Mixing
Square root of fractional part integral
$begingroup$
Does the following integral have a closed form ?
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx$$
Where ${x}$ denotes the fractional part of $x$.
integration zeta-functions fractional-part
edited Jan 29 at 17:38
YuiTo Cheng
2,1863937
asked Jul 19 '18 at 16:26
Kays Tomy Kays Tomy
19717
$endgroup$
add a comment |
$begingroup$
Does the following integral have a closed form ?
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx$$
Where ${x}$ denotes the fractional part of $x$.
integration zeta-functions fractional-part
edited Jan 29 at 17:38
YuiTo Cheng
2,1863937
asked Jul 19 '18 at 16:26
Kays Tomy Kays Tomy
19717
$endgroup$
6
$begingroup$
Have you tried any methods of evaluating it? Perhaps in series form?
$endgroup$
– B. Mehta
Jul 19 '18 at 16:31
add a comment |
$begingroup$
Does the following integral have a closed form ?
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx$$
Where ${x}$ denotes the fractional part of $x$.
integration zeta-functions fractional-part
edited Jan 29 at 17:38
YuiTo Cheng
2,1863937
asked Jul 19 '18 at 16:26
Kays Tomy Kays Tomy
19717
$endgroup$
Does the following integral have a closed form ?
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx$$
Where ${x}$ denotes the fractional part of $x$.
integration zeta-functions fractional-part
integration zeta-functions fractional-part
edited Jan 29 at 17:38
YuiTo Cheng
2,1863937
asked Jul 19 '18 at 16:26
Kays Tomy Kays Tomy
19717
edited Jan 29 at 17:38
YuiTo Cheng
2,1863937
asked Jul 19 '18 at 16:26
Kays Tomy Kays Tomy
19717
edited Jan 29 at 17:38
YuiTo Cheng
2,1863937
edited Jan 29 at 17:38
YuiTo Cheng
2,1863937
edited Jan 29 at 17:38
YuiTo Cheng
2,1863937
2,1863937
asked Jul 19 '18 at 16:26
Kays Tomy Kays Tomy
19717
asked Jul 19 '18 at 16:26
Kays Tomy Kays Tomy
19717
asked Jul 19 '18 at 16:26
Kays Tomy Kays Tomy
19717
19717
6
$begingroup$
Have you tried any methods of evaluating it? Perhaps in series form?
$endgroup$
– B. Mehta
Jul 19 '18 at 16:31
add a comment |
6
$begingroup$
Have you tried any methods of evaluating it? Perhaps in series form?
$endgroup$
– B. Mehta
Jul 19 '18 at 16:31
6
6
$begingroup$
Have you tried any methods of evaluating it? Perhaps in series form?
$endgroup$
– B. Mehta
Jul 19 '18 at 16:31
$begingroup$
Have you tried any methods of evaluating it? Perhaps in series form?
$endgroup$
– B. Mehta
Jul 19 '18 at 16:31
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Due to the complexity of the computation (see other answers), we consider integrating $x,dy$ instead of $y,dx$.
The function has discontinuities at every $x=dfrac1n$, which are jumps from $y=0$ to $y=1$. We can invert the function between $dfrac1{n+1}$ and $dfrac1n$ using
$$y=sqrt{frac1x-n}iff x=frac1{y^2+n}.$$
From this, the area under a tooth is given by
$$I_n=int_0^1frac{dy}{y^2+n}-frac1{n+1}=frac1{sqrt n}arctanfrac1{sqrt n}-frac1{n+1}.$$
The total area is the sum of the $I_n$, which doesn't seem to have a closed form.
If we use the Taylor development of the arc tangent,
$$frac1{sqrt n}left(frac1{sqrt n}-frac1{3nsqrt n}+frac1{5n^2sqrt n}-cdotsright)-frac1{n+1}.$$
Now if we sum on $n$, noticing that the first term will telescope with the last one, we get
$$1-frac13zeta(2)+frac15zeta(3)-frac17zeta(4)+cdots$$
answered Jul 19 '18 at 20:59
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
This is the same solution as that of Marty Cohen, but explained my way.
$endgroup$
– Yves Daoust
Jul 19 '18 at 21:00
add a comment |
$begingroup$
You want
$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx
$.
Consider
$I
=int_{0}^{1}f({1/x})dx
$
where
$f(0) = 0.
f(1) = 1,
f'(x) ge 0
$.
Since
$n le 1/x
le n+1
$
for
$1/(n+1)
le x
le 1/n
$,
$begin{array}\
I
&=int_{0}^{1}f({1/x})dx\
&=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f({1/x})dx\
&=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f(1/x-n)dx\
&=sum_{n=1}^{infty}int_{n}^{n+1}f(y-n)dy/y^2
qquad y=1/x, x=1/y, dx=-dy/y^2\
&=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
&=int_{0}^{1}f(y)dysum_{n=1}^{infty}dfrac1{(y+n)^2}\
end{array}
$
so your result is
$begin{array}
I
&=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
&=sum_{n=1}^{infty}int_{0}^{1}sqrt{y}dy/(y+n)^2\
&=sum_{n=1}^{infty}left(dfrac{cot(sqrt{n})}{sqrt{n}} - dfrac1{n + 1}right)
qquadtext{(according to Wolfy)}\
end{array}
$
Note:
You can do this
for any function
(instead of $1/x$)
such that
$g(1) = 1,
g'(x) > 0,
g(x) to infty$
as $x to 0$.
You have to look
at $g^{(-1)}(x)$.
answered Jul 19 '18 at 18:56
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}}sqrt{bigg{frac{1}{x}bigg}}
, dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}} sqrt{frac{1}{x}-n} , dx=\sum _{n=1}^{infty } frac{-4 sqrt{n}+pi +n pi -2 (1+n) sin
^{-1}left(frac{-1+n}{1+n}right)}{4 sqrt{n} (1+n)}$$
With the aid of computer algebra, we obtain integral and approximate series with value.
Probably closed form of the sum does not exist.
Supplement:
Borrowing sum from user: Yves Daoust we can write:
$$sum _{n=1}^{infty } left(frac{cot ^{-1}left(sqrt{n}right)}{sqrt{n}}-frac{1}{n+1}right)=1-gamma -int_0^1 frac{psi (1+x)}{2 sqrt{x}} , dxapprox 0.60204991$$
where: $psi left(1+xright)$ is PolyGamma function
answered Jul 19 '18 at 20:23
Mariusz IwaniukMariusz Iwaniuk
1,9771718
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Due to the complexity of the computation (see other answers), we consider integrating $x,dy$ instead of $y,dx$.
The function has discontinuities at every $x=dfrac1n$, which are jumps from $y=0$ to $y=1$. We can invert the function between $dfrac1{n+1}$ and $dfrac1n$ using
$$y=sqrt{frac1x-n}iff x=frac1{y^2+n}.$$
From this, the area under a tooth is given by
$$I_n=int_0^1frac{dy}{y^2+n}-frac1{n+1}=frac1{sqrt n}arctanfrac1{sqrt n}-frac1{n+1}.$$
The total area is the sum of the $I_n$, which doesn't seem to have a closed form.
If we use the Taylor development of the arc tangent,
$$frac1{sqrt n}left(frac1{sqrt n}-frac1{3nsqrt n}+frac1{5n^2sqrt n}-cdotsright)-frac1{n+1}.$$
Now if we sum on $n$, noticing that the first term will telescope with the last one, we get
$$1-frac13zeta(2)+frac15zeta(3)-frac17zeta(4)+cdots$$
answered Jul 19 '18 at 20:59
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
This is the same solution as that of Marty Cohen, but explained my way.
$endgroup$
– Yves Daoust
Jul 19 '18 at 21:00
add a comment |
$begingroup$
Due to the complexity of the computation (see other answers), we consider integrating $x,dy$ instead of $y,dx$.
The function has discontinuities at every $x=dfrac1n$, which are jumps from $y=0$ to $y=1$. We can invert the function between $dfrac1{n+1}$ and $dfrac1n$ using
$$y=sqrt{frac1x-n}iff x=frac1{y^2+n}.$$
From this, the area under a tooth is given by
$$I_n=int_0^1frac{dy}{y^2+n}-frac1{n+1}=frac1{sqrt n}arctanfrac1{sqrt n}-frac1{n+1}.$$
The total area is the sum of the $I_n$, which doesn't seem to have a closed form.
If we use the Taylor development of the arc tangent,
$$frac1{sqrt n}left(frac1{sqrt n}-frac1{3nsqrt n}+frac1{5n^2sqrt n}-cdotsright)-frac1{n+1}.$$
Now if we sum on $n$, noticing that the first term will telescope with the last one, we get
$$1-frac13zeta(2)+frac15zeta(3)-frac17zeta(4)+cdots$$
answered Jul 19 '18 at 20:59
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
This is the same solution as that of Marty Cohen, but explained my way.
$endgroup$
– Yves Daoust
Jul 19 '18 at 21:00
add a comment |
$begingroup$
Due to the complexity of the computation (see other answers), we consider integrating $x,dy$ instead of $y,dx$.
The function has discontinuities at every $x=dfrac1n$, which are jumps from $y=0$ to $y=1$. We can invert the function between $dfrac1{n+1}$ and $dfrac1n$ using
$$y=sqrt{frac1x-n}iff x=frac1{y^2+n}.$$
From this, the area under a tooth is given by
$$I_n=int_0^1frac{dy}{y^2+n}-frac1{n+1}=frac1{sqrt n}arctanfrac1{sqrt n}-frac1{n+1}.$$
The total area is the sum of the $I_n$, which doesn't seem to have a closed form.
If we use the Taylor development of the arc tangent,
$$frac1{sqrt n}left(frac1{sqrt n}-frac1{3nsqrt n}+frac1{5n^2sqrt n}-cdotsright)-frac1{n+1}.$$
Now if we sum on $n$, noticing that the first term will telescope with the last one, we get
$$1-frac13zeta(2)+frac15zeta(3)-frac17zeta(4)+cdots$$
answered Jul 19 '18 at 20:59
Yves DaoustYves Daoust
131k676229
$endgroup$
Due to the complexity of the computation (see other answers), we consider integrating $x,dy$ instead of $y,dx$.
The function has discontinuities at every $x=dfrac1n$, which are jumps from $y=0$ to $y=1$. We can invert the function between $dfrac1{n+1}$ and $dfrac1n$ using
$$y=sqrt{frac1x-n}iff x=frac1{y^2+n}.$$
From this, the area under a tooth is given by
$$I_n=int_0^1frac{dy}{y^2+n}-frac1{n+1}=frac1{sqrt n}arctanfrac1{sqrt n}-frac1{n+1}.$$
The total area is the sum of the $I_n$, which doesn't seem to have a closed form.
If we use the Taylor development of the arc tangent,
$$frac1{sqrt n}left(frac1{sqrt n}-frac1{3nsqrt n}+frac1{5n^2sqrt n}-cdotsright)-frac1{n+1}.$$
Now if we sum on $n$, noticing that the first term will telescope with the last one, we get
$$1-frac13zeta(2)+frac15zeta(3)-frac17zeta(4)+cdots$$
answered Jul 19 '18 at 20:59
Yves DaoustYves Daoust
131k676229
edited Jul 19 '18 at 21:09
answered Jul 19 '18 at 20:59
Yves DaoustYves Daoust
131k676229
answered Jul 19 '18 at 20:59
Yves DaoustYves Daoust
131k676229
answered Jul 19 '18 at 20:59
Yves DaoustYves Daoust
131k676229
131k676229
$begingroup$
This is the same solution as that of Marty Cohen, but explained my way.
$endgroup$
– Yves Daoust
Jul 19 '18 at 21:00
add a comment |
$begingroup$
This is the same solution as that of Marty Cohen, but explained my way.
$endgroup$
– Yves Daoust
Jul 19 '18 at 21:00
$begingroup$
This is the same solution as that of Marty Cohen, but explained my way.
$endgroup$
– Yves Daoust
Jul 19 '18 at 21:00
$begingroup$
This is the same solution as that of Marty Cohen, but explained my way.
$endgroup$
– Yves Daoust
Jul 19 '18 at 21:00
add a comment |
$begingroup$
You want
$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx
$.
Consider
$I
=int_{0}^{1}f({1/x})dx
$
where
$f(0) = 0.
f(1) = 1,
f'(x) ge 0
$.
Since
$n le 1/x
le n+1
$
for
$1/(n+1)
le x
le 1/n
$,
$begin{array}\
I
&=int_{0}^{1}f({1/x})dx\
&=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f({1/x})dx\
&=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f(1/x-n)dx\
&=sum_{n=1}^{infty}int_{n}^{n+1}f(y-n)dy/y^2
qquad y=1/x, x=1/y, dx=-dy/y^2\
&=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
&=int_{0}^{1}f(y)dysum_{n=1}^{infty}dfrac1{(y+n)^2}\
end{array}
$
so your result is
$begin{array}
I
&=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
&=sum_{n=1}^{infty}int_{0}^{1}sqrt{y}dy/(y+n)^2\
&=sum_{n=1}^{infty}left(dfrac{cot(sqrt{n})}{sqrt{n}} - dfrac1{n + 1}right)
qquadtext{(according to Wolfy)}\
end{array}
$
Note:
You can do this
for any function
(instead of $1/x$)
such that
$g(1) = 1,
g'(x) > 0,
g(x) to infty$
as $x to 0$.
You have to look
at $g^{(-1)}(x)$.
answered Jul 19 '18 at 18:56
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
You want
$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx
$.
Consider
$I
=int_{0}^{1}f({1/x})dx
$
where
$f(0) = 0.
f(1) = 1,
f'(x) ge 0
$.
Since
$n le 1/x
le n+1
$
for
$1/(n+1)
le x
le 1/n
$,
$begin{array}\
I
&=int_{0}^{1}f({1/x})dx\
&=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f({1/x})dx\
&=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f(1/x-n)dx\
&=sum_{n=1}^{infty}int_{n}^{n+1}f(y-n)dy/y^2
qquad y=1/x, x=1/y, dx=-dy/y^2\
&=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
&=int_{0}^{1}f(y)dysum_{n=1}^{infty}dfrac1{(y+n)^2}\
end{array}
$
so your result is
$begin{array}
I
&=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
&=sum_{n=1}^{infty}int_{0}^{1}sqrt{y}dy/(y+n)^2\
&=sum_{n=1}^{infty}left(dfrac{cot(sqrt{n})}{sqrt{n}} - dfrac1{n + 1}right)
qquadtext{(according to Wolfy)}\
end{array}
$
Note:
You can do this
for any function
(instead of $1/x$)
such that
$g(1) = 1,
g'(x) > 0,
g(x) to infty$
as $x to 0$.
You have to look
at $g^{(-1)}(x)$.
answered Jul 19 '18 at 18:56
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
You want
$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx
$.
Consider
$I
=int_{0}^{1}f({1/x})dx
$
where
$f(0) = 0.
f(1) = 1,
f'(x) ge 0
$.
Since
$n le 1/x
le n+1
$
for
$1/(n+1)
le x
le 1/n
$,
$begin{array}\
I
&=int_{0}^{1}f({1/x})dx\
&=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f({1/x})dx\
&=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f(1/x-n)dx\
&=sum_{n=1}^{infty}int_{n}^{n+1}f(y-n)dy/y^2
qquad y=1/x, x=1/y, dx=-dy/y^2\
&=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
&=int_{0}^{1}f(y)dysum_{n=1}^{infty}dfrac1{(y+n)^2}\
end{array}
$
so your result is
$begin{array}
I
&=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
&=sum_{n=1}^{infty}int_{0}^{1}sqrt{y}dy/(y+n)^2\
&=sum_{n=1}^{infty}left(dfrac{cot(sqrt{n})}{sqrt{n}} - dfrac1{n + 1}right)
qquadtext{(according to Wolfy)}\
end{array}
$
Note:
You can do this
for any function
(instead of $1/x$)
such that
$g(1) = 1,
g'(x) > 0,
g(x) to infty$
as $x to 0$.
You have to look
at $g^{(-1)}(x)$.
answered Jul 19 '18 at 18:56
marty cohenmarty cohen
74.9k549130
$endgroup$
You want
$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx
$.
Consider
$I
=int_{0}^{1}f({1/x})dx
$
where
$f(0) = 0.
f(1) = 1,
f'(x) ge 0
$.
Since
$n le 1/x
le n+1
$
for
$1/(n+1)
le x
le 1/n
$,
$begin{array}\
I
&=int_{0}^{1}f({1/x})dx\
&=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f({1/x})dx\
&=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f(1/x-n)dx\
&=sum_{n=1}^{infty}int_{n}^{n+1}f(y-n)dy/y^2
qquad y=1/x, x=1/y, dx=-dy/y^2\
&=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
&=int_{0}^{1}f(y)dysum_{n=1}^{infty}dfrac1{(y+n)^2}\
end{array}
$
so your result is
$begin{array}
I
&=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
&=sum_{n=1}^{infty}int_{0}^{1}sqrt{y}dy/(y+n)^2\
&=sum_{n=1}^{infty}left(dfrac{cot(sqrt{n})}{sqrt{n}} - dfrac1{n + 1}right)
qquadtext{(according to Wolfy)}\
end{array}
$
Note:
You can do this
for any function
(instead of $1/x$)
such that
$g(1) = 1,
g'(x) > 0,
g(x) to infty$
as $x to 0$.
You have to look
at $g^{(-1)}(x)$.
answered Jul 19 '18 at 18:56
marty cohenmarty cohen
74.9k549130
answered Jul 19 '18 at 18:56
marty cohenmarty cohen
74.9k549130
answered Jul 19 '18 at 18:56
marty cohenmarty cohen
74.9k549130
answered Jul 19 '18 at 18:56
marty cohenmarty cohen
74.9k549130
74.9k549130
add a comment |
add a comment |
$begingroup$
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}}sqrt{bigg{frac{1}{x}bigg}}
, dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}} sqrt{frac{1}{x}-n} , dx=\sum _{n=1}^{infty } frac{-4 sqrt{n}+pi +n pi -2 (1+n) sin
^{-1}left(frac{-1+n}{1+n}right)}{4 sqrt{n} (1+n)}$$
With the aid of computer algebra, we obtain integral and approximate series with value.
Probably closed form of the sum does not exist.
Supplement:
Borrowing sum from user: Yves Daoust we can write:
$$sum _{n=1}^{infty } left(frac{cot ^{-1}left(sqrt{n}right)}{sqrt{n}}-frac{1}{n+1}right)=1-gamma -int_0^1 frac{psi (1+x)}{2 sqrt{x}} , dxapprox 0.60204991$$
where: $psi left(1+xright)$ is PolyGamma function
answered Jul 19 '18 at 20:23
Mariusz IwaniukMariusz Iwaniuk
1,9771718
$endgroup$
add a comment |
$begingroup$
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}}sqrt{bigg{frac{1}{x}bigg}}
, dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}} sqrt{frac{1}{x}-n} , dx=\sum _{n=1}^{infty } frac{-4 sqrt{n}+pi +n pi -2 (1+n) sin
^{-1}left(frac{-1+n}{1+n}right)}{4 sqrt{n} (1+n)}$$
With the aid of computer algebra, we obtain integral and approximate series with value.
Probably closed form of the sum does not exist.
Supplement:
Borrowing sum from user: Yves Daoust we can write:
$$sum _{n=1}^{infty } left(frac{cot ^{-1}left(sqrt{n}right)}{sqrt{n}}-frac{1}{n+1}right)=1-gamma -int_0^1 frac{psi (1+x)}{2 sqrt{x}} , dxapprox 0.60204991$$
where: $psi left(1+xright)$ is PolyGamma function
answered Jul 19 '18 at 20:23
Mariusz IwaniukMariusz Iwaniuk
1,9771718
$endgroup$
add a comment |
$begingroup$
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}}sqrt{bigg{frac{1}{x}bigg}}
, dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}} sqrt{frac{1}{x}-n} , dx=\sum _{n=1}^{infty } frac{-4 sqrt{n}+pi +n pi -2 (1+n) sin
^{-1}left(frac{-1+n}{1+n}right)}{4 sqrt{n} (1+n)}$$
With the aid of computer algebra, we obtain integral and approximate series with value.
Probably closed form of the sum does not exist.
Supplement:
Borrowing sum from user: Yves Daoust we can write:
$$sum _{n=1}^{infty } left(frac{cot ^{-1}left(sqrt{n}right)}{sqrt{n}}-frac{1}{n+1}right)=1-gamma -int_0^1 frac{psi (1+x)}{2 sqrt{x}} , dxapprox 0.60204991$$
where: $psi left(1+xright)$ is PolyGamma function
answered Jul 19 '18 at 20:23
Mariusz IwaniukMariusz Iwaniuk
1,9771718
$endgroup$
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}}sqrt{bigg{frac{1}{x}bigg}}
, dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}} sqrt{frac{1}{x}-n} , dx=\sum _{n=1}^{infty } frac{-4 sqrt{n}+pi +n pi -2 (1+n) sin
^{-1}left(frac{-1+n}{1+n}right)}{4 sqrt{n} (1+n)}$$
With the aid of computer algebra, we obtain integral and approximate series with value.
Probably closed form of the sum does not exist.
Supplement:
Borrowing sum from user: Yves Daoust we can write:
$$sum _{n=1}^{infty } left(frac{cot ^{-1}left(sqrt{n}right)}{sqrt{n}}-frac{1}{n+1}right)=1-gamma -int_0^1 frac{psi (1+x)}{2 sqrt{x}} , dxapprox 0.60204991$$
where: $psi left(1+xright)$ is PolyGamma function
answered Jul 19 '18 at 20:23
Mariusz IwaniukMariusz Iwaniuk
1,9771718
edited Jul 20 '18 at 7:29
answered Jul 19 '18 at 20:23
Mariusz IwaniukMariusz Iwaniuk
1,9771718
answered Jul 19 '18 at 20:23
Mariusz IwaniukMariusz Iwaniuk
1,9771718
answered Jul 19 '18 at 20:23
Mariusz IwaniukMariusz Iwaniuk
1,9771718
1,9771718
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6
$begingroup$
Have you tried any methods of evaluating it? Perhaps in series form?
$endgroup$
– B. Mehta
Jul 19 '18 at 16:31