what is integral of this function?
$begingroup$
Let $f(x)$ be an arbitrary continuous function, $nin mathbb{N}$ and
$$g(x) = frac{1}{1+ncdot f(x)^2}$$
then what is anti-derivative of this:
$$
int left(frac{d}{dx}g(x)right)cdottanhleft(ncdot f(x)cdotfrac{d}{dx}f(x)right)dx = ?
$$
or
$$
int left(frac{d}{dx}g(x)right)cdottanhleft(ncdot f(x)right)cdottanhleft(ncdotfrac{d}{dx}f(x)right)dx = ?
$$
Both integral equals(in the limit when n goes to infinity).
Also we know that $int left(frac{d}{dx}g(x)right)dx = g(x)$ and $int left(n.f(x).frac{d}{dx}f(x)right)dx = ncdotfrac{f(x)^2}{2}$
calculus integration complex-analysis limits special-functions
asked Dec 28 '18 at 15:59
Michael BrockhausenMichael Brockhausen
122
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ be an arbitrary continuous function, $nin mathbb{N}$ and
$$g(x) = frac{1}{1+ncdot f(x)^2}$$
then what is anti-derivative of this:
$$
int left(frac{d}{dx}g(x)right)cdottanhleft(ncdot f(x)cdotfrac{d}{dx}f(x)right)dx = ?
$$
or
$$
int left(frac{d}{dx}g(x)right)cdottanhleft(ncdot f(x)right)cdottanhleft(ncdotfrac{d}{dx}f(x)right)dx = ?
$$
Both integral equals(in the limit when n goes to infinity).
Also we know that $int left(frac{d}{dx}g(x)right)dx = g(x)$ and $int left(n.f(x).frac{d}{dx}f(x)right)dx = ncdotfrac{f(x)^2}{2}$
calculus integration complex-analysis limits special-functions
asked Dec 28 '18 at 15:59
Michael BrockhausenMichael Brockhausen
122
$endgroup$
$begingroup$
Let $u=2[1+nf(x)^2]$. Observe that $g(x)=2/u$ so that $g'(x)dx=-2du/u^2$. Note also that $u'=nf(x)f'(x)$. We can now write the first integral as $-2intfrac{tanh(u')}{u^2};du$. Not sure that helps though.
$endgroup$
– Ben W
Dec 28 '18 at 16:19
add a comment |
$begingroup$
Let $f(x)$ be an arbitrary continuous function, $nin mathbb{N}$ and
$$g(x) = frac{1}{1+ncdot f(x)^2}$$
then what is anti-derivative of this:
$$
int left(frac{d}{dx}g(x)right)cdottanhleft(ncdot f(x)cdotfrac{d}{dx}f(x)right)dx = ?
$$
or
$$
int left(frac{d}{dx}g(x)right)cdottanhleft(ncdot f(x)right)cdottanhleft(ncdotfrac{d}{dx}f(x)right)dx = ?
$$
Both integral equals(in the limit when n goes to infinity).
Also we know that $int left(frac{d}{dx}g(x)right)dx = g(x)$ and $int left(n.f(x).frac{d}{dx}f(x)right)dx = ncdotfrac{f(x)^2}{2}$
calculus integration complex-analysis limits special-functions
asked Dec 28 '18 at 15:59
Michael BrockhausenMichael Brockhausen
122
$endgroup$
Let $f(x)$ be an arbitrary continuous function, $nin mathbb{N}$ and
$$g(x) = frac{1}{1+ncdot f(x)^2}$$
then what is anti-derivative of this:
$$
int left(frac{d}{dx}g(x)right)cdottanhleft(ncdot f(x)cdotfrac{d}{dx}f(x)right)dx = ?
$$
or
$$
int left(frac{d}{dx}g(x)right)cdottanhleft(ncdot f(x)right)cdottanhleft(ncdotfrac{d}{dx}f(x)right)dx = ?
$$
Both integral equals(in the limit when n goes to infinity).
Also we know that $int left(frac{d}{dx}g(x)right)dx = g(x)$ and $int left(n.f(x).frac{d}{dx}f(x)right)dx = ncdotfrac{f(x)^2}{2}$
calculus integration complex-analysis limits special-functions
calculus integration complex-analysis limits special-functions
asked Dec 28 '18 at 15:59
Michael BrockhausenMichael Brockhausen
122
asked Dec 28 '18 at 15:59
Michael BrockhausenMichael Brockhausen
122
asked Dec 28 '18 at 15:59
Michael BrockhausenMichael Brockhausen
122
asked Dec 28 '18 at 15:59
Michael BrockhausenMichael Brockhausen
122
asked Dec 28 '18 at 15:59
Michael BrockhausenMichael Brockhausen
122
122
$begingroup$
Let $u=2[1+nf(x)^2]$. Observe that $g(x)=2/u$ so that $g'(x)dx=-2du/u^2$. Note also that $u'=nf(x)f'(x)$. We can now write the first integral as $-2intfrac{tanh(u')}{u^2};du$. Not sure that helps though.
$endgroup$
– Ben W
Dec 28 '18 at 16:19
add a comment |
$begingroup$
Let $u=2[1+nf(x)^2]$. Observe that $g(x)=2/u$ so that $g'(x)dx=-2du/u^2$. Note also that $u'=nf(x)f'(x)$. We can now write the first integral as $-2intfrac{tanh(u')}{u^2};du$. Not sure that helps though.
$endgroup$
– Ben W
Dec 28 '18 at 16:19
$begingroup$
Let $u=2[1+nf(x)^2]$. Observe that $g(x)=2/u$ so that $g'(x)dx=-2du/u^2$. Note also that $u'=nf(x)f'(x)$. We can now write the first integral as $-2intfrac{tanh(u')}{u^2};du$. Not sure that helps though.
$endgroup$
– Ben W
Dec 28 '18 at 16:19
$begingroup$
Let $u=2[1+nf(x)^2]$. Observe that $g(x)=2/u$ so that $g'(x)dx=-2du/u^2$. Note also that $u'=nf(x)f'(x)$. We can now write the first integral as $-2intfrac{tanh(u')}{u^2};du$. Not sure that helps though.
$endgroup$
– Ben W
Dec 28 '18 at 16:19
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
There is no reason to expect a closed form answer for this problem; take $n=1$ and $f(x)=x^2$, and ask your favorite CAS to do your first antiderivative. Wolfram Alpha states that there is no closed form, and while that is not a proof, it is good enough for me (at least until you can demonstrate that there should be a closed form somehow).
answered Dec 28 '18 at 16:08
DudeManDudeMan
1113
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is no reason to expect a closed form answer for this problem; take $n=1$ and $f(x)=x^2$, and ask your favorite CAS to do your first antiderivative. Wolfram Alpha states that there is no closed form, and while that is not a proof, it is good enough for me (at least until you can demonstrate that there should be a closed form somehow).
answered Dec 28 '18 at 16:08
DudeManDudeMan
1113
$endgroup$
add a comment |
$begingroup$
There is no reason to expect a closed form answer for this problem; take $n=1$ and $f(x)=x^2$, and ask your favorite CAS to do your first antiderivative. Wolfram Alpha states that there is no closed form, and while that is not a proof, it is good enough for me (at least until you can demonstrate that there should be a closed form somehow).
answered Dec 28 '18 at 16:08
DudeManDudeMan
1113
$endgroup$
add a comment |
$begingroup$
There is no reason to expect a closed form answer for this problem; take $n=1$ and $f(x)=x^2$, and ask your favorite CAS to do your first antiderivative. Wolfram Alpha states that there is no closed form, and while that is not a proof, it is good enough for me (at least until you can demonstrate that there should be a closed form somehow).
answered Dec 28 '18 at 16:08
DudeManDudeMan
1113
$endgroup$
There is no reason to expect a closed form answer for this problem; take $n=1$ and $f(x)=x^2$, and ask your favorite CAS to do your first antiderivative. Wolfram Alpha states that there is no closed form, and while that is not a proof, it is good enough for me (at least until you can demonstrate that there should be a closed form somehow).
answered Dec 28 '18 at 16:08
DudeManDudeMan
1113
answered Dec 28 '18 at 16:08
DudeManDudeMan
1113
answered Dec 28 '18 at 16:08
DudeManDudeMan
1113
answered Dec 28 '18 at 16:08
DudeManDudeMan
1113
1113
add a comment |
add a comment |
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$begingroup$
Let $u=2[1+nf(x)^2]$. Observe that $g(x)=2/u$ so that $g'(x)dx=-2du/u^2$. Note also that $u'=nf(x)f'(x)$. We can now write the first integral as $-2intfrac{tanh(u')}{u^2};du$. Not sure that helps though.
$endgroup$
– Ben W
Dec 28 '18 at 16:19