Limits and Infinite Integration by Parts The Next CEO of Stack OverflowDelta 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
The Next CEO of Stack OverflowDelta 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?$intfracsin(x)arcsin(-x)dx$
$begingroup$
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!
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
$$int fracsin(x)x ,dx$$
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
asked Mar 22 at 21:07
HyperionHyperion
702111
$endgroup$
add a comment |
$begingroup$
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!
real-analysis calculus integration sequences-and-series
asked Mar 22 at 21:07
HyperionHyperion
702111
$endgroup$
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
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
1
active
oldest
votes
$begingroup$
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.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$
answered Mar 22 at 21:18
Robert IsraelRobert Israel
329k23218473
$endgroup$
add a comment |
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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,
$$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$
i.e.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$
answered Mar 22 at 21:18
Robert IsraelRobert Israel
329k23218473
$endgroup$
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,
$$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$
i.e.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$
answered Mar 22 at 21:18
Robert IsraelRobert Israel
329k23218473
$endgroup$
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,
$$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$
i.e.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$
answered Mar 22 at 21:18
Robert IsraelRobert Israel
329k23218473
$endgroup$
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.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$
answered Mar 22 at 21:18
Robert IsraelRobert Israel
329k23218473
answered Mar 22 at 21:18
Robert IsraelRobert Israel
329k23218473
answered Mar 22 at 21:18
Robert IsraelRobert Israel
329k23218473
answered Mar 22 at 21:18
Robert IsraelRobert Israel
329k23218473
329k23218473
add a comment |
add a comment |
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