(Self-Answered) A Nice Substitution for an Improper Integral












3












$begingroup$


Evaluate



$$
int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx , , , (phi>0)
$$






calculus improper-integrals







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$endgroup$

















    3












    $begingroup$


    Evaluate



    $$
    int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx , , , (phi>0)
    $$






    calculus improper-integrals







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      2



      $begingroup$


      Evaluate



      $$
      int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx , , , (phi>0)
      $$






      calculus improper-integrals







      share|cite|improve this question









      $endgroup$




      Evaluate



      $$
      int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx , , , (phi>0)
      $$






      calculus improper-integrals




      calculus improper-integrals






      share|cite|improve this question













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      asked Jan 14 at 12:41









      NetUser5y62NetUser5y62

      525215




      525215






















          2 Answers
          2






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          6












          $begingroup$

          Solution



          $$ small
          int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx = int_0^{1} frac{1}{(1+x^2)(1+x^{phi})}, dx + int_1^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx
          $$



          Notice that with the substitution $y=frac{1}{x}$,
          $$begin{aligned}
          int_1^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx &= int_1^0 frac{1}{left( 1+frac{1}{y^2}right)left(1+frac{1}{y^{phi}} right)}cdot left(frac{-1}{y^2} right), dy\
          &=int_0^1frac{y^{phi}}{(1+y^2)(1+y^{phi})}, dy
          end{aligned}$$



          Therefore



          $$begin{aligned}
          int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx &= int_0^1frac{1+x^{phi}}{(1+x^2)(1+x^{phi})}, dx \ &= int_0^1 frac{1}{1+x^2}, dx \ &= arctan x|_0^1 \ &= frac{pi}{4}
          end{aligned}$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Another way would be to make the substitution $xmapstotan x$ so that$$begin{align*}mathfrak{I} & =intlimits_0^{pi/2}frac {mathrm dx}{1+tan^nx}\ & =intlimits_0^{pi/2}mathrm dx-intlimits_0^{pi/2}mathrm dx,frac {tan^n x}{1+tan^nx}\ & =frac {pi}2-intlimits_0^{pi/2}mathrm dx,frac 1{1+cot^nx}end{align*}$$Now use the identity that$$cot x=tanleft(frac {pi}2-xright)$$to get that$$begin{align*}mathfrak{I} & =frac {pi}2-intlimits_0^{pi/2}frac {mathrm dx}{1+tan^nx}\ & =frac {pi}2-mathfrak{I}end{align*}$$



            Now isolate $mathfrak{I}$ to get that$$intlimits_0^{infty}frac {mathrm dx}{(1+x^2)(1+x^n)}color{blue}{=frac {pi}4}$$






            share|cite|improve this answer









            $endgroup$













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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

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              active

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              active

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              6












              $begingroup$

              Solution



              $$ small
              int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx = int_0^{1} frac{1}{(1+x^2)(1+x^{phi})}, dx + int_1^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx
              $$



              Notice that with the substitution $y=frac{1}{x}$,
              $$begin{aligned}
              int_1^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx &= int_1^0 frac{1}{left( 1+frac{1}{y^2}right)left(1+frac{1}{y^{phi}} right)}cdot left(frac{-1}{y^2} right), dy\
              &=int_0^1frac{y^{phi}}{(1+y^2)(1+y^{phi})}, dy
              end{aligned}$$



              Therefore



              $$begin{aligned}
              int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx &= int_0^1frac{1+x^{phi}}{(1+x^2)(1+x^{phi})}, dx \ &= int_0^1 frac{1}{1+x^2}, dx \ &= arctan x|_0^1 \ &= frac{pi}{4}
              end{aligned}$$






              share|cite|improve this answer









              $endgroup$


















                6












                $begingroup$

                Solution



                $$ small
                int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx = int_0^{1} frac{1}{(1+x^2)(1+x^{phi})}, dx + int_1^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx
                $$



                Notice that with the substitution $y=frac{1}{x}$,
                $$begin{aligned}
                int_1^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx &= int_1^0 frac{1}{left( 1+frac{1}{y^2}right)left(1+frac{1}{y^{phi}} right)}cdot left(frac{-1}{y^2} right), dy\
                &=int_0^1frac{y^{phi}}{(1+y^2)(1+y^{phi})}, dy
                end{aligned}$$



                Therefore



                $$begin{aligned}
                int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx &= int_0^1frac{1+x^{phi}}{(1+x^2)(1+x^{phi})}, dx \ &= int_0^1 frac{1}{1+x^2}, dx \ &= arctan x|_0^1 \ &= frac{pi}{4}
                end{aligned}$$






                share|cite|improve this answer









                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  Solution



                  $$ small
                  int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx = int_0^{1} frac{1}{(1+x^2)(1+x^{phi})}, dx + int_1^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx
                  $$



                  Notice that with the substitution $y=frac{1}{x}$,
                  $$begin{aligned}
                  int_1^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx &= int_1^0 frac{1}{left( 1+frac{1}{y^2}right)left(1+frac{1}{y^{phi}} right)}cdot left(frac{-1}{y^2} right), dy\
                  &=int_0^1frac{y^{phi}}{(1+y^2)(1+y^{phi})}, dy
                  end{aligned}$$



                  Therefore



                  $$begin{aligned}
                  int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx &= int_0^1frac{1+x^{phi}}{(1+x^2)(1+x^{phi})}, dx \ &= int_0^1 frac{1}{1+x^2}, dx \ &= arctan x|_0^1 \ &= frac{pi}{4}
                  end{aligned}$$






                  share|cite|improve this answer









                  $endgroup$



                  Solution



                  $$ small
                  int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx = int_0^{1} frac{1}{(1+x^2)(1+x^{phi})}, dx + int_1^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx
                  $$



                  Notice that with the substitution $y=frac{1}{x}$,
                  $$begin{aligned}
                  int_1^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx &= int_1^0 frac{1}{left( 1+frac{1}{y^2}right)left(1+frac{1}{y^{phi}} right)}cdot left(frac{-1}{y^2} right), dy\
                  &=int_0^1frac{y^{phi}}{(1+y^2)(1+y^{phi})}, dy
                  end{aligned}$$



                  Therefore



                  $$begin{aligned}
                  int_0^{+infty} frac{1}{(1+x^2)(1+x^{phi})}, dx &= int_0^1frac{1+x^{phi}}{(1+x^2)(1+x^{phi})}, dx \ &= int_0^1 frac{1}{1+x^2}, dx \ &= arctan x|_0^1 \ &= frac{pi}{4}
                  end{aligned}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 14 at 12:41









                  NetUser5y62NetUser5y62

                  525215




                  525215























                      1












                      $begingroup$

                      Another way would be to make the substitution $xmapstotan x$ so that$$begin{align*}mathfrak{I} & =intlimits_0^{pi/2}frac {mathrm dx}{1+tan^nx}\ & =intlimits_0^{pi/2}mathrm dx-intlimits_0^{pi/2}mathrm dx,frac {tan^n x}{1+tan^nx}\ & =frac {pi}2-intlimits_0^{pi/2}mathrm dx,frac 1{1+cot^nx}end{align*}$$Now use the identity that$$cot x=tanleft(frac {pi}2-xright)$$to get that$$begin{align*}mathfrak{I} & =frac {pi}2-intlimits_0^{pi/2}frac {mathrm dx}{1+tan^nx}\ & =frac {pi}2-mathfrak{I}end{align*}$$



                      Now isolate $mathfrak{I}$ to get that$$intlimits_0^{infty}frac {mathrm dx}{(1+x^2)(1+x^n)}color{blue}{=frac {pi}4}$$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Another way would be to make the substitution $xmapstotan x$ so that$$begin{align*}mathfrak{I} & =intlimits_0^{pi/2}frac {mathrm dx}{1+tan^nx}\ & =intlimits_0^{pi/2}mathrm dx-intlimits_0^{pi/2}mathrm dx,frac {tan^n x}{1+tan^nx}\ & =frac {pi}2-intlimits_0^{pi/2}mathrm dx,frac 1{1+cot^nx}end{align*}$$Now use the identity that$$cot x=tanleft(frac {pi}2-xright)$$to get that$$begin{align*}mathfrak{I} & =frac {pi}2-intlimits_0^{pi/2}frac {mathrm dx}{1+tan^nx}\ & =frac {pi}2-mathfrak{I}end{align*}$$



                        Now isolate $mathfrak{I}$ to get that$$intlimits_0^{infty}frac {mathrm dx}{(1+x^2)(1+x^n)}color{blue}{=frac {pi}4}$$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Another way would be to make the substitution $xmapstotan x$ so that$$begin{align*}mathfrak{I} & =intlimits_0^{pi/2}frac {mathrm dx}{1+tan^nx}\ & =intlimits_0^{pi/2}mathrm dx-intlimits_0^{pi/2}mathrm dx,frac {tan^n x}{1+tan^nx}\ & =frac {pi}2-intlimits_0^{pi/2}mathrm dx,frac 1{1+cot^nx}end{align*}$$Now use the identity that$$cot x=tanleft(frac {pi}2-xright)$$to get that$$begin{align*}mathfrak{I} & =frac {pi}2-intlimits_0^{pi/2}frac {mathrm dx}{1+tan^nx}\ & =frac {pi}2-mathfrak{I}end{align*}$$



                          Now isolate $mathfrak{I}$ to get that$$intlimits_0^{infty}frac {mathrm dx}{(1+x^2)(1+x^n)}color{blue}{=frac {pi}4}$$






                          share|cite|improve this answer









                          $endgroup$



                          Another way would be to make the substitution $xmapstotan x$ so that$$begin{align*}mathfrak{I} & =intlimits_0^{pi/2}frac {mathrm dx}{1+tan^nx}\ & =intlimits_0^{pi/2}mathrm dx-intlimits_0^{pi/2}mathrm dx,frac {tan^n x}{1+tan^nx}\ & =frac {pi}2-intlimits_0^{pi/2}mathrm dx,frac 1{1+cot^nx}end{align*}$$Now use the identity that$$cot x=tanleft(frac {pi}2-xright)$$to get that$$begin{align*}mathfrak{I} & =frac {pi}2-intlimits_0^{pi/2}frac {mathrm dx}{1+tan^nx}\ & =frac {pi}2-mathfrak{I}end{align*}$$



                          Now isolate $mathfrak{I}$ to get that$$intlimits_0^{infty}frac {mathrm dx}{(1+x^2)(1+x^n)}color{blue}{=frac {pi}4}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 16 at 3:45









                          Frank W.Frank W.

                          3,8101321




                          3,8101321






























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