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?$intfracsin(x)arcsin(-x)dx$
$begingroup$
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 57 mins ago
HyperionHyperion
653110
$endgroup$
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 57 mins ago
HyperionHyperion
653110
$endgroup$
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 57 mins ago
HyperionHyperion
653110
$endgroup$
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 57 mins ago
HyperionHyperion
653110
asked 57 mins ago
HyperionHyperion
653110
asked 57 mins ago
HyperionHyperion
653110
asked 57 mins ago
HyperionHyperion
653110
asked 57 mins ago
HyperionHyperion
653110
653110
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 46 mins ago
Robert IsraelRobert Israel
328k23216469
$endgroup$
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 46 mins ago
Robert IsraelRobert Israel
328k23216469
$endgroup$
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 46 mins ago
Robert IsraelRobert Israel
328k23216469
$endgroup$
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 46 mins ago
Robert IsraelRobert Israel
328k23216469
$endgroup$
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 46 mins ago
Robert IsraelRobert Israel
328k23216469
answered 46 mins ago
Robert IsraelRobert Israel
328k23216469
answered 46 mins ago
Robert IsraelRobert Israel
328k23216469
answered 46 mins ago
Robert IsraelRobert Israel
328k23216469
328k23216469
add a comment |
add a comment |
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