finding $int frac{(1-x^2)^{n-1}}{(1+x^2)^n}mathrm{d}x$












3














While trying to evaluate a different integral, I met the following integral:
$$I_n=intfrac{(1-x^2)^{n-1}}{(1+x^2)^n}mathrm{d}x$$
I was lost as to what to do with it, so I looked it up on wolfram alpha, and I got a result in terms of The Appell Hypergeometric Function $F_1$:
$$I_n=xF_1bigg(frac12;1-n,n;frac32;x^2,-x^2bigg)+C$$
Where, for $|x|<1$, and $|y|<1$,
$$F_1(a;b_1,b_2;c;x,y)=sum_{mgeq0}sum_{kgeq0}frac{(a)_{m+k}(b_1)_m(b_2)_k}{m!k!(c)_{m+k}}x^my^k$$
With $$(x)_k=frac{Gamma(x+k)}{Gamma(x)}$$
I found on Wikipedia that
$$F_1(a;b1,b2;c;x,y)=\sum_{kgeq0}frac{(a)_k(b_1)_k(b_2)_k(c-a)_k}{(c+k-1)_k(c)_{2k}}frac{x^ky^k}{k!},_2F_1(a+k,b_1+k;c+2k;x),_2F_1(a+k,b_2+k;c+2k;y)$$
Which I thought might be useful.



I also know that
$$,_1F_0(1-n;;x^2)=(1-x^2)^{n-1}$$
Which gives $$I_n=sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{x^{2m}}{(1+x^2)^n}mathrm{d}x$$
One could preform the substitution $u=1+x^2$:
$$I_n=frac12sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{(u-1)^{m-1/2}}{u^n}mathrm{d}u$$
But this doesn't look any better.



Could anyone provide an alternative solution or give a proof of the result given by Wolfram?






integration special-functions







share|cite|improve this question



























    3














    While trying to evaluate a different integral, I met the following integral:
    $$I_n=intfrac{(1-x^2)^{n-1}}{(1+x^2)^n}mathrm{d}x$$
    I was lost as to what to do with it, so I looked it up on wolfram alpha, and I got a result in terms of The Appell Hypergeometric Function $F_1$:
    $$I_n=xF_1bigg(frac12;1-n,n;frac32;x^2,-x^2bigg)+C$$
    Where, for $|x|<1$, and $|y|<1$,
    $$F_1(a;b_1,b_2;c;x,y)=sum_{mgeq0}sum_{kgeq0}frac{(a)_{m+k}(b_1)_m(b_2)_k}{m!k!(c)_{m+k}}x^my^k$$
    With $$(x)_k=frac{Gamma(x+k)}{Gamma(x)}$$
    I found on Wikipedia that
    $$F_1(a;b1,b2;c;x,y)=\sum_{kgeq0}frac{(a)_k(b_1)_k(b_2)_k(c-a)_k}{(c+k-1)_k(c)_{2k}}frac{x^ky^k}{k!},_2F_1(a+k,b_1+k;c+2k;x),_2F_1(a+k,b_2+k;c+2k;y)$$
    Which I thought might be useful.



    I also know that
    $$,_1F_0(1-n;;x^2)=(1-x^2)^{n-1}$$
    Which gives $$I_n=sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{x^{2m}}{(1+x^2)^n}mathrm{d}x$$
    One could preform the substitution $u=1+x^2$:
    $$I_n=frac12sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{(u-1)^{m-1/2}}{u^n}mathrm{d}u$$
    But this doesn't look any better.



    Could anyone provide an alternative solution or give a proof of the result given by Wolfram?






    integration special-functions







    share|cite|improve this question

























      3












      3








      3


      1





      While trying to evaluate a different integral, I met the following integral:
      $$I_n=intfrac{(1-x^2)^{n-1}}{(1+x^2)^n}mathrm{d}x$$
      I was lost as to what to do with it, so I looked it up on wolfram alpha, and I got a result in terms of The Appell Hypergeometric Function $F_1$:
      $$I_n=xF_1bigg(frac12;1-n,n;frac32;x^2,-x^2bigg)+C$$
      Where, for $|x|<1$, and $|y|<1$,
      $$F_1(a;b_1,b_2;c;x,y)=sum_{mgeq0}sum_{kgeq0}frac{(a)_{m+k}(b_1)_m(b_2)_k}{m!k!(c)_{m+k}}x^my^k$$
      With $$(x)_k=frac{Gamma(x+k)}{Gamma(x)}$$
      I found on Wikipedia that
      $$F_1(a;b1,b2;c;x,y)=\sum_{kgeq0}frac{(a)_k(b_1)_k(b_2)_k(c-a)_k}{(c+k-1)_k(c)_{2k}}frac{x^ky^k}{k!},_2F_1(a+k,b_1+k;c+2k;x),_2F_1(a+k,b_2+k;c+2k;y)$$
      Which I thought might be useful.



      I also know that
      $$,_1F_0(1-n;;x^2)=(1-x^2)^{n-1}$$
      Which gives $$I_n=sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{x^{2m}}{(1+x^2)^n}mathrm{d}x$$
      One could preform the substitution $u=1+x^2$:
      $$I_n=frac12sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{(u-1)^{m-1/2}}{u^n}mathrm{d}u$$
      But this doesn't look any better.



      Could anyone provide an alternative solution or give a proof of the result given by Wolfram?






      integration special-functions







      share|cite|improve this question













      While trying to evaluate a different integral, I met the following integral:
      $$I_n=intfrac{(1-x^2)^{n-1}}{(1+x^2)^n}mathrm{d}x$$
      I was lost as to what to do with it, so I looked it up on wolfram alpha, and I got a result in terms of The Appell Hypergeometric Function $F_1$:
      $$I_n=xF_1bigg(frac12;1-n,n;frac32;x^2,-x^2bigg)+C$$
      Where, for $|x|<1$, and $|y|<1$,
      $$F_1(a;b_1,b_2;c;x,y)=sum_{mgeq0}sum_{kgeq0}frac{(a)_{m+k}(b_1)_m(b_2)_k}{m!k!(c)_{m+k}}x^my^k$$
      With $$(x)_k=frac{Gamma(x+k)}{Gamma(x)}$$
      I found on Wikipedia that
      $$F_1(a;b1,b2;c;x,y)=\sum_{kgeq0}frac{(a)_k(b_1)_k(b_2)_k(c-a)_k}{(c+k-1)_k(c)_{2k}}frac{x^ky^k}{k!},_2F_1(a+k,b_1+k;c+2k;x),_2F_1(a+k,b_2+k;c+2k;y)$$
      Which I thought might be useful.



      I also know that
      $$,_1F_0(1-n;;x^2)=(1-x^2)^{n-1}$$
      Which gives $$I_n=sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{x^{2m}}{(1+x^2)^n}mathrm{d}x$$
      One could preform the substitution $u=1+x^2$:
      $$I_n=frac12sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{(u-1)^{m-1/2}}{u^n}mathrm{d}u$$
      But this doesn't look any better.



      Could anyone provide an alternative solution or give a proof of the result given by Wolfram?






      integration special-functions




      integration special-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 25 at 5:52









      clathratus

      2,862328




      2,862328






















          1 Answer
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          5














          Hint:
          With substitution $x=tan t$ the integral is
          $$I_n=int cos^{n-1}2t dt$$
          and it can be solved with integration by parts and recursive formula.






          share|cite|improve this answer





















          • I feel so stupid right now. Thanks (+1)
            – clathratus
            Nov 25 at 5:58











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          5














          Hint:
          With substitution $x=tan t$ the integral is
          $$I_n=int cos^{n-1}2t dt$$
          and it can be solved with integration by parts and recursive formula.






          share|cite|improve this answer





















          • I feel so stupid right now. Thanks (+1)
            – clathratus
            Nov 25 at 5:58
















          5














          Hint:
          With substitution $x=tan t$ the integral is
          $$I_n=int cos^{n-1}2t dt$$
          and it can be solved with integration by parts and recursive formula.






          share|cite|improve this answer





















          • I feel so stupid right now. Thanks (+1)
            – clathratus
            Nov 25 at 5:58














          5












          5








          5






          Hint:
          With substitution $x=tan t$ the integral is
          $$I_n=int cos^{n-1}2t dt$$
          and it can be solved with integration by parts and recursive formula.






          share|cite|improve this answer












          Hint:
          With substitution $x=tan t$ the integral is
          $$I_n=int cos^{n-1}2t dt$$
          and it can be solved with integration by parts and recursive formula.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 5:56









          Nosrati

          26.4k62353




          26.4k62353












          • I feel so stupid right now. Thanks (+1)
            – clathratus
            Nov 25 at 5:58


















          • I feel so stupid right now. Thanks (+1)
            – clathratus
            Nov 25 at 5:58
















          I feel so stupid right now. Thanks (+1)
          – clathratus
          Nov 25 at 5:58




          I feel so stupid right now. Thanks (+1)
          – clathratus
          Nov 25 at 5:58


















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