Finding convergence of integral $int_0^1 frac{x^n}{1+x}dx$
Test the convergence $$int_0^1 frac{x^n}{1+x}dx$$
I have used comparison test for improper integrals..by comparing with $1/(1+x)$...
so I found it convergent ..
But the solution set says that it is convergent if $n> -1$.
calculus
edited Nov 24 at 18:05
gimusi
1
asked Nov 24 at 17:58
Kashmira
463
add a comment |
Test the convergence $$int_0^1 frac{x^n}{1+x}dx$$
I have used comparison test for improper integrals..by comparing with $1/(1+x)$...
so I found it convergent ..
But the solution set says that it is convergent if $n> -1$.
calculus
edited Nov 24 at 18:05
gimusi
1
asked Nov 24 at 17:58
Kashmira
463
add a comment |
Test the convergence $$int_0^1 frac{x^n}{1+x}dx$$
I have used comparison test for improper integrals..by comparing with $1/(1+x)$...
so I found it convergent ..
But the solution set says that it is convergent if $n> -1$.
calculus
edited Nov 24 at 18:05
gimusi
1
asked Nov 24 at 17:58
Kashmira
463
Test the convergence $$int_0^1 frac{x^n}{1+x}dx$$
I have used comparison test for improper integrals..by comparing with $1/(1+x)$...
so I found it convergent ..
But the solution set says that it is convergent if $n> -1$.
calculus
calculus
edited Nov 24 at 18:05
gimusi
1
asked Nov 24 at 17:58
Kashmira
463
edited Nov 24 at 18:05
gimusi
1
asked Nov 24 at 17:58
Kashmira
463
edited Nov 24 at 18:05
gimusi
1
edited Nov 24 at 18:05
gimusi
1
edited Nov 24 at 18:05
gimusi
1
1
asked Nov 24 at 17:58
Kashmira
463
asked Nov 24 at 17:58
Kashmira
463
asked Nov 24 at 17:58
Kashmira
463
463
add a comment |
add a comment |
3 Answers
3
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oldest
votes
We have that for $nge 0$ the integral is a proper integral, then consider $n<0$ and by $m=-n>0$ we have
$$int_0^1 frac{1}{x^m+x^{m+1}}dx$$
and since as $xto 0^+$
$$frac{1}{x^m+x^{m+1}} sim frac{1}{x^m}$$
the integral converges for $m<1$ that is $n>-1$.
As an alternative by $y=frac1x$ we have
$$int_0^1 frac{x^n}{1+x}dx=int_1^infty frac{1}{y^{n+1}+y^{n+2}}dx$$
and since as $xto infty$
$$ frac{1}{y^{n+1}+y^{n+2}}sim frac{1}{y^{n+2}}$$
the integral converges for $n+2>1$ that is $n>-1$.
answered Nov 24 at 18:05
gimusi
1
add a comment |
Let $a_n = displaystyle int_{0}^1 dfrac{x^n}{1+x}dx$. Since $dfrac{x^n}{1+x} = x^{n-1}left(1-dfrac{1}{1+x}right)=x^{n-1}-dfrac{x^{n-1}}{1+x}$, taking the integral $displaystyle int_{0}^1$ both sides give: $a_n = dfrac{1}{n}- a_{n-1}=dfrac{1}{n}-dfrac{1}{n-1}+a_{n-2}= dfrac{1}{n}-dfrac{1}{n-1}+dfrac{1}{n-2}-a_{n-3}=...=-ln 2+ 1-dfrac{1}{2}+dfrac{1}{3}-dfrac{1}{4}+dfrac{1}{5}-dfrac{1}{6}+cdots + dfrac{1}{n-2}-dfrac{1}{n-1}+dfrac{1}{n}$. $a_n$ converges when considered as a partial sum of an alternating harmonic series.
answered Nov 24 at 18:24
DeepSea
70.9k54487
add a comment |
Since $x^n$ is continuous on $[0,1]$ for $nge 0,$ the integral converges for $nge 0.$
For $n<0,$ $x^n$ blows up at $0.$ So we need to consider the integral over $[a,1]$ for small $a>0.$ Notice that for $xin (0,1],$
$$frac{x^n}{2}le frac{x^n}{1+x} le x^n.$$
It follows that the integral of interest converges iff $int_0^1 x^n,dx $ converges. Things are easy now, so I'll stop here. Ask questions if you like.
answered Nov 24 at 19:04
zhw.
71.6k43075
add a comment |
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3 Answers
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active
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3 Answers
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oldest
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We have that for $nge 0$ the integral is a proper integral, then consider $n<0$ and by $m=-n>0$ we have
$$int_0^1 frac{1}{x^m+x^{m+1}}dx$$
and since as $xto 0^+$
$$frac{1}{x^m+x^{m+1}} sim frac{1}{x^m}$$
the integral converges for $m<1$ that is $n>-1$.
As an alternative by $y=frac1x$ we have
$$int_0^1 frac{x^n}{1+x}dx=int_1^infty frac{1}{y^{n+1}+y^{n+2}}dx$$
and since as $xto infty$
$$ frac{1}{y^{n+1}+y^{n+2}}sim frac{1}{y^{n+2}}$$
the integral converges for $n+2>1$ that is $n>-1$.
answered Nov 24 at 18:05
gimusi
1
add a comment |
We have that for $nge 0$ the integral is a proper integral, then consider $n<0$ and by $m=-n>0$ we have
$$int_0^1 frac{1}{x^m+x^{m+1}}dx$$
and since as $xto 0^+$
$$frac{1}{x^m+x^{m+1}} sim frac{1}{x^m}$$
the integral converges for $m<1$ that is $n>-1$.
As an alternative by $y=frac1x$ we have
$$int_0^1 frac{x^n}{1+x}dx=int_1^infty frac{1}{y^{n+1}+y^{n+2}}dx$$
and since as $xto infty$
$$ frac{1}{y^{n+1}+y^{n+2}}sim frac{1}{y^{n+2}}$$
the integral converges for $n+2>1$ that is $n>-1$.
answered Nov 24 at 18:05
gimusi
1
add a comment |
We have that for $nge 0$ the integral is a proper integral, then consider $n<0$ and by $m=-n>0$ we have
$$int_0^1 frac{1}{x^m+x^{m+1}}dx$$
and since as $xto 0^+$
$$frac{1}{x^m+x^{m+1}} sim frac{1}{x^m}$$
the integral converges for $m<1$ that is $n>-1$.
As an alternative by $y=frac1x$ we have
$$int_0^1 frac{x^n}{1+x}dx=int_1^infty frac{1}{y^{n+1}+y^{n+2}}dx$$
and since as $xto infty$
$$ frac{1}{y^{n+1}+y^{n+2}}sim frac{1}{y^{n+2}}$$
the integral converges for $n+2>1$ that is $n>-1$.
answered Nov 24 at 18:05
gimusi
1
We have that for $nge 0$ the integral is a proper integral, then consider $n<0$ and by $m=-n>0$ we have
$$int_0^1 frac{1}{x^m+x^{m+1}}dx$$
and since as $xto 0^+$
$$frac{1}{x^m+x^{m+1}} sim frac{1}{x^m}$$
the integral converges for $m<1$ that is $n>-1$.
As an alternative by $y=frac1x$ we have
$$int_0^1 frac{x^n}{1+x}dx=int_1^infty frac{1}{y^{n+1}+y^{n+2}}dx$$
and since as $xto infty$
$$ frac{1}{y^{n+1}+y^{n+2}}sim frac{1}{y^{n+2}}$$
the integral converges for $n+2>1$ that is $n>-1$.
answered Nov 24 at 18:05
gimusi
1
answered Nov 24 at 18:05
gimusi
1
answered Nov 24 at 18:05
gimusi
1
answered Nov 24 at 18:05
gimusi
1
1
add a comment |
add a comment |
Let $a_n = displaystyle int_{0}^1 dfrac{x^n}{1+x}dx$. Since $dfrac{x^n}{1+x} = x^{n-1}left(1-dfrac{1}{1+x}right)=x^{n-1}-dfrac{x^{n-1}}{1+x}$, taking the integral $displaystyle int_{0}^1$ both sides give: $a_n = dfrac{1}{n}- a_{n-1}=dfrac{1}{n}-dfrac{1}{n-1}+a_{n-2}= dfrac{1}{n}-dfrac{1}{n-1}+dfrac{1}{n-2}-a_{n-3}=...=-ln 2+ 1-dfrac{1}{2}+dfrac{1}{3}-dfrac{1}{4}+dfrac{1}{5}-dfrac{1}{6}+cdots + dfrac{1}{n-2}-dfrac{1}{n-1}+dfrac{1}{n}$. $a_n$ converges when considered as a partial sum of an alternating harmonic series.
answered Nov 24 at 18:24
DeepSea
70.9k54487
add a comment |
Let $a_n = displaystyle int_{0}^1 dfrac{x^n}{1+x}dx$. Since $dfrac{x^n}{1+x} = x^{n-1}left(1-dfrac{1}{1+x}right)=x^{n-1}-dfrac{x^{n-1}}{1+x}$, taking the integral $displaystyle int_{0}^1$ both sides give: $a_n = dfrac{1}{n}- a_{n-1}=dfrac{1}{n}-dfrac{1}{n-1}+a_{n-2}= dfrac{1}{n}-dfrac{1}{n-1}+dfrac{1}{n-2}-a_{n-3}=...=-ln 2+ 1-dfrac{1}{2}+dfrac{1}{3}-dfrac{1}{4}+dfrac{1}{5}-dfrac{1}{6}+cdots + dfrac{1}{n-2}-dfrac{1}{n-1}+dfrac{1}{n}$. $a_n$ converges when considered as a partial sum of an alternating harmonic series.
answered Nov 24 at 18:24
DeepSea
70.9k54487
add a comment |
Let $a_n = displaystyle int_{0}^1 dfrac{x^n}{1+x}dx$. Since $dfrac{x^n}{1+x} = x^{n-1}left(1-dfrac{1}{1+x}right)=x^{n-1}-dfrac{x^{n-1}}{1+x}$, taking the integral $displaystyle int_{0}^1$ both sides give: $a_n = dfrac{1}{n}- a_{n-1}=dfrac{1}{n}-dfrac{1}{n-1}+a_{n-2}= dfrac{1}{n}-dfrac{1}{n-1}+dfrac{1}{n-2}-a_{n-3}=...=-ln 2+ 1-dfrac{1}{2}+dfrac{1}{3}-dfrac{1}{4}+dfrac{1}{5}-dfrac{1}{6}+cdots + dfrac{1}{n-2}-dfrac{1}{n-1}+dfrac{1}{n}$. $a_n$ converges when considered as a partial sum of an alternating harmonic series.
answered Nov 24 at 18:24
DeepSea
70.9k54487
Let $a_n = displaystyle int_{0}^1 dfrac{x^n}{1+x}dx$. Since $dfrac{x^n}{1+x} = x^{n-1}left(1-dfrac{1}{1+x}right)=x^{n-1}-dfrac{x^{n-1}}{1+x}$, taking the integral $displaystyle int_{0}^1$ both sides give: $a_n = dfrac{1}{n}- a_{n-1}=dfrac{1}{n}-dfrac{1}{n-1}+a_{n-2}= dfrac{1}{n}-dfrac{1}{n-1}+dfrac{1}{n-2}-a_{n-3}=...=-ln 2+ 1-dfrac{1}{2}+dfrac{1}{3}-dfrac{1}{4}+dfrac{1}{5}-dfrac{1}{6}+cdots + dfrac{1}{n-2}-dfrac{1}{n-1}+dfrac{1}{n}$. $a_n$ converges when considered as a partial sum of an alternating harmonic series.
answered Nov 24 at 18:24
DeepSea
70.9k54487
edited Nov 24 at 18:46
answered Nov 24 at 18:24
DeepSea
70.9k54487
answered Nov 24 at 18:24
DeepSea
70.9k54487
answered Nov 24 at 18:24
DeepSea
70.9k54487
70.9k54487
add a comment |
add a comment |
Since $x^n$ is continuous on $[0,1]$ for $nge 0,$ the integral converges for $nge 0.$
For $n<0,$ $x^n$ blows up at $0.$ So we need to consider the integral over $[a,1]$ for small $a>0.$ Notice that for $xin (0,1],$
$$frac{x^n}{2}le frac{x^n}{1+x} le x^n.$$
It follows that the integral of interest converges iff $int_0^1 x^n,dx $ converges. Things are easy now, so I'll stop here. Ask questions if you like.
answered Nov 24 at 19:04
zhw.
71.6k43075
add a comment |
Since $x^n$ is continuous on $[0,1]$ for $nge 0,$ the integral converges for $nge 0.$
For $n<0,$ $x^n$ blows up at $0.$ So we need to consider the integral over $[a,1]$ for small $a>0.$ Notice that for $xin (0,1],$
$$frac{x^n}{2}le frac{x^n}{1+x} le x^n.$$
It follows that the integral of interest converges iff $int_0^1 x^n,dx $ converges. Things are easy now, so I'll stop here. Ask questions if you like.
answered Nov 24 at 19:04
zhw.
71.6k43075
add a comment |
Since $x^n$ is continuous on $[0,1]$ for $nge 0,$ the integral converges for $nge 0.$
For $n<0,$ $x^n$ blows up at $0.$ So we need to consider the integral over $[a,1]$ for small $a>0.$ Notice that for $xin (0,1],$
$$frac{x^n}{2}le frac{x^n}{1+x} le x^n.$$
It follows that the integral of interest converges iff $int_0^1 x^n,dx $ converges. Things are easy now, so I'll stop here. Ask questions if you like.
answered Nov 24 at 19:04
zhw.
71.6k43075
Since $x^n$ is continuous on $[0,1]$ for $nge 0,$ the integral converges for $nge 0.$
For $n<0,$ $x^n$ blows up at $0.$ So we need to consider the integral over $[a,1]$ for small $a>0.$ Notice that for $xin (0,1],$
$$frac{x^n}{2}le frac{x^n}{1+x} le x^n.$$
It follows that the integral of interest converges iff $int_0^1 x^n,dx $ converges. Things are easy now, so I'll stop here. Ask questions if you like.
answered Nov 24 at 19:04
zhw.
71.6k43075
answered Nov 24 at 19:04
zhw.
71.6k43075
answered Nov 24 at 19:04
zhw.
71.6k43075
answered Nov 24 at 19:04
zhw.
71.6k43075
71.6k43075
add a comment |
add a comment |
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