follows



$$int_{0_{+}}^{infty}dfrac{tan x}{x}dx=int_{0_{+}}^{infty}sec x dfrac{sin x}{x}dx$$



Now using identity (a) because $sec(-x)= sec x$ and $sec(x+pi)=-sec x$ we get:
$$int_{0_{+}}^{infty}sec x dfrac{sin x}{x}dx=int_{0_{+}}^{pi/2}sec xcos x dx=dfrac{pi}{2}$$



But I don't know how to prove identity (a) and how can I use identity to evaluate some famous definite integrals.










share|cite|improve this question















Question: prove



(a) If $f(x)=f(-x)$ and $f(x+pi)=-f(x)$ then,



$$int_{0_{+}}^{infty} f(x)dfrac{sin x}{x}dx=int_{0_{+}}^{dfrac{pi}{2}}f(x) cos x dx$$



Then, use this identity in proving result (b):



$$int_{0_{+}}^{infty}dfrac{tan x}{x}dx=dfrac{pi}{2}$$



My attempt: I easily solved part (b) as follows



$$int_{0_{+}}^{infty}dfrac{tan x}{x}dx=int_{0_{+}}^{infty}sec x dfrac{sin x}{x}dx$$



Now using identity (a) because $sec(-x)= sec x$ and $sec(x+pi)=-sec x$ we get:
$$int_{0_{+}}^{infty}sec x dfrac{sin x}{x}dx=int_{0_{+}}^{pi/2}sec xcos x dx=dfrac{pi}{2}$$



But I don't know how to prove identity (a) and how can I use identity to evaluate some famous definite integrals.







real-analysis integration definite-integrals improper-integrals contest-math






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edited yesterday









Mutantoe

582412




582412










asked 2 days ago









deleteprofiledeleteprofile

1,148316




1,148316








  • 2




    How do you even define $int_0^infty frac{tan x}{x},dx$?
    – Zachary
    2 days ago












  • @Zachary What's wrong with that?
    – Andrei
    2 days ago










  • yes $dfrac{tanx}{x} $is undefined at $x=0$
    – deleteprofile
    2 days ago






  • 2




    $f(x)=frac{tan(x)}{x}$ is undefined at $x=frac{pi}{2}+kpi, kinmathbb{Z}$ and $x=0$.
    – Ixion
    2 days ago






  • 4




    Well, it has limit $1$ there, but I think @Zachary was more worried about the infinite areas of alternating sign either side of all the other asymptotes.
    – J.G.
    2 days ago














  • 2




    How do you even define $int_0^infty frac{tan x}{x},dx$?
    – Zachary
    2 days ago












  • @Zachary What's wrong with that?
    – Andrei
    2 days ago










  • yes $dfrac{tanx}{x} $is undefined at $x=0$
    – deleteprofile
    2 days ago






  • 2




    $f(x)=frac{tan(x)}{x}$ is undefined at $x=frac{pi}{2}+kpi, kinmathbb{Z}$ and $x=0$.
    – Ixion
    2 days ago






  • 4




    Well, it has limit $1$ there, but I think @Zachary was more worried about the infinite areas of alternating sign either side of all the other asymptotes.
    – J.G.
    2 days ago








2




2




How do you even define $int_0^infty frac{tan x}{x},dx$?
– Zachary
2 days ago






How do you even define $int_0^infty frac{tan x}{x},dx$?
– Zachary
2 days ago














@Zachary What's wrong with that?
– Andrei
2 days ago




@Zachary What's wrong with that?
– Andrei
2 days ago












yes $dfrac{tanx}{x} $is undefined at $x=0$
– deleteprofile
2 days ago




yes $dfrac{tanx}{x} $is undefined at $x=0$
– deleteprofile
2 days ago




2




2




$f(x)=frac{tan(x)}{x}$ is undefined at $x=frac{pi}{2}+kpi, kinmathbb{Z}$ and $x=0$.
– Ixion
2 days ago




$f(x)=frac{tan(x)}{x}$ is undefined at $x=frac{pi}{2}+kpi, kinmathbb{Z}$ and $x=0$.
– Ixion
2 days ago




4




4




Well, it has limit $1$ there, but I think @Zachary was more worried about the infinite areas of alternating sign either side of all the other asymptotes.
– J.G.
2 days ago




Well, it has limit $1$ there, but I think @Zachary was more worried about the infinite areas of alternating sign either side of all the other asymptotes.
– J.G.
2 days ago










2 Answers
2






active

oldest

votes


















6














The conditions imposed on $f$ imply its Fourier series is a linear combination of the $cos nx$ with $n$ odd, so by linearity we need only check that case. Note for $nne 1$ that $$int_0^inftyfrac{sin xcos nx}{x}dx=int_0^inftyfrac{sin (n+1)x-sin (n-1)x}{2x}dx=0$$ and $$int_0^{pi/2}cos xcos nxdx=frac{1}{2}int_0^{pi/2} (cos(n+1)x+cos(n-1)x)dx=0.$$By contrast, in the $n=1$ case the first integral reduces to $int_0^inftyfrac{sin 2x dx}{2x}=frac{pi}{4}$ and the second to $int_0^{pi/2}cos^2 xdx=frac{pi}{4}$.






share|cite|improve this answer





















  • +1.. ..can we say that since $f(x)$ is an even function and also posses half wave symmetry so, it's fourier series contains only odd harmonics with zero dc component .....i think you used this ...
    – deleteprofile
    2 days ago






  • 1




    @deleteprofile Yes. The second property implies the period is $2pi$, and in the most general series satisfying that said property also eliminates the even-$n$ case, and parity removes sines.
    – J.G.
    2 days ago



















3














HINTS rather than a complete solution



It appears from the question that small issues like whether the integral converges at all, domains of definition, etc., are not your concern. So employing every possible form of what a friend used to call "engineer's prerogative" (i.e., swapping integrals, taking sums inside and outside integrals, rearranging series, etc.), here's a suggestion or two:



We have
$$
f(x)=f(-x)\
f(x+pi)=-f(x)
$$

so the value of $f$ is entirely determined by its values on the interval $[0, pi/2]$. So let's take the big integral and break it into pieces:



begin{align}
int_{0_{+}}^{infty} f(x)dfrac{sin x}{x}~dx
&= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
+ int_{frac{pi}{2}}^pi f(x)dfrac{sin x}{x}~dx
+ int_pi^{frac{3pi}{2}} f(x)dfrac{sin x}{x}~dx + ldots\
&= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
- int_{-frac{pi}{2}}^0 f(x)dfrac{sin x}{(x+pi)}~dx
- int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
&= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
+ int_{frac{pi}{2}}^0 f(x)dfrac{sin (-x)}{(-x+pi)}~dx
- int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
&= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
- int_0^{frac{pi}{2}} f(x)dfrac{sin (-x)}{(-x+pi)}~dx
- int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
&= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
+ int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(-x+pi)}~dx
- int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
end{align}

by subsituting $x = u - pi$ in the 2nd and third integrals of the first line, and then $u = -x$ in the second integral of the second line, and then swapping limits on the integral, and replacing $sin -x$ with $-sin x$. If you keep at this, you end up with something where all terms integrate from $0$ to $pi/2$.



So then you combine them all under one big integral (!), and factor out the $f(x) sin x$ part to get something that looks like
$$
int_0^frac{pi}{2} f(x) sin x sum frac{1}{pm x pm kpi} ~ dx
$$

where you'll now have to fiddle to get the signs and the values for $k$ right, and then observe that the stuff in the sum turns out to be $cot x$. Indeed, maybe it's obvious from some power series that I won't know off the top of my head.



Anyhow, that'll get you started. When you've done all the algebra, you can go back and worry about convergence and whether the stuff you've written down is well-defined or not.






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    2 Answers
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    active

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    2 Answers
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    active

    oldest

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    active

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    active

    oldest

    votes









    6














    The conditions imposed on $f$ imply its Fourier series is a linear combination of the $cos nx$ with $n$ odd, so by linearity we need only check that case. Note for $nne 1$ that $$int_0^inftyfrac{sin xcos nx}{x}dx=int_0^inftyfrac{sin (n+1)x-sin (n-1)x}{2x}dx=0$$ and $$int_0^{pi/2}cos xcos nxdx=frac{1}{2}int_0^{pi/2} (cos(n+1)x+cos(n-1)x)dx=0.$$By contrast, in the $n=1$ case the first integral reduces to $int_0^inftyfrac{sin 2x dx}{2x}=frac{pi}{4}$ and the second to $int_0^{pi/2}cos^2 xdx=frac{pi}{4}$.






    share|cite|improve this answer





















    • +1.. ..can we say that since $f(x)$ is an even function and also posses half wave symmetry so, it's fourier series contains only odd harmonics with zero dc component .....i think you used this ...
      – deleteprofile
      2 days ago






    • 1




      @deleteprofile Yes. The second property implies the period is $2pi$, and in the most general series satisfying that said property also eliminates the even-$n$ case, and parity removes sines.
      – J.G.
      2 days ago
















    6














    The conditions imposed on $f$ imply its Fourier series is a linear combination of the $cos nx$ with $n$ odd, so by linearity we need only check that case. Note for $nne 1$ that $$int_0^inftyfrac{sin xcos nx}{x}dx=int_0^inftyfrac{sin (n+1)x-sin (n-1)x}{2x}dx=0$$ and $$int_0^{pi/2}cos xcos nxdx=frac{1}{2}int_0^{pi/2} (cos(n+1)x+cos(n-1)x)dx=0.$$By contrast, in the $n=1$ case the first integral reduces to $int_0^inftyfrac{sin 2x dx}{2x}=frac{pi}{4}$ and the second to $int_0^{pi/2}cos^2 xdx=frac{pi}{4}$.






    share|cite|improve this answer





















    • +1.. ..can we say that since $f(x)$ is an even function and also posses half wave symmetry so, it's fourier series contains only odd harmonics with zero dc component .....i think you used this ...
      – deleteprofile
      2 days ago






    • 1




      @deleteprofile Yes. The second property implies the period is $2pi$, and in the most general series satisfying that said property also eliminates the even-$n$ case, and parity removes sines.
      – J.G.
      2 days ago














    6












    6








    6






    The conditions imposed on $f$ imply its Fourier series is a linear combination of the $cos nx$ with $n$ odd, so by linearity we need only check that case. Note for $nne 1$ that $$int_0^inftyfrac{sin xcos nx}{x}dx=int_0^inftyfrac{sin (n+1)x-sin (n-1)x}{2x}dx=0$$ and $$int_0^{pi/2}cos xcos nxdx=frac{1}{2}int_0^{pi/2} (cos(n+1)x+cos(n-1)x)dx=0.$$By contrast, in the $n=1$ case the first integral reduces to $int_0^inftyfrac{sin 2x dx}{2x}=frac{pi}{4}$ and the second to $int_0^{pi/2}cos^2 xdx=frac{pi}{4}$.






    share|cite|improve this answer












    The conditions imposed on $f$ imply its Fourier series is a linear combination of the $cos nx$ with $n$ odd, so by linearity we need only check that case. Note for $nne 1$ that $$int_0^inftyfrac{sin xcos nx}{x}dx=int_0^inftyfrac{sin (n+1)x-sin (n-1)x}{2x}dx=0$$ and $$int_0^{pi/2}cos xcos nxdx=frac{1}{2}int_0^{pi/2} (cos(n+1)x+cos(n-1)x)dx=0.$$By contrast, in the $n=1$ case the first integral reduces to $int_0^inftyfrac{sin 2x dx}{2x}=frac{pi}{4}$ and the second to $int_0^{pi/2}cos^2 xdx=frac{pi}{4}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 days ago









    J.G.J.G.

    23.4k22237




    23.4k22237












    • +1.. ..can we say that since $f(x)$ is an even function and also posses half wave symmetry so, it's fourier series contains only odd harmonics with zero dc component .....i think you used this ...
      – deleteprofile
      2 days ago






    • 1




      @deleteprofile Yes. The second property implies the period is $2pi$, and in the most general series satisfying that said property also eliminates the even-$n$ case, and parity removes sines.
      – J.G.
      2 days ago


















    • +1.. ..can we say that since $f(x)$ is an even function and also posses half wave symmetry so, it's fourier series contains only odd harmonics with zero dc component .....i think you used this ...
      – deleteprofile
      2 days ago






    • 1




      @deleteprofile Yes. The second property implies the period is $2pi$, and in the most general series satisfying that said property also eliminates the even-$n$ case, and parity removes sines.
      – J.G.
      2 days ago
















    +1.. ..can we say that since $f(x)$ is an even function and also posses half wave symmetry so, it's fourier series contains only odd harmonics with zero dc component .....i think you used this ...
    – deleteprofile
    2 days ago




    +1.. ..can we say that since $f(x)$ is an even function and also posses half wave symmetry so, it's fourier series contains only odd harmonics with zero dc component .....i think you used this ...
    – deleteprofile
    2 days ago




    1




    1




    @deleteprofile Yes. The second property implies the period is $2pi$, and in the most general series satisfying that said property also eliminates the even-$n$ case, and parity removes sines.
    – J.G.
    2 days ago




    @deleteprofile Yes. The second property implies the period is $2pi$, and in the most general series satisfying that said property also eliminates the even-$n$ case, and parity removes sines.
    – J.G.
    2 days ago











    3














    HINTS rather than a complete solution



    It appears from the question that small issues like whether the integral converges at all, domains of definition, etc., are not your concern. So employing every possible form of what a friend used to call "engineer's prerogative" (i.e., swapping integrals, taking sums inside and outside integrals, rearranging series, etc.), here's a suggestion or two:



    We have
    $$
    f(x)=f(-x)\
    f(x+pi)=-f(x)
    $$

    so the value of $f$ is entirely determined by its values on the interval $[0, pi/2]$. So let's take the big integral and break it into pieces:



    begin{align}
    int_{0_{+}}^{infty} f(x)dfrac{sin x}{x}~dx
    &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
    + int_{frac{pi}{2}}^pi f(x)dfrac{sin x}{x}~dx
    + int_pi^{frac{3pi}{2}} f(x)dfrac{sin x}{x}~dx + ldots\
    &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
    - int_{-frac{pi}{2}}^0 f(x)dfrac{sin x}{(x+pi)}~dx
    - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
    &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
    + int_{frac{pi}{2}}^0 f(x)dfrac{sin (-x)}{(-x+pi)}~dx
    - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
    &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
    - int_0^{frac{pi}{2}} f(x)dfrac{sin (-x)}{(-x+pi)}~dx
    - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
    &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
    + int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(-x+pi)}~dx
    - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
    end{align}

    by subsituting $x = u - pi$ in the 2nd and third integrals of the first line, and then $u = -x$ in the second integral of the second line, and then swapping limits on the integral, and replacing $sin -x$ with $-sin x$. If you keep at this, you end up with something where all terms integrate from $0$ to $pi/2$.



    So then you combine them all under one big integral (!), and factor out the $f(x) sin x$ part to get something that looks like
    $$
    int_0^frac{pi}{2} f(x) sin x sum frac{1}{pm x pm kpi} ~ dx
    $$

    where you'll now have to fiddle to get the signs and the values for $k$ right, and then observe that the stuff in the sum turns out to be $cot x$. Indeed, maybe it's obvious from some power series that I won't know off the top of my head.



    Anyhow, that'll get you started. When you've done all the algebra, you can go back and worry about convergence and whether the stuff you've written down is well-defined or not.






    share|cite|improve this answer


























      3














      HINTS rather than a complete solution



      It appears from the question that small issues like whether the integral converges at all, domains of definition, etc., are not your concern. So employing every possible form of what a friend used to call "engineer's prerogative" (i.e., swapping integrals, taking sums inside and outside integrals, rearranging series, etc.), here's a suggestion or two:



      We have
      $$
      f(x)=f(-x)\
      f(x+pi)=-f(x)
      $$

      so the value of $f$ is entirely determined by its values on the interval $[0, pi/2]$. So let's take the big integral and break it into pieces:



      begin{align}
      int_{0_{+}}^{infty} f(x)dfrac{sin x}{x}~dx
      &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
      + int_{frac{pi}{2}}^pi f(x)dfrac{sin x}{x}~dx
      + int_pi^{frac{3pi}{2}} f(x)dfrac{sin x}{x}~dx + ldots\
      &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
      - int_{-frac{pi}{2}}^0 f(x)dfrac{sin x}{(x+pi)}~dx
      - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
      &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
      + int_{frac{pi}{2}}^0 f(x)dfrac{sin (-x)}{(-x+pi)}~dx
      - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
      &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
      - int_0^{frac{pi}{2}} f(x)dfrac{sin (-x)}{(-x+pi)}~dx
      - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
      &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
      + int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(-x+pi)}~dx
      - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
      end{align}

      by subsituting $x = u - pi$ in the 2nd and third integrals of the first line, and then $u = -x$ in the second integral of the second line, and then swapping limits on the integral, and replacing $sin -x$ with $-sin x$. If you keep at this, you end up with something where all terms integrate from $0$ to $pi/2$.



      So then you combine them all under one big integral (!), and factor out the $f(x) sin x$ part to get something that looks like
      $$
      int_0^frac{pi}{2} f(x) sin x sum frac{1}{pm x pm kpi} ~ dx
      $$

      where you'll now have to fiddle to get the signs and the values for $k$ right, and then observe that the stuff in the sum turns out to be $cot x$. Indeed, maybe it's obvious from some power series that I won't know off the top of my head.



      Anyhow, that'll get you started. When you've done all the algebra, you can go back and worry about convergence and whether the stuff you've written down is well-defined or not.






      share|cite|improve this answer
























        3












        3








        3






        HINTS rather than a complete solution



        It appears from the question that small issues like whether the integral converges at all, domains of definition, etc., are not your concern. So employing every possible form of what a friend used to call "engineer's prerogative" (i.e., swapping integrals, taking sums inside and outside integrals, rearranging series, etc.), here's a suggestion or two:



        We have
        $$
        f(x)=f(-x)\
        f(x+pi)=-f(x)
        $$

        so the value of $f$ is entirely determined by its values on the interval $[0, pi/2]$. So let's take the big integral and break it into pieces:



        begin{align}
        int_{0_{+}}^{infty} f(x)dfrac{sin x}{x}~dx
        &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
        + int_{frac{pi}{2}}^pi f(x)dfrac{sin x}{x}~dx
        + int_pi^{frac{3pi}{2}} f(x)dfrac{sin x}{x}~dx + ldots\
        &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
        - int_{-frac{pi}{2}}^0 f(x)dfrac{sin x}{(x+pi)}~dx
        - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
        &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
        + int_{frac{pi}{2}}^0 f(x)dfrac{sin (-x)}{(-x+pi)}~dx
        - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
        &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
        - int_0^{frac{pi}{2}} f(x)dfrac{sin (-x)}{(-x+pi)}~dx
        - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
        &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
        + int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(-x+pi)}~dx
        - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
        end{align}

        by subsituting $x = u - pi$ in the 2nd and third integrals of the first line, and then $u = -x$ in the second integral of the second line, and then swapping limits on the integral, and replacing $sin -x$ with $-sin x$. If you keep at this, you end up with something where all terms integrate from $0$ to $pi/2$.



        So then you combine them all under one big integral (!), and factor out the $f(x) sin x$ part to get something that looks like
        $$
        int_0^frac{pi}{2} f(x) sin x sum frac{1}{pm x pm kpi} ~ dx
        $$

        where you'll now have to fiddle to get the signs and the values for $k$ right, and then observe that the stuff in the sum turns out to be $cot x$. Indeed, maybe it's obvious from some power series that I won't know off the top of my head.



        Anyhow, that'll get you started. When you've done all the algebra, you can go back and worry about convergence and whether the stuff you've written down is well-defined or not.






        share|cite|improve this answer












        HINTS rather than a complete solution



        It appears from the question that small issues like whether the integral converges at all, domains of definition, etc., are not your concern. So employing every possible form of what a friend used to call "engineer's prerogative" (i.e., swapping integrals, taking sums inside and outside integrals, rearranging series, etc.), here's a suggestion or two:



        We have
        $$
        f(x)=f(-x)\
        f(x+pi)=-f(x)
        $$

        so the value of $f$ is entirely determined by its values on the interval $[0, pi/2]$. So let's take the big integral and break it into pieces:



        begin{align}
        int_{0_{+}}^{infty} f(x)dfrac{sin x}{x}~dx
        &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
        + int_{frac{pi}{2}}^pi f(x)dfrac{sin x}{x}~dx
        + int_pi^{frac{3pi}{2}} f(x)dfrac{sin x}{x}~dx + ldots\
        &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
        - int_{-frac{pi}{2}}^0 f(x)dfrac{sin x}{(x+pi)}~dx
        - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
        &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
        + int_{frac{pi}{2}}^0 f(x)dfrac{sin (-x)}{(-x+pi)}~dx
        - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
        &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
        - int_0^{frac{pi}{2}} f(x)dfrac{sin (-x)}{(-x+pi)}~dx
        - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
        &= int_{0}^{frac{pi}{2}} f(x)dfrac{sin x}{x}~dx
        + int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(-x+pi)}~dx
        - int_0^{frac{pi}{2}} f(x)dfrac{sin x}{(x+pi)}~dx + ldots\
        end{align}

        by subsituting $x = u - pi$ in the 2nd and third integrals of the first line, and then $u = -x$ in the second integral of the second line, and then swapping limits on the integral, and replacing $sin -x$ with $-sin x$. If you keep at this, you end up with something where all terms integrate from $0$ to $pi/2$.



        So then you combine them all under one big integral (!), and factor out the $f(x) sin x$ part to get something that looks like
        $$
        int_0^frac{pi}{2} f(x) sin x sum frac{1}{pm x pm kpi} ~ dx
        $$

        where you'll now have to fiddle to get the signs and the values for $k$ right, and then observe that the stuff in the sum turns out to be $cot x$. Indeed, maybe it's obvious from some power series that I won't know off the top of my head.



        Anyhow, that'll get you started. When you've done all the algebra, you can go back and worry about convergence and whether the stuff you've written down is well-defined or not.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        John HughesJohn Hughes

        62.6k24090




        62.6k24090






























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