Limits and Infinite Integration by PartsDelta distribution - integration by parts of its differentiationMethods of constructing rapidly convergent seriesGauss Hermite Integration of 1/(1+x^2)Infinite sum: $sum_n=-infty^infty frac110^(n/100)^2$Infinite Integration by PartsIntegral of $sin(1/x)$ from $0$ to $infty$Evaluating $limlimits_btoinftybint_0^1cos(b x) cosh^-1(frac1x)dx$Formula obtained by repeated integration by parts.When do Taylor series converge quickly?$intfracsin(x)arcsin(-x)dx$

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Limits and Infinite Integration by Parts


Delta distribution - integration by parts of its differentiationMethods of constructing rapidly convergent seriesGauss Hermite Integration of 1/(1+x^2)Infinite sum: $sum_n=-infty^infty frac110^(n/100)^2$Infinite Integration by PartsIntegral of $sin(1/x)$ from $0$ to $infty$Evaluating $limlimits_btoinftybint_0^1cos(b x) cosh^-1(frac1x)dx$Formula obtained by repeated integration by parts.When do Taylor series converge quickly?$intfracsin(x)arcsin(-x)dx$













4












$begingroup$


It is well known that
$$int fracsin(x)x ,dx$$
cannot be expressed in terms of elementary functions. However, if we repeatedly use integration by parts, we seem to be able to at least approximate the integral through the formula
$$int f(x) ,dx approx sum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
where $a in mathbbN$. When plugging this in to a graphing calculator, it converges, but very slowly. It also tends to converge more quickly for functions that tend to $0$ as $x to infty$. My guess is that
$$int f(x) ,dx = lim_atoinftysum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
at least on a certain interval, but I am uncertain where to look to learn more about these series. Any ideas? Thanks!










share|cite|improve this question









$endgroup$
















    4












    $begingroup$


    It is well known that
    $$int fracsin(x)x ,dx$$
    cannot be expressed in terms of elementary functions. However, if we repeatedly use integration by parts, we seem to be able to at least approximate the integral through the formula
    $$int f(x) ,dx approx sum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
    where $a in mathbbN$. When plugging this in to a graphing calculator, it converges, but very slowly. It also tends to converge more quickly for functions that tend to $0$ as $x to infty$. My guess is that
    $$int f(x) ,dx = lim_atoinftysum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
    at least on a certain interval, but I am uncertain where to look to learn more about these series. Any ideas? Thanks!










    share|cite|improve this question









    $endgroup$














      4












      4








      4


      1



      $begingroup$


      It is well known that
      $$int fracsin(x)x ,dx$$
      cannot be expressed in terms of elementary functions. However, if we repeatedly use integration by parts, we seem to be able to at least approximate the integral through the formula
      $$int f(x) ,dx approx sum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
      where $a in mathbbN$. When plugging this in to a graphing calculator, it converges, but very slowly. It also tends to converge more quickly for functions that tend to $0$ as $x to infty$. My guess is that
      $$int f(x) ,dx = lim_atoinftysum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
      at least on a certain interval, but I am uncertain where to look to learn more about these series. Any ideas? Thanks!










      share|cite|improve this question









      $endgroup$




      It is well known that
      $$int fracsin(x)x ,dx$$
      cannot be expressed in terms of elementary functions. However, if we repeatedly use integration by parts, we seem to be able to at least approximate the integral through the formula
      $$int f(x) ,dx approx sum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
      where $a in mathbbN$. When plugging this in to a graphing calculator, it converges, but very slowly. It also tends to converge more quickly for functions that tend to $0$ as $x to infty$. My guess is that
      $$int f(x) ,dx = lim_atoinftysum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
      at least on a certain interval, but I am uncertain where to look to learn more about these series. Any ideas? Thanks!







      real-analysis calculus integration sequences-and-series






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      asked 1 hour ago









      HyperionHyperion

      658110




      658110




















          1 Answer
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          7












          $begingroup$

          You've essentially rediscovered Taylor series.
          Let $G(x)$ be an antiderivative of $f(x)$, so $f^(k)(x) = G^(k+1)(x)$ for $k ge 0$. If $G$ is analytic in a neighbourhood of $0$, and $x$ is in that neighbourhood,



          $$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$



          i.e.
          $$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$






          share|cite|improve this answer









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            7












            $begingroup$

            You've essentially rediscovered Taylor series.
            Let $G(x)$ be an antiderivative of $f(x)$, so $f^(k)(x) = G^(k+1)(x)$ for $k ge 0$. If $G$ is analytic in a neighbourhood of $0$, and $x$ is in that neighbourhood,



            $$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$



            i.e.
            $$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$






            share|cite|improve this answer









            $endgroup$

















              7












              $begingroup$

              You've essentially rediscovered Taylor series.
              Let $G(x)$ be an antiderivative of $f(x)$, so $f^(k)(x) = G^(k+1)(x)$ for $k ge 0$. If $G$ is analytic in a neighbourhood of $0$, and $x$ is in that neighbourhood,



              $$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$



              i.e.
              $$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$






              share|cite|improve this answer









              $endgroup$















                7












                7








                7





                $begingroup$

                You've essentially rediscovered Taylor series.
                Let $G(x)$ be an antiderivative of $f(x)$, so $f^(k)(x) = G^(k+1)(x)$ for $k ge 0$. If $G$ is analytic in a neighbourhood of $0$, and $x$ is in that neighbourhood,



                $$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$



                i.e.
                $$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$






                share|cite|improve this answer









                $endgroup$



                You've essentially rediscovered Taylor series.
                Let $G(x)$ be an antiderivative of $f(x)$, so $f^(k)(x) = G^(k+1)(x)$ for $k ge 0$. If $G$ is analytic in a neighbourhood of $0$, and $x$ is in that neighbourhood,



                $$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$



                i.e.
                $$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Robert IsraelRobert Israel

                328k23216469




                328k23216469



























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