Limits and Infinite Integration by Parts
$begingroup$
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
asked Mar 22 at 21:07
HyperionHyperion
702111
$endgroup$
add a comment |
$begingroup$
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
asked Mar 22 at 21:07
HyperionHyperion
702111
$endgroup$
add a comment |
$begingroup$
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
asked Mar 22 at 21:07
HyperionHyperion
702111
$endgroup$
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
asked Mar 22 at 21:07
HyperionHyperion
702111
asked Mar 22 at 21:07
HyperionHyperion
702111
asked Mar 22 at 21:07
HyperionHyperion
702111
asked Mar 22 at 21:07
HyperionHyperion
702111
asked Mar 22 at 21:07
HyperionHyperion
702111
702111
add a comment |
add a comment |
1 Answer
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$begingroup$
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.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_{k=1}^infty f^{(k-1)}(x) frac{(-1)^{k-1} x^k}{k!} $$
answered Mar 22 at 21:18
Robert IsraelRobert Israel
330k23218473
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_{k=1}^infty f^{(k-1)}(x) frac{(-1)^{k-1} x^k}{k!} $$
answered Mar 22 at 21:18
Robert IsraelRobert Israel
330k23218473
$endgroup$
add a comment |
$begingroup$
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.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_{k=1}^infty f^{(k-1)}(x) frac{(-1)^{k-1} x^k}{k!} $$
answered Mar 22 at 21:18
Robert IsraelRobert Israel
330k23218473
$endgroup$
add a comment |
$begingroup$
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.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_{k=1}^infty f^{(k-1)}(x) frac{(-1)^{k-1} x^k}{k!} $$
answered Mar 22 at 21:18
Robert IsraelRobert Israel
330k23218473
$endgroup$
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.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_{k=1}^infty f^{(k-1)}(x) frac{(-1)^{k-1} x^k}{k!} $$
answered Mar 22 at 21:18
Robert IsraelRobert Israel
330k23218473
answered Mar 22 at 21:18
Robert IsraelRobert Israel
330k23218473
answered Mar 22 at 21:18
Robert IsraelRobert Israel
330k23218473
answered Mar 22 at 21:18
Robert IsraelRobert Israel
330k23218473
330k23218473
add a comment |
add a comment |
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