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Evaluation of Integration using limit as a sum
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Evaluation of $displaystyle int^2_1frac1xdx$ using limit as a sum
Try: Using The formula $$int^b_af(x)dx = lim_hrightarrow 0htimes sum^n-1_r=1f(a+rh)$$
where $nh=b-a$
above $a=1,b=2$ and $displaystyle f(x)=frac1x$ and $nh=1$
$$int^2_1frac1xdx = lim_hrightarrow 0 sum^n-1_r=1f(a+rh)=lim_hrightarrow 0hcdot sum^n-1_r=0f(1+rh) $$
$$lim_hrightarrow 0hcdot sum^n-1_r=0frac11+rh=lim_hrightarrow 0bigg[frach1+h+frach1+2h+cdots cdots +frach1+(r-1)hbigg]$$
i did not know how i proceed, struck here
could some help me. thanks
calculus
asked 3 hours ago
Durgesh Tiwari
5,0432626
add a comment |Â
up vote
2
down vote
favorite
Evaluation of $displaystyle int^2_1frac1xdx$ using limit as a sum
Try: Using The formula $$int^b_af(x)dx = lim_hrightarrow 0htimes sum^n-1_r=1f(a+rh)$$
where $nh=b-a$
above $a=1,b=2$ and $displaystyle f(x)=frac1x$ and $nh=1$
$$int^2_1frac1xdx = lim_hrightarrow 0 sum^n-1_r=1f(a+rh)=lim_hrightarrow 0hcdot sum^n-1_r=0f(1+rh) $$
$$lim_hrightarrow 0hcdot sum^n-1_r=0frac11+rh=lim_hrightarrow 0bigg[frach1+h+frach1+2h+cdots cdots +frach1+(r-1)hbigg]$$
i did not know how i proceed, struck here
could some help me. thanks
calculus
asked 3 hours ago
Durgesh Tiwari
5,0432626
1
Well the answer is $log 2$ and you need to know something about $log 2$ before you can actually deal with the problem.
– Paramanand Singh
1 hour ago
1
If you accept that $log 2=lim_ntoinftyn(2^1/n-1)$ then you need to use a partition of $[1,2]$ via points $x_k=2^k/n$. If you accept that $log 2=sum_i=1^infty (-1)^i+1(1/i)$ then your approach can be used to show that the integral is $log 2$.
– Paramanand Singh
1 hour ago
In particular see method 3 of this answer math.stackexchange.com/a/155248/72031 which is helpful for your approach.
– Paramanand Singh
1 hour ago
I did not understand the line $x_k=2^frackn$ for $[1,2].$ thanks parmanand singh, Can you please explain me in detail. thanks
– Durgesh Tiwari
1 hour ago
Wait for my answer.
– Paramanand Singh
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Evaluation of $displaystyle int^2_1frac1xdx$ using limit as a sum
Try: Using The formula $$int^b_af(x)dx = lim_hrightarrow 0htimes sum^n-1_r=1f(a+rh)$$
where $nh=b-a$
above $a=1,b=2$ and $displaystyle f(x)=frac1x$ and $nh=1$
$$int^2_1frac1xdx = lim_hrightarrow 0 sum^n-1_r=1f(a+rh)=lim_hrightarrow 0hcdot sum^n-1_r=0f(1+rh) $$
$$lim_hrightarrow 0hcdot sum^n-1_r=0frac11+rh=lim_hrightarrow 0bigg[frach1+h+frach1+2h+cdots cdots +frach1+(r-1)hbigg]$$
i did not know how i proceed, struck here
could some help me. thanks
calculus
asked 3 hours ago
Durgesh Tiwari
5,0432626
Evaluation of $displaystyle int^2_1frac1xdx$ using limit as a sum
Try: Using The formula $$int^b_af(x)dx = lim_hrightarrow 0htimes sum^n-1_r=1f(a+rh)$$
where $nh=b-a$
above $a=1,b=2$ and $displaystyle f(x)=frac1x$ and $nh=1$
$$int^2_1frac1xdx = lim_hrightarrow 0 sum^n-1_r=1f(a+rh)=lim_hrightarrow 0hcdot sum^n-1_r=0f(1+rh) $$
$$lim_hrightarrow 0hcdot sum^n-1_r=0frac11+rh=lim_hrightarrow 0bigg[frach1+h+frach1+2h+cdots cdots +frach1+(r-1)hbigg]$$
i did not know how i proceed, struck here
could some help me. thanks
calculus
calculus
asked 3 hours ago
Durgesh Tiwari
5,0432626
asked 3 hours ago
Durgesh Tiwari
5,0432626
asked 3 hours ago
Durgesh Tiwari
5,0432626
asked 3 hours ago
Durgesh Tiwari
5,0432626
asked 3 hours ago
Durgesh Tiwari
5,0432626
5,0432626
1
Well the answer is $log 2$ and you need to know something about $log 2$ before you can actually deal with the problem.
– Paramanand Singh
1 hour ago
1
If you accept that $log 2=lim_ntoinftyn(2^1/n-1)$ then you need to use a partition of $[1,2]$ via points $x_k=2^k/n$. If you accept that $log 2=sum_i=1^infty (-1)^i+1(1/i)$ then your approach can be used to show that the integral is $log 2$.
– Paramanand Singh
1 hour ago
In particular see method 3 of this answer math.stackexchange.com/a/155248/72031 which is helpful for your approach.
– Paramanand Singh
1 hour ago
I did not understand the line $x_k=2^frackn$ for $[1,2].$ thanks parmanand singh, Can you please explain me in detail. thanks
– Durgesh Tiwari
1 hour ago
Wait for my answer.
– Paramanand Singh
1 hour ago
add a comment |Â
1
Well the answer is $log 2$ and you need to know something about $log 2$ before you can actually deal with the problem.
– Paramanand Singh
1 hour ago
1
If you accept that $log 2=lim_ntoinftyn(2^1/n-1)$ then you need to use a partition of $[1,2]$ via points $x_k=2^k/n$. If you accept that $log 2=sum_i=1^infty (-1)^i+1(1/i)$ then your approach can be used to show that the integral is $log 2$.
– Paramanand Singh
1 hour ago
In particular see method 3 of this answer math.stackexchange.com/a/155248/72031 which is helpful for your approach.
– Paramanand Singh
1 hour ago
I did not understand the line $x_k=2^frackn$ for $[1,2].$ thanks parmanand singh, Can you please explain me in detail. thanks
– Durgesh Tiwari
1 hour ago
Wait for my answer.
– Paramanand Singh
1 hour ago
1
1
Well the answer is $log 2$ and you need to know something about $log 2$ before you can actually deal with the problem.
– Paramanand Singh
1 hour ago
Well the answer is $log 2$ and you need to know something about $log 2$ before you can actually deal with the problem.
– Paramanand Singh
1 hour ago
1
1
If you accept that $log 2=lim_ntoinftyn(2^1/n-1)$ then you need to use a partition of $[1,2]$ via points $x_k=2^k/n$. If you accept that $log 2=sum_i=1^infty (-1)^i+1(1/i)$ then your approach can be used to show that the integral is $log 2$.
– Paramanand Singh
1 hour ago
If you accept that $log 2=lim_ntoinftyn(2^1/n-1)$ then you need to use a partition of $[1,2]$ via points $x_k=2^k/n$. If you accept that $log 2=sum_i=1^infty (-1)^i+1(1/i)$ then your approach can be used to show that the integral is $log 2$.
– Paramanand Singh
1 hour ago
In particular see method 3 of this answer math.stackexchange.com/a/155248/72031 which is helpful for your approach.
– Paramanand Singh
1 hour ago
In particular see method 3 of this answer math.stackexchange.com/a/155248/72031 which is helpful for your approach.
– Paramanand Singh
1 hour ago
I did not understand the line $x_k=2^frackn$ for $[1,2].$ thanks parmanand singh, Can you please explain me in detail. thanks
– Durgesh Tiwari
1 hour ago
I did not understand the line $x_k=2^frackn$ for $[1,2].$ thanks parmanand singh, Can you please explain me in detail. thanks
– Durgesh Tiwari
1 hour ago
Wait for my answer.
– Paramanand Singh
1 hour ago
Wait for my answer.
– Paramanand Singh
1 hour ago
add a comment |Â
3 Answers
3
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oldest
votes
up vote
2
down vote
Consider the points $x=c^k$ with $c^n=2$.
$$int_1^2fracdxxapproxsum_k=0^n+1 fracDelta c^kc^k=sum_k=1^n frac c^k+1-c^kc^k=n(c-1)=nleft(sqrt[n]2-1right).$$
Then,
$$lim_ntoinftynleft(sqrt[n]2-1right)=lim_hto0frac2^h-1h=left.(2^h)'right|_h=0=log2.$$
Note that this is in fact a discrete version of an exponential change of variable, $x=e^t$, giving
$$int_1^2fracdxx=int_log1^log2frace^t,dte^t=int_0^log2dt.$$
The latter integral can be trivially computed as a sum.
answered 53 mins ago
Yves Daoust
117k667213
add a comment |Â
up vote
2
down vote
$$frac1nsum_k=n+1^2nleft(dfrac knright)^-1=sum_k=n+1^2nfrac1k=H_2n-H_n$$ where $H_n$ denotes an harmonic number.
Now,
$$lim_ntoinfty(H_2n-H_n)=lim_ntoinfty(log2n+gamma+o(1)-log n-gamma+o(1))=log2.$$
It is questionable whether this approach makes sense, as deriving the asymptotic expression of the harmonic numbers is much harder than the initial problem.
answered 39 mins ago
Yves Daoust
117k667213
+1 especially for the remark at the end.
– Paramanand Singh
15 mins ago
add a comment |Â
up vote
1
down vote
Let's assume the following equation as given $$log 2=lim_ntoinfty n(2^1/n-1)tag1$$ By definition of Riemann integral if $f:[a, b] tomathbb R $ is Riemann integrable on $[a, b] $ with integral $I$ then for every $epsilon >0$ there is a $delta>0$ such that for all partitions $P$ of $[a, b] $ with norm less than $delta$ we have $|S(f, P) - I|<epsilon $ and we write $lim_PS(f,P)=I$.
To explain notation and terms we say that a set $P=x_0,x_1,dots, x_n $ is a partition of interval $[a, b] $ if $$a=x_0<x_1<dots<x_n=b$$ The norm of $P$ (denoted by $||P||$) is defined as $max_i=1^n(x_i-x_i-1)$. A Riemann sum for $f$ over a partition $P$ of $[a, b] $ (denoted by $S(f, P) $) is a sum of the form $$sum_i=1^nf(t_i)(x_i-x_i-1)$$ where points $t_i$ called tags are arbitrary points of the intervals $[x_i-1,x_i]$ respectively.
Now it is well known that function $f$ defined by $f(x) =1/x$ is Riemann integrable on any interval $[a, b] $ if $0notin[a,b]$ (because of continuity of $f$ in that interval) and in particular the Riemann integral $$I=int_1^2fracdxx$$ exists.
Let's choose partition $P$ of $[1,2]$ using points $x_k=2^k/n$. Then norm of $P$ is $max_k=1^n(2^k/n-2^(k-1)/n)$ and this tends to $0$ if $nto infty $. Also let's choose tags $t_k$ as $t_k=x_k-1=2^(k-1)/n$. Then the Riemann sum $$S(f, P) =sum_k=1^nf(t_k)(x_k-x_k-1)=sum_k=1^nfrac2^k/n-2^(k-1)/n2^(k-1)/n=n(2^1/n-1)$$ and by our starting equation this tends to $log 2$ as $ntoinfty$. Hence the value of the integral is $log 2$.
Note: This is an expansion of my comment to the question and another answer based on same idea has been given by another user. I have tried here to give some more detail about Riemann integration and in particular highlighted the arbitrary nature of points of a partition. Typical introductory calculus texts almost exclusively use partition points in arithmetic progression and this answer in contrast uses points in geometric progression for creating a partition. The same technique can be used to evaluate the integral of $f(x) =x^p$ for any $p$.
answered 33 mins ago
Paramanand Singh
46.1k555146
Expanded version of mine, it seems ;-)
– Yves Daoust
27 mins ago
1
@YvesDaoust: see updated note at the end.
– Paramanand Singh
23 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Consider the points $x=c^k$ with $c^n=2$.
$$int_1^2fracdxxapproxsum_k=0^n+1 fracDelta c^kc^k=sum_k=1^n frac c^k+1-c^kc^k=n(c-1)=nleft(sqrt[n]2-1right).$$
Then,
$$lim_ntoinftynleft(sqrt[n]2-1right)=lim_hto0frac2^h-1h=left.(2^h)'right|_h=0=log2.$$
Note that this is in fact a discrete version of an exponential change of variable, $x=e^t$, giving
$$int_1^2fracdxx=int_log1^log2frace^t,dte^t=int_0^log2dt.$$
The latter integral can be trivially computed as a sum.
answered 53 mins ago
Yves Daoust
117k667213
add a comment |Â
up vote
2
down vote
Consider the points $x=c^k$ with $c^n=2$.
$$int_1^2fracdxxapproxsum_k=0^n+1 fracDelta c^kc^k=sum_k=1^n frac c^k+1-c^kc^k=n(c-1)=nleft(sqrt[n]2-1right).$$
Then,
$$lim_ntoinftynleft(sqrt[n]2-1right)=lim_hto0frac2^h-1h=left.(2^h)'right|_h=0=log2.$$
Note that this is in fact a discrete version of an exponential change of variable, $x=e^t$, giving
$$int_1^2fracdxx=int_log1^log2frace^t,dte^t=int_0^log2dt.$$
The latter integral can be trivially computed as a sum.
answered 53 mins ago
Yves Daoust
117k667213
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Consider the points $x=c^k$ with $c^n=2$.
$$int_1^2fracdxxapproxsum_k=0^n+1 fracDelta c^kc^k=sum_k=1^n frac c^k+1-c^kc^k=n(c-1)=nleft(sqrt[n]2-1right).$$
Then,
$$lim_ntoinftynleft(sqrt[n]2-1right)=lim_hto0frac2^h-1h=left.(2^h)'right|_h=0=log2.$$
Note that this is in fact a discrete version of an exponential change of variable, $x=e^t$, giving
$$int_1^2fracdxx=int_log1^log2frace^t,dte^t=int_0^log2dt.$$
The latter integral can be trivially computed as a sum.
answered 53 mins ago
Yves Daoust
117k667213
Consider the points $x=c^k$ with $c^n=2$.
$$int_1^2fracdxxapproxsum_k=0^n+1 fracDelta c^kc^k=sum_k=1^n frac c^k+1-c^kc^k=n(c-1)=nleft(sqrt[n]2-1right).$$
Then,
$$lim_ntoinftynleft(sqrt[n]2-1right)=lim_hto0frac2^h-1h=left.(2^h)'right|_h=0=log2.$$
Note that this is in fact a discrete version of an exponential change of variable, $x=e^t$, giving
$$int_1^2fracdxx=int_log1^log2frace^t,dte^t=int_0^log2dt.$$
The latter integral can be trivially computed as a sum.
answered 53 mins ago
Yves Daoust
117k667213
edited 47 mins ago
answered 53 mins ago
Yves Daoust
117k667213
answered 53 mins ago
Yves Daoust
117k667213
answered 53 mins ago
Yves Daoust
117k667213
117k667213
add a comment |Â
add a comment |Â
up vote
2
down vote
$$frac1nsum_k=n+1^2nleft(dfrac knright)^-1=sum_k=n+1^2nfrac1k=H_2n-H_n$$ where $H_n$ denotes an harmonic number.
Now,
$$lim_ntoinfty(H_2n-H_n)=lim_ntoinfty(log2n+gamma+o(1)-log n-gamma+o(1))=log2.$$
It is questionable whether this approach makes sense, as deriving the asymptotic expression of the harmonic numbers is much harder than the initial problem.
answered 39 mins ago
Yves Daoust
117k667213
+1 especially for the remark at the end.
– Paramanand Singh
15 mins ago
add a comment |Â
up vote
2
down vote
$$frac1nsum_k=n+1^2nleft(dfrac knright)^-1=sum_k=n+1^2nfrac1k=H_2n-H_n$$ where $H_n$ denotes an harmonic number.
Now,
$$lim_ntoinfty(H_2n-H_n)=lim_ntoinfty(log2n+gamma+o(1)-log n-gamma+o(1))=log2.$$
It is questionable whether this approach makes sense, as deriving the asymptotic expression of the harmonic numbers is much harder than the initial problem.
answered 39 mins ago
Yves Daoust
117k667213
+1 especially for the remark at the end.
– Paramanand Singh
15 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$frac1nsum_k=n+1^2nleft(dfrac knright)^-1=sum_k=n+1^2nfrac1k=H_2n-H_n$$ where $H_n$ denotes an harmonic number.
Now,
$$lim_ntoinfty(H_2n-H_n)=lim_ntoinfty(log2n+gamma+o(1)-log n-gamma+o(1))=log2.$$
It is questionable whether this approach makes sense, as deriving the asymptotic expression of the harmonic numbers is much harder than the initial problem.
answered 39 mins ago
Yves Daoust
117k667213
$$frac1nsum_k=n+1^2nleft(dfrac knright)^-1=sum_k=n+1^2nfrac1k=H_2n-H_n$$ where $H_n$ denotes an harmonic number.
Now,
$$lim_ntoinfty(H_2n-H_n)=lim_ntoinfty(log2n+gamma+o(1)-log n-gamma+o(1))=log2.$$
It is questionable whether this approach makes sense, as deriving the asymptotic expression of the harmonic numbers is much harder than the initial problem.
answered 39 mins ago
Yves Daoust
117k667213
edited 34 mins ago
answered 39 mins ago
Yves Daoust
117k667213
answered 39 mins ago
Yves Daoust
117k667213
answered 39 mins ago
Yves Daoust
117k667213
117k667213
+1 especially for the remark at the end.
– Paramanand Singh
15 mins ago
add a comment |Â
+1 especially for the remark at the end.
– Paramanand Singh
15 mins ago
+1 especially for the remark at the end.
– Paramanand Singh
15 mins ago
+1 especially for the remark at the end.
– Paramanand Singh
15 mins ago
add a comment |Â
up vote
1
down vote
Let's assume the following equation as given $$log 2=lim_ntoinfty n(2^1/n-1)tag1$$ By definition of Riemann integral if $f:[a, b] tomathbb R $ is Riemann integrable on $[a, b] $ with integral $I$ then for every $epsilon >0$ there is a $delta>0$ such that for all partitions $P$ of $[a, b] $ with norm less than $delta$ we have $|S(f, P) - I|<epsilon $ and we write $lim_PS(f,P)=I$.
To explain notation and terms we say that a set $P=x_0,x_1,dots, x_n $ is a partition of interval $[a, b] $ if $$a=x_0<x_1<dots<x_n=b$$ The norm of $P$ (denoted by $||P||$) is defined as $max_i=1^n(x_i-x_i-1)$. A Riemann sum for $f$ over a partition $P$ of $[a, b] $ (denoted by $S(f, P) $) is a sum of the form $$sum_i=1^nf(t_i)(x_i-x_i-1)$$ where points $t_i$ called tags are arbitrary points of the intervals $[x_i-1,x_i]$ respectively.
Now it is well known that function $f$ defined by $f(x) =1/x$ is Riemann integrable on any interval $[a, b] $ if $0notin[a,b]$ (because of continuity of $f$ in that interval) and in particular the Riemann integral $$I=int_1^2fracdxx$$ exists.
Let's choose partition $P$ of $[1,2]$ using points $x_k=2^k/n$. Then norm of $P$ is $max_k=1^n(2^k/n-2^(k-1)/n)$ and this tends to $0$ if $nto infty $. Also let's choose tags $t_k$ as $t_k=x_k-1=2^(k-1)/n$. Then the Riemann sum $$S(f, P) =sum_k=1^nf(t_k)(x_k-x_k-1)=sum_k=1^nfrac2^k/n-2^(k-1)/n2^(k-1)/n=n(2^1/n-1)$$ and by our starting equation this tends to $log 2$ as $ntoinfty$. Hence the value of the integral is $log 2$.
Note: This is an expansion of my comment to the question and another answer based on same idea has been given by another user. I have tried here to give some more detail about Riemann integration and in particular highlighted the arbitrary nature of points of a partition. Typical introductory calculus texts almost exclusively use partition points in arithmetic progression and this answer in contrast uses points in geometric progression for creating a partition. The same technique can be used to evaluate the integral of $f(x) =x^p$ for any $p$.
answered 33 mins ago
Paramanand Singh
46.1k555146
Expanded version of mine, it seems ;-)
– Yves Daoust
27 mins ago
1
@YvesDaoust: see updated note at the end.
– Paramanand Singh
23 mins ago
add a comment |Â
up vote
1
down vote
Let's assume the following equation as given $$log 2=lim_ntoinfty n(2^1/n-1)tag1$$ By definition of Riemann integral if $f:[a, b] tomathbb R $ is Riemann integrable on $[a, b] $ with integral $I$ then for every $epsilon >0$ there is a $delta>0$ such that for all partitions $P$ of $[a, b] $ with norm less than $delta$ we have $|S(f, P) - I|<epsilon $ and we write $lim_PS(f,P)=I$.
To explain notation and terms we say that a set $P=x_0,x_1,dots, x_n $ is a partition of interval $[a, b] $ if $$a=x_0<x_1<dots<x_n=b$$ The norm of $P$ (denoted by $||P||$) is defined as $max_i=1^n(x_i-x_i-1)$. A Riemann sum for $f$ over a partition $P$ of $[a, b] $ (denoted by $S(f, P) $) is a sum of the form $$sum_i=1^nf(t_i)(x_i-x_i-1)$$ where points $t_i$ called tags are arbitrary points of the intervals $[x_i-1,x_i]$ respectively.
Now it is well known that function $f$ defined by $f(x) =1/x$ is Riemann integrable on any interval $[a, b] $ if $0notin[a,b]$ (because of continuity of $f$ in that interval) and in particular the Riemann integral $$I=int_1^2fracdxx$$ exists.
Let's choose partition $P$ of $[1,2]$ using points $x_k=2^k/n$. Then norm of $P$ is $max_k=1^n(2^k/n-2^(k-1)/n)$ and this tends to $0$ if $nto infty $. Also let's choose tags $t_k$ as $t_k=x_k-1=2^(k-1)/n$. Then the Riemann sum $$S(f, P) =sum_k=1^nf(t_k)(x_k-x_k-1)=sum_k=1^nfrac2^k/n-2^(k-1)/n2^(k-1)/n=n(2^1/n-1)$$ and by our starting equation this tends to $log 2$ as $ntoinfty$. Hence the value of the integral is $log 2$.
Note: This is an expansion of my comment to the question and another answer based on same idea has been given by another user. I have tried here to give some more detail about Riemann integration and in particular highlighted the arbitrary nature of points of a partition. Typical introductory calculus texts almost exclusively use partition points in arithmetic progression and this answer in contrast uses points in geometric progression for creating a partition. The same technique can be used to evaluate the integral of $f(x) =x^p$ for any $p$.
answered 33 mins ago
Paramanand Singh
46.1k555146
Expanded version of mine, it seems ;-)
– Yves Daoust
27 mins ago
1
@YvesDaoust: see updated note at the end.
– Paramanand Singh
23 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let's assume the following equation as given $$log 2=lim_ntoinfty n(2^1/n-1)tag1$$ By definition of Riemann integral if $f:[a, b] tomathbb R $ is Riemann integrable on $[a, b] $ with integral $I$ then for every $epsilon >0$ there is a $delta>0$ such that for all partitions $P$ of $[a, b] $ with norm less than $delta$ we have $|S(f, P) - I|<epsilon $ and we write $lim_PS(f,P)=I$.
To explain notation and terms we say that a set $P=x_0,x_1,dots, x_n $ is a partition of interval $[a, b] $ if $$a=x_0<x_1<dots<x_n=b$$ The norm of $P$ (denoted by $||P||$) is defined as $max_i=1^n(x_i-x_i-1)$. A Riemann sum for $f$ over a partition $P$ of $[a, b] $ (denoted by $S(f, P) $) is a sum of the form $$sum_i=1^nf(t_i)(x_i-x_i-1)$$ where points $t_i$ called tags are arbitrary points of the intervals $[x_i-1,x_i]$ respectively.
Now it is well known that function $f$ defined by $f(x) =1/x$ is Riemann integrable on any interval $[a, b] $ if $0notin[a,b]$ (because of continuity of $f$ in that interval) and in particular the Riemann integral $$I=int_1^2fracdxx$$ exists.
Let's choose partition $P$ of $[1,2]$ using points $x_k=2^k/n$. Then norm of $P$ is $max_k=1^n(2^k/n-2^(k-1)/n)$ and this tends to $0$ if $nto infty $. Also let's choose tags $t_k$ as $t_k=x_k-1=2^(k-1)/n$. Then the Riemann sum $$S(f, P) =sum_k=1^nf(t_k)(x_k-x_k-1)=sum_k=1^nfrac2^k/n-2^(k-1)/n2^(k-1)/n=n(2^1/n-1)$$ and by our starting equation this tends to $log 2$ as $ntoinfty$. Hence the value of the integral is $log 2$.
Note: This is an expansion of my comment to the question and another answer based on same idea has been given by another user. I have tried here to give some more detail about Riemann integration and in particular highlighted the arbitrary nature of points of a partition. Typical introductory calculus texts almost exclusively use partition points in arithmetic progression and this answer in contrast uses points in geometric progression for creating a partition. The same technique can be used to evaluate the integral of $f(x) =x^p$ for any $p$.
answered 33 mins ago
Paramanand Singh
46.1k555146
Let's assume the following equation as given $$log 2=lim_ntoinfty n(2^1/n-1)tag1$$ By definition of Riemann integral if $f:[a, b] tomathbb R $ is Riemann integrable on $[a, b] $ with integral $I$ then for every $epsilon >0$ there is a $delta>0$ such that for all partitions $P$ of $[a, b] $ with norm less than $delta$ we have $|S(f, P) - I|<epsilon $ and we write $lim_PS(f,P)=I$.
To explain notation and terms we say that a set $P=x_0,x_1,dots, x_n $ is a partition of interval $[a, b] $ if $$a=x_0<x_1<dots<x_n=b$$ The norm of $P$ (denoted by $||P||$) is defined as $max_i=1^n(x_i-x_i-1)$. A Riemann sum for $f$ over a partition $P$ of $[a, b] $ (denoted by $S(f, P) $) is a sum of the form $$sum_i=1^nf(t_i)(x_i-x_i-1)$$ where points $t_i$ called tags are arbitrary points of the intervals $[x_i-1,x_i]$ respectively.
Now it is well known that function $f$ defined by $f(x) =1/x$ is Riemann integrable on any interval $[a, b] $ if $0notin[a,b]$ (because of continuity of $f$ in that interval) and in particular the Riemann integral $$I=int_1^2fracdxx$$ exists.
Let's choose partition $P$ of $[1,2]$ using points $x_k=2^k/n$. Then norm of $P$ is $max_k=1^n(2^k/n-2^(k-1)/n)$ and this tends to $0$ if $nto infty $. Also let's choose tags $t_k$ as $t_k=x_k-1=2^(k-1)/n$. Then the Riemann sum $$S(f, P) =sum_k=1^nf(t_k)(x_k-x_k-1)=sum_k=1^nfrac2^k/n-2^(k-1)/n2^(k-1)/n=n(2^1/n-1)$$ and by our starting equation this tends to $log 2$ as $ntoinfty$. Hence the value of the integral is $log 2$.
Note: This is an expansion of my comment to the question and another answer based on same idea has been given by another user. I have tried here to give some more detail about Riemann integration and in particular highlighted the arbitrary nature of points of a partition. Typical introductory calculus texts almost exclusively use partition points in arithmetic progression and this answer in contrast uses points in geometric progression for creating a partition. The same technique can be used to evaluate the integral of $f(x) =x^p$ for any $p$.
answered 33 mins ago
Paramanand Singh
46.1k555146
edited 24 mins ago
answered 33 mins ago
Paramanand Singh
46.1k555146
answered 33 mins ago
Paramanand Singh
46.1k555146
answered 33 mins ago
Paramanand Singh
46.1k555146
46.1k555146
Expanded version of mine, it seems ;-)
– Yves Daoust
27 mins ago
1
@YvesDaoust: see updated note at the end.
– Paramanand Singh
23 mins ago
add a comment |Â
Expanded version of mine, it seems ;-)
– Yves Daoust
27 mins ago
1
@YvesDaoust: see updated note at the end.
– Paramanand Singh
23 mins ago
Expanded version of mine, it seems ;-)
– Yves Daoust
27 mins ago
Expanded version of mine, it seems ;-)
– Yves Daoust
27 mins ago
1
1
@YvesDaoust: see updated note at the end.
– Paramanand Singh
23 mins ago
@YvesDaoust: see updated note at the end.
– Paramanand Singh
23 mins ago
add a comment |Â
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1
Well the answer is $log 2$ and you need to know something about $log 2$ before you can actually deal with the problem.
– Paramanand Singh
1 hour ago
1
If you accept that $log 2=lim_ntoinftyn(2^1/n-1)$ then you need to use a partition of $[1,2]$ via points $x_k=2^k/n$. If you accept that $log 2=sum_i=1^infty (-1)^i+1(1/i)$ then your approach can be used to show that the integral is $log 2$.
– Paramanand Singh
1 hour ago
In particular see method 3 of this answer math.stackexchange.com/a/155248/72031 which is helpful for your approach.
– Paramanand Singh
1 hour ago
I did not understand the line $x_k=2^frackn$ for $[1,2].$ thanks parmanand singh, Can you please explain me in detail. thanks
– Durgesh Tiwari
1 hour ago
Wait for my answer.
– Paramanand Singh
1 hour ago