Integrate $frac{dx}{x^{2n}+1}$ for $n in mathbb{N}$ from $-infty$ to $infty$ [duplicate]












0












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This question already has an answer here:




  • Prove $int_0^{infty}! frac{mathbb{d}x}{1+x^n}=frac{pi}{n sinfrac{pi}{n}}$ using real analysis techniques only

    4 answers




How can one calculate



$$
int_{- infty}^{infty} frac{dx}{x^{2n} + 1}, ;n in mathbb{N}
$$



without using complex plane and Residue theorem?






real-analysis calculus integration definite-integrals







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marked as duplicate by Jean Marie, mrtaurho, RRL real-analysis
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 Jan 31 at 3:00


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    See here: math.stackexchange.com/q/110457/515527
    $endgroup$
    – Zacky
    Jan 30 at 19:39


















0












$begingroup$



This question already has an answer here:




  • Prove $int_0^{infty}! frac{mathbb{d}x}{1+x^n}=frac{pi}{n sinfrac{pi}{n}}$ using real analysis techniques only

    4 answers




How can one calculate



$$
int_{- infty}^{infty} frac{dx}{x^{2n} + 1}, ;n in mathbb{N}
$$



without using complex plane and Residue theorem?






real-analysis calculus integration definite-integrals







share|cite|improve this question











$endgroup$



marked as duplicate by Jean Marie, mrtaurho, RRL real-analysis
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 Jan 31 at 3:00


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    See here: math.stackexchange.com/q/110457/515527
    $endgroup$
    – Zacky
    Jan 30 at 19:39
















0












0








0





$begingroup$



This question already has an answer here:




  • Prove $int_0^{infty}! frac{mathbb{d}x}{1+x^n}=frac{pi}{n sinfrac{pi}{n}}$ using real analysis techniques only

    4 answers




How can one calculate



$$
int_{- infty}^{infty} frac{dx}{x^{2n} + 1}, ;n in mathbb{N}
$$



without using complex plane and Residue theorem?






real-analysis calculus integration definite-integrals







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Prove $int_0^{infty}! frac{mathbb{d}x}{1+x^n}=frac{pi}{n sinfrac{pi}{n}}$ using real analysis techniques only

    4 answers




How can one calculate



$$
int_{- infty}^{infty} frac{dx}{x^{2n} + 1}, ;n in mathbb{N}
$$



without using complex plane and Residue theorem?





This question already has an answer here:




  • Prove $int_0^{infty}! frac{mathbb{d}x}{1+x^n}=frac{pi}{n sinfrac{pi}{n}}$ using real analysis techniques only

    4 answers







real-analysis calculus integration definite-integrals




real-analysis calculus integration definite-integrals






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share|cite|improve this question













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edited Jan 30 at 19:41









Adrian Keister

5,26971933




5,26971933










asked Jan 30 at 19:38









NikromNikrom

1376




1376




marked as duplicate by Jean Marie, mrtaurho, RRL real-analysis
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 Jan 31 at 3:00


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 Jan 31 at 3:00


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    See here: math.stackexchange.com/q/110457/515527
    $endgroup$
    – Zacky
    Jan 30 at 19:39
















  • 1




    $begingroup$
    See here: math.stackexchange.com/q/110457/515527
    $endgroup$
    – Zacky
    Jan 30 at 19:39










1




1




$begingroup$
See here: math.stackexchange.com/q/110457/515527
$endgroup$
– Zacky
Jan 30 at 19:39






$begingroup$
See here: math.stackexchange.com/q/110457/515527
$endgroup$
– Zacky
Jan 30 at 19:39












2 Answers
2






active

oldest

votes


















1












$begingroup$

With $x=tan^{1/n} t$ and some well-known properties of the Gamma function, you can prove the answer is $2/operatorname{sinc}frac{pi}{2n}$. The case $n=1$ is an easy sanity check.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Consider the integral
    $$F(a,b,c)=int_0^inftyfrac{t^a}{(t^b+1)^c}mathrm dt$$
    Set $x^{1/b}=t$:
    $$F(a,b,c)=frac1bint_0^inftyfrac{x^{
    frac{a+1}b}}{(x+1)^c}mathrm dx$$
    Then $u=frac1{x+1}$ gives
    $$begin{align}
    F(a,b,c)&=frac1bint_0^1 left[frac{1-u}{u}right]^{
    frac{a+1}b}u^cfrac{mathrm du}{u^2}\
    &=frac1bint_0^1u^{c-2-frac{a+1}{b}}(1-u)^{frac{a+1}b}mathrm du
    end{align}$$
    Then recall the definition of the Beta function:
    $$mathrm{B}(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm dt=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
    Where $Gamma(s)$ is the gamma function. So of course we have
    $$F(a,b,c)=frac1bmathrm{B}left(c-1-frac{a+1}b,1+frac{a+1}bright)$$
    We then have your integral as
    $$int_0^infty frac{mathrm dt}{t^{2n}+1}=F(0,2n,1)$$






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      With $x=tan^{1/n} t$ and some well-known properties of the Gamma function, you can prove the answer is $2/operatorname{sinc}frac{pi}{2n}$. The case $n=1$ is an easy sanity check.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        With $x=tan^{1/n} t$ and some well-known properties of the Gamma function, you can prove the answer is $2/operatorname{sinc}frac{pi}{2n}$. The case $n=1$ is an easy sanity check.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          With $x=tan^{1/n} t$ and some well-known properties of the Gamma function, you can prove the answer is $2/operatorname{sinc}frac{pi}{2n}$. The case $n=1$ is an easy sanity check.






          share|cite|improve this answer









          $endgroup$



          With $x=tan^{1/n} t$ and some well-known properties of the Gamma function, you can prove the answer is $2/operatorname{sinc}frac{pi}{2n}$. The case $n=1$ is an easy sanity check.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 30 at 19:40









          J.G.J.G.

          32.6k23250




          32.6k23250























              0












              $begingroup$

              Consider the integral
              $$F(a,b,c)=int_0^inftyfrac{t^a}{(t^b+1)^c}mathrm dt$$
              Set $x^{1/b}=t$:
              $$F(a,b,c)=frac1bint_0^inftyfrac{x^{
              frac{a+1}b}}{(x+1)^c}mathrm dx$$
              Then $u=frac1{x+1}$ gives
              $$begin{align}
              F(a,b,c)&=frac1bint_0^1 left[frac{1-u}{u}right]^{
              frac{a+1}b}u^cfrac{mathrm du}{u^2}\
              &=frac1bint_0^1u^{c-2-frac{a+1}{b}}(1-u)^{frac{a+1}b}mathrm du
              end{align}$$
              Then recall the definition of the Beta function:
              $$mathrm{B}(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm dt=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
              Where $Gamma(s)$ is the gamma function. So of course we have
              $$F(a,b,c)=frac1bmathrm{B}left(c-1-frac{a+1}b,1+frac{a+1}bright)$$
              We then have your integral as
              $$int_0^infty frac{mathrm dt}{t^{2n}+1}=F(0,2n,1)$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Consider the integral
                $$F(a,b,c)=int_0^inftyfrac{t^a}{(t^b+1)^c}mathrm dt$$
                Set $x^{1/b}=t$:
                $$F(a,b,c)=frac1bint_0^inftyfrac{x^{
                frac{a+1}b}}{(x+1)^c}mathrm dx$$
                Then $u=frac1{x+1}$ gives
                $$begin{align}
                F(a,b,c)&=frac1bint_0^1 left[frac{1-u}{u}right]^{
                frac{a+1}b}u^cfrac{mathrm du}{u^2}\
                &=frac1bint_0^1u^{c-2-frac{a+1}{b}}(1-u)^{frac{a+1}b}mathrm du
                end{align}$$
                Then recall the definition of the Beta function:
                $$mathrm{B}(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm dt=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
                Where $Gamma(s)$ is the gamma function. So of course we have
                $$F(a,b,c)=frac1bmathrm{B}left(c-1-frac{a+1}b,1+frac{a+1}bright)$$
                We then have your integral as
                $$int_0^infty frac{mathrm dt}{t^{2n}+1}=F(0,2n,1)$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Consider the integral
                  $$F(a,b,c)=int_0^inftyfrac{t^a}{(t^b+1)^c}mathrm dt$$
                  Set $x^{1/b}=t$:
                  $$F(a,b,c)=frac1bint_0^inftyfrac{x^{
                  frac{a+1}b}}{(x+1)^c}mathrm dx$$
                  Then $u=frac1{x+1}$ gives
                  $$begin{align}
                  F(a,b,c)&=frac1bint_0^1 left[frac{1-u}{u}right]^{
                  frac{a+1}b}u^cfrac{mathrm du}{u^2}\
                  &=frac1bint_0^1u^{c-2-frac{a+1}{b}}(1-u)^{frac{a+1}b}mathrm du
                  end{align}$$
                  Then recall the definition of the Beta function:
                  $$mathrm{B}(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm dt=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
                  Where $Gamma(s)$ is the gamma function. So of course we have
                  $$F(a,b,c)=frac1bmathrm{B}left(c-1-frac{a+1}b,1+frac{a+1}bright)$$
                  We then have your integral as
                  $$int_0^infty frac{mathrm dt}{t^{2n}+1}=F(0,2n,1)$$






                  share|cite|improve this answer









                  $endgroup$



                  Consider the integral
                  $$F(a,b,c)=int_0^inftyfrac{t^a}{(t^b+1)^c}mathrm dt$$
                  Set $x^{1/b}=t$:
                  $$F(a,b,c)=frac1bint_0^inftyfrac{x^{
                  frac{a+1}b}}{(x+1)^c}mathrm dx$$
                  Then $u=frac1{x+1}$ gives
                  $$begin{align}
                  F(a,b,c)&=frac1bint_0^1 left[frac{1-u}{u}right]^{
                  frac{a+1}b}u^cfrac{mathrm du}{u^2}\
                  &=frac1bint_0^1u^{c-2-frac{a+1}{b}}(1-u)^{frac{a+1}b}mathrm du
                  end{align}$$
                  Then recall the definition of the Beta function:
                  $$mathrm{B}(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm dt=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
                  Where $Gamma(s)$ is the gamma function. So of course we have
                  $$F(a,b,c)=frac1bmathrm{B}left(c-1-frac{a+1}b,1+frac{a+1}bright)$$
                  We then have your integral as
                  $$int_0^infty frac{mathrm dt}{t^{2n}+1}=F(0,2n,1)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 31 at 4:23









                  clathratusclathratus

                  5,0891439




                  5,0891439















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