Mixing
Convergent integral of divergent function
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On one of my calculus lectures I was told that there exist convergent improper integrals (in infinity) of divergent function. I was searching for an example in the internet, but I didn't find any. Has any of you idea how would this kind of function look like? Thanks in advance
calculus functions convergence improper-integrals
edited Jan 19 at 4:32
BigM
2,57411530
asked Mar 12 '14 at 21:26
bartopbartop
1135
$endgroup$
add a comment |
$begingroup$
On one of my calculus lectures I was told that there exist convergent improper integrals (in infinity) of divergent function. I was searching for an example in the internet, but I didn't find any. Has any of you idea how would this kind of function look like? Thanks in advance
calculus functions convergence improper-integrals
edited Jan 19 at 4:32
BigM
2,57411530
asked Mar 12 '14 at 21:26
bartopbartop
1135
$endgroup$
add a comment |
$begingroup$
On one of my calculus lectures I was told that there exist convergent improper integrals (in infinity) of divergent function. I was searching for an example in the internet, but I didn't find any. Has any of you idea how would this kind of function look like? Thanks in advance
calculus functions convergence improper-integrals
edited Jan 19 at 4:32
BigM
2,57411530
asked Mar 12 '14 at 21:26
bartopbartop
1135
$endgroup$
On one of my calculus lectures I was told that there exist convergent improper integrals (in infinity) of divergent function. I was searching for an example in the internet, but I didn't find any. Has any of you idea how would this kind of function look like? Thanks in advance
calculus functions convergence improper-integrals
calculus functions convergence improper-integrals
edited Jan 19 at 4:32
BigM
2,57411530
asked Mar 12 '14 at 21:26
bartopbartop
1135
edited Jan 19 at 4:32
BigM
2,57411530
asked Mar 12 '14 at 21:26
bartopbartop
1135
edited Jan 19 at 4:32
BigM
2,57411530
edited Jan 19 at 4:32
BigM
2,57411530
edited Jan 19 at 4:32
BigM
2,57411530
2,57411530
asked Mar 12 '14 at 21:26
bartopbartop
1135
asked Mar 12 '14 at 21:26
bartopbartop
1135
asked Mar 12 '14 at 21:26
bartopbartop
1135
1135
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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For any positive integer $n$, let $f(x)=0$ at $x=n-frac{1}{2^{n}}$. Let it climb linearly to $1$ at $x=n$, and then fall linearly to $0$ at $x=n+frac{1}{2^{n}}$. Let $f(x)=0$ everywhere else.
So $f$ has fast-narrowing "spikes" of height $1$ around every positive integer. The combined area of these spikes is small.
We can modify the construction to make the maximum height of the spikes become arbitrarily large as $ntoinfty$, for example by replacing $f(n)=1$ by $f(n)=n$.
$endgroup$
$begingroup$
If I may ask, what would be integral of such a function?
$endgroup$
– bartop
Mar 12 '14 at 22:16
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For the first function I gave, the integral from $0$ to $infty$ is the sum of the areas of the triangles. The isosceles triangle centered on $n$ has area $frac{1}{2^n}$. Sum from $n=1$ to $infty$. This is a geometric series, sum $1$. So the definite integral is $1$. The second one I suggested (height $n$ at $n$) turns out to be also a doable sum, it is $2$.
$endgroup$
– André Nicolas
Mar 12 '14 at 23:21
add a comment |
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How about $int_0^1ln(x)dx=-1$;)
answered Mar 12 '14 at 21:42
BigMBigM
2,57411530
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add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
For any positive integer $n$, let $f(x)=0$ at $x=n-frac{1}{2^{n}}$. Let it climb linearly to $1$ at $x=n$, and then fall linearly to $0$ at $x=n+frac{1}{2^{n}}$. Let $f(x)=0$ everywhere else.
So $f$ has fast-narrowing "spikes" of height $1$ around every positive integer. The combined area of these spikes is small.
We can modify the construction to make the maximum height of the spikes become arbitrarily large as $ntoinfty$, for example by replacing $f(n)=1$ by $f(n)=n$.
$endgroup$
$begingroup$
If I may ask, what would be integral of such a function?
$endgroup$
– bartop
Mar 12 '14 at 22:16
$begingroup$
For the first function I gave, the integral from $0$ to $infty$ is the sum of the areas of the triangles. The isosceles triangle centered on $n$ has area $frac{1}{2^n}$. Sum from $n=1$ to $infty$. This is a geometric series, sum $1$. So the definite integral is $1$. The second one I suggested (height $n$ at $n$) turns out to be also a doable sum, it is $2$.
$endgroup$
– André Nicolas
Mar 12 '14 at 23:21
add a comment |
$begingroup$
For any positive integer $n$, let $f(x)=0$ at $x=n-frac{1}{2^{n}}$. Let it climb linearly to $1$ at $x=n$, and then fall linearly to $0$ at $x=n+frac{1}{2^{n}}$. Let $f(x)=0$ everywhere else.
So $f$ has fast-narrowing "spikes" of height $1$ around every positive integer. The combined area of these spikes is small.
We can modify the construction to make the maximum height of the spikes become arbitrarily large as $ntoinfty$, for example by replacing $f(n)=1$ by $f(n)=n$.
$endgroup$
$begingroup$
If I may ask, what would be integral of such a function?
$endgroup$
– bartop
Mar 12 '14 at 22:16
$begingroup$
For the first function I gave, the integral from $0$ to $infty$ is the sum of the areas of the triangles. The isosceles triangle centered on $n$ has area $frac{1}{2^n}$. Sum from $n=1$ to $infty$. This is a geometric series, sum $1$. So the definite integral is $1$. The second one I suggested (height $n$ at $n$) turns out to be also a doable sum, it is $2$.
$endgroup$
– André Nicolas
Mar 12 '14 at 23:21
add a comment |
$begingroup$
For any positive integer $n$, let $f(x)=0$ at $x=n-frac{1}{2^{n}}$. Let it climb linearly to $1$ at $x=n$, and then fall linearly to $0$ at $x=n+frac{1}{2^{n}}$. Let $f(x)=0$ everywhere else.
So $f$ has fast-narrowing "spikes" of height $1$ around every positive integer. The combined area of these spikes is small.
We can modify the construction to make the maximum height of the spikes become arbitrarily large as $ntoinfty$, for example by replacing $f(n)=1$ by $f(n)=n$.
$endgroup$
For any positive integer $n$, let $f(x)=0$ at $x=n-frac{1}{2^{n}}$. Let it climb linearly to $1$ at $x=n$, and then fall linearly to $0$ at $x=n+frac{1}{2^{n}}$. Let $f(x)=0$ everywhere else.
So $f$ has fast-narrowing "spikes" of height $1$ around every positive integer. The combined area of these spikes is small.
We can modify the construction to make the maximum height of the spikes become arbitrarily large as $ntoinfty$, for example by replacing $f(n)=1$ by $f(n)=n$.
answered Mar 12 '14 at 21:38
community wiki
André Nicolas
$begingroup$
If I may ask, what would be integral of such a function?
$endgroup$
– bartop
Mar 12 '14 at 22:16
$begingroup$
For the first function I gave, the integral from $0$ to $infty$ is the sum of the areas of the triangles. The isosceles triangle centered on $n$ has area $frac{1}{2^n}$. Sum from $n=1$ to $infty$. This is a geometric series, sum $1$. So the definite integral is $1$. The second one I suggested (height $n$ at $n$) turns out to be also a doable sum, it is $2$.
$endgroup$
– André Nicolas
Mar 12 '14 at 23:21
add a comment |
$begingroup$
If I may ask, what would be integral of such a function?
$endgroup$
– bartop
Mar 12 '14 at 22:16
$begingroup$
For the first function I gave, the integral from $0$ to $infty$ is the sum of the areas of the triangles. The isosceles triangle centered on $n$ has area $frac{1}{2^n}$. Sum from $n=1$ to $infty$. This is a geometric series, sum $1$. So the definite integral is $1$. The second one I suggested (height $n$ at $n$) turns out to be also a doable sum, it is $2$.
$endgroup$
– André Nicolas
Mar 12 '14 at 23:21
$begingroup$
If I may ask, what would be integral of such a function?
$endgroup$
– bartop
Mar 12 '14 at 22:16
$begingroup$
If I may ask, what would be integral of such a function?
$endgroup$
– bartop
Mar 12 '14 at 22:16
$begingroup$
For the first function I gave, the integral from $0$ to $infty$ is the sum of the areas of the triangles. The isosceles triangle centered on $n$ has area $frac{1}{2^n}$. Sum from $n=1$ to $infty$. This is a geometric series, sum $1$. So the definite integral is $1$. The second one I suggested (height $n$ at $n$) turns out to be also a doable sum, it is $2$.
$endgroup$
– André Nicolas
Mar 12 '14 at 23:21
$begingroup$
For the first function I gave, the integral from $0$ to $infty$ is the sum of the areas of the triangles. The isosceles triangle centered on $n$ has area $frac{1}{2^n}$. Sum from $n=1$ to $infty$. This is a geometric series, sum $1$. So the definite integral is $1$. The second one I suggested (height $n$ at $n$) turns out to be also a doable sum, it is $2$.
$endgroup$
– André Nicolas
Mar 12 '14 at 23:21
add a comment |
$begingroup$
How about $int_0^1ln(x)dx=-1$;)
answered Mar 12 '14 at 21:42
BigMBigM
2,57411530
$endgroup$
add a comment |
$begingroup$
How about $int_0^1ln(x)dx=-1$;)
answered Mar 12 '14 at 21:42
BigMBigM
2,57411530
$endgroup$
add a comment |
$begingroup$
How about $int_0^1ln(x)dx=-1$;)
answered Mar 12 '14 at 21:42
BigMBigM
2,57411530
$endgroup$
How about $int_0^1ln(x)dx=-1$;)
answered Mar 12 '14 at 21:42
BigMBigM
2,57411530
answered Mar 12 '14 at 21:42
BigMBigM
2,57411530
answered Mar 12 '14 at 21:42
BigMBigM
2,57411530
answered Mar 12 '14 at 21:42
BigMBigM
2,57411530
2,57411530
add a comment |
add a comment |
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