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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: $sum_{n=-infty}^{infty} frac{1}{10^{{(n/100)}^2}}$Infinite Integration by PartsIntegral of $sin(1/x)$ from $0$ to $infty$Evaluating $limlimits_{btoinfty}bint_0^1cos(b x) cosh^{-1}(frac{1}{x})dx$Formula obtained by repeated integration by parts.When do Taylor series converge quickly?$intfrac{sin(x)}{arcsin(-x)}dx$













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$begingroup$


It is well known that
$$int frac{sin(x)}{x} ,dx$$
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
$$int f(x) ,dx approx sum_{n=1}^a frac{(-1)^{n-1}cdot f^{(n-1)}(x)cdot x^n}{n!}$$
where $a in mathbb{N}$. 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
$$int f(x) ,dx = lim_{atoinfty}sum_{n=1}^a frac{(-1)^{n-1}cdot f^{(n-1)}(x)cdot x^n}{n!}$$
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







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    It is well known that
    $$int frac{sin(x)}{x} ,dx$$
    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
    $$int f(x) ,dx approx sum_{n=1}^a frac{(-1)^{n-1}cdot f^{(n-1)}(x)cdot x^n}{n!}$$
    where $a in mathbb{N}$. 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
    $$int f(x) ,dx = lim_{atoinfty}sum_{n=1}^a frac{(-1)^{n-1}cdot f^{(n-1)}(x)cdot x^n}{n!}$$
    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







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      2



      $begingroup$


      It is well known that
      $$int frac{sin(x)}{x} ,dx$$
      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
      $$int f(x) ,dx approx sum_{n=1}^a frac{(-1)^{n-1}cdot f^{(n-1)}(x)cdot x^n}{n!}$$
      where $a in mathbb{N}$. 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
      $$int f(x) ,dx = lim_{atoinfty}sum_{n=1}^a frac{(-1)^{n-1}cdot f^{(n-1)}(x)cdot x^n}{n!}$$
      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







      share|cite|improve this question









      $endgroup$




      It is well known that
      $$int frac{sin(x)}{x} ,dx$$
      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
      $$int f(x) ,dx approx sum_{n=1}^a frac{(-1)^{n-1}cdot f^{(n-1)}(x)cdot x^n}{n!}$$
      where $a in mathbb{N}$. 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
      $$int f(x) ,dx = lim_{atoinfty}sum_{n=1}^a frac{(-1)^{n-1}cdot f^{(n-1)}(x)cdot x^n}{n!}$$
      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




      real-analysis calculus integration sequences-and-series






      share|cite|improve this question













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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,



          $$ G(0) = sum_{k=0}^infty G^{(k)}(x) frac{(-x)^k}{k!} = G(x) + sum_{k=1}^infty f^{(k-1)}(x) frac{(-x)^k}{k!}$$



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






          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,



            $$ G(0) = sum_{k=0}^infty G^{(k)}(x) frac{(-x)^k}{k!} = G(x) + sum_{k=1}^infty f^{(k-1)}(x) frac{(-x)^k}{k!}$$



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






            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,



              $$ G(0) = sum_{k=0}^infty G^{(k)}(x) frac{(-x)^k}{k!} = G(x) + sum_{k=1}^infty f^{(k-1)}(x) frac{(-x)^k}{k!}$$



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






              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,



                $$ G(0) = sum_{k=0}^infty G^{(k)}(x) frac{(-x)^k}{k!} = G(x) + sum_{k=1}^infty f^{(k-1)}(x) frac{(-x)^k}{k!}$$



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






                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,



                $$ G(0) = sum_{k=0}^infty G^{(k)}(x) frac{(-x)^k}{k!} = G(x) + sum_{k=1}^infty f^{(k-1)}(x) frac{(-x)^k}{k!}$$



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







                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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