Proof integration identity $int_{0}^{1}dxint_{0}^{x}e^{x^2}dy=int_{0}^{1}dyint_{y}^{1}e^{x^2}dx$












0












$begingroup$


I have to prove this identity:
$$int_{0}^{1}dxint_{0}^{x}e^{x^2}dy=int_{0}^{1}dyint_{y}^{1}e^{x^2}dx$$
I've shown that:
$$int_{0}^{1}dxint_{0}^{x}e^{x^2}dy=int_{0}^{1}xe^{x^2}dx=frac{1}{2}(e-1)$$



After i tried to solve the second member of identity in this way:



The first part is done by substituting $x^2$ in series
$$ e^{x} = 1 + x + frac{x^2}{2!} + frac{x^3}{3!} + cdots $$
yielding
$$e^{x^2} = sum_{n=0}^{infty} frac{x^{2n}}{n!} $$
Through some theorems as uniform continuity, then we can switch the order of integration and summation, that is:
$$int_{y}^{1} sum_{n=0}^{infty} frac{x^{2n}}{n!}dx= sum_{n=0}^{infty}frac{1}{n!}int_{y}^{1} x^{2n}dx=sum_{n=0}^{infty}frac{1}{n!}*Big(1-frac{y^{2n+1}}{2n+1}Big)dx$$
And from here i'm not sure how to proceed. What can i do? Thanks for the help in advance!!






real-analysis integration derivatives power-series







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  • 1




    $begingroup$
    Note that $int_{0}^{1}dx=int_{0}^{1}dy$, so you only need to show $int_{0}^{x}e^{t^2}dt=int_{y}^{1}e^{t^2}dt$. Now, they are not equal as functions of $x$ and $y$
    $endgroup$
    – alexp9
    Jan 23 at 11:43










  • $begingroup$
    Beware in your notations. What you wrote in first line are products of 1-dim integrals but on second line you consider a 2-dim integral. Use parenthesis and let the $dx,dy$ at the end for 2-dim integrals. Thus what you are asking is not clear.
    $endgroup$
    – zwim
    Jan 23 at 12:01
















0












$begingroup$


I have to prove this identity:
$$int_{0}^{1}dxint_{0}^{x}e^{x^2}dy=int_{0}^{1}dyint_{y}^{1}e^{x^2}dx$$
I've shown that:
$$int_{0}^{1}dxint_{0}^{x}e^{x^2}dy=int_{0}^{1}xe^{x^2}dx=frac{1}{2}(e-1)$$



After i tried to solve the second member of identity in this way:



The first part is done by substituting $x^2$ in series
$$ e^{x} = 1 + x + frac{x^2}{2!} + frac{x^3}{3!} + cdots $$
yielding
$$e^{x^2} = sum_{n=0}^{infty} frac{x^{2n}}{n!} $$
Through some theorems as uniform continuity, then we can switch the order of integration and summation, that is:
$$int_{y}^{1} sum_{n=0}^{infty} frac{x^{2n}}{n!}dx= sum_{n=0}^{infty}frac{1}{n!}int_{y}^{1} x^{2n}dx=sum_{n=0}^{infty}frac{1}{n!}*Big(1-frac{y^{2n+1}}{2n+1}Big)dx$$
And from here i'm not sure how to proceed. What can i do? Thanks for the help in advance!!






real-analysis integration derivatives power-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that $int_{0}^{1}dx=int_{0}^{1}dy$, so you only need to show $int_{0}^{x}e^{t^2}dt=int_{y}^{1}e^{t^2}dt$. Now, they are not equal as functions of $x$ and $y$
    $endgroup$
    – alexp9
    Jan 23 at 11:43










  • $begingroup$
    Beware in your notations. What you wrote in first line are products of 1-dim integrals but on second line you consider a 2-dim integral. Use parenthesis and let the $dx,dy$ at the end for 2-dim integrals. Thus what you are asking is not clear.
    $endgroup$
    – zwim
    Jan 23 at 12:01














0












0








0





$begingroup$


I have to prove this identity:
$$int_{0}^{1}dxint_{0}^{x}e^{x^2}dy=int_{0}^{1}dyint_{y}^{1}e^{x^2}dx$$
I've shown that:
$$int_{0}^{1}dxint_{0}^{x}e^{x^2}dy=int_{0}^{1}xe^{x^2}dx=frac{1}{2}(e-1)$$



After i tried to solve the second member of identity in this way:



The first part is done by substituting $x^2$ in series
$$ e^{x} = 1 + x + frac{x^2}{2!} + frac{x^3}{3!} + cdots $$
yielding
$$e^{x^2} = sum_{n=0}^{infty} frac{x^{2n}}{n!} $$
Through some theorems as uniform continuity, then we can switch the order of integration and summation, that is:
$$int_{y}^{1} sum_{n=0}^{infty} frac{x^{2n}}{n!}dx= sum_{n=0}^{infty}frac{1}{n!}int_{y}^{1} x^{2n}dx=sum_{n=0}^{infty}frac{1}{n!}*Big(1-frac{y^{2n+1}}{2n+1}Big)dx$$
And from here i'm not sure how to proceed. What can i do? Thanks for the help in advance!!






real-analysis integration derivatives power-series







share|cite|improve this question











$endgroup$




I have to prove this identity:
$$int_{0}^{1}dxint_{0}^{x}e^{x^2}dy=int_{0}^{1}dyint_{y}^{1}e^{x^2}dx$$
I've shown that:
$$int_{0}^{1}dxint_{0}^{x}e^{x^2}dy=int_{0}^{1}xe^{x^2}dx=frac{1}{2}(e-1)$$



After i tried to solve the second member of identity in this way:



The first part is done by substituting $x^2$ in series
$$ e^{x} = 1 + x + frac{x^2}{2!} + frac{x^3}{3!} + cdots $$
yielding
$$e^{x^2} = sum_{n=0}^{infty} frac{x^{2n}}{n!} $$
Through some theorems as uniform continuity, then we can switch the order of integration and summation, that is:
$$int_{y}^{1} sum_{n=0}^{infty} frac{x^{2n}}{n!}dx= sum_{n=0}^{infty}frac{1}{n!}int_{y}^{1} x^{2n}dx=sum_{n=0}^{infty}frac{1}{n!}*Big(1-frac{y^{2n+1}}{2n+1}Big)dx$$
And from here i'm not sure how to proceed. What can i do? Thanks for the help in advance!!






real-analysis integration derivatives power-series




real-analysis integration derivatives power-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 11:37







andrew

















asked Jan 23 at 11:32









andrewandrew

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698








  • 1




    $begingroup$
    Note that $int_{0}^{1}dx=int_{0}^{1}dy$, so you only need to show $int_{0}^{x}e^{t^2}dt=int_{y}^{1}e^{t^2}dt$. Now, they are not equal as functions of $x$ and $y$
    $endgroup$
    – alexp9
    Jan 23 at 11:43










  • $begingroup$
    Beware in your notations. What you wrote in first line are products of 1-dim integrals but on second line you consider a 2-dim integral. Use parenthesis and let the $dx,dy$ at the end for 2-dim integrals. Thus what you are asking is not clear.
    $endgroup$
    – zwim
    Jan 23 at 12:01














  • 1




    $begingroup$
    Note that $int_{0}^{1}dx=int_{0}^{1}dy$, so you only need to show $int_{0}^{x}e^{t^2}dt=int_{y}^{1}e^{t^2}dt$. Now, they are not equal as functions of $x$ and $y$
    $endgroup$
    – alexp9
    Jan 23 at 11:43










  • $begingroup$
    Beware in your notations. What you wrote in first line are products of 1-dim integrals but on second line you consider a 2-dim integral. Use parenthesis and let the $dx,dy$ at the end for 2-dim integrals. Thus what you are asking is not clear.
    $endgroup$
    – zwim
    Jan 23 at 12:01








1




1




$begingroup$
Note that $int_{0}^{1}dx=int_{0}^{1}dy$, so you only need to show $int_{0}^{x}e^{t^2}dt=int_{y}^{1}e^{t^2}dt$. Now, they are not equal as functions of $x$ and $y$
$endgroup$
– alexp9
Jan 23 at 11:43




$begingroup$
Note that $int_{0}^{1}dx=int_{0}^{1}dy$, so you only need to show $int_{0}^{x}e^{t^2}dt=int_{y}^{1}e^{t^2}dt$. Now, they are not equal as functions of $x$ and $y$
$endgroup$
– alexp9
Jan 23 at 11:43












$begingroup$
Beware in your notations. What you wrote in first line are products of 1-dim integrals but on second line you consider a 2-dim integral. Use parenthesis and let the $dx,dy$ at the end for 2-dim integrals. Thus what you are asking is not clear.
$endgroup$
– zwim
Jan 23 at 12:01




$begingroup$
Beware in your notations. What you wrote in first line are products of 1-dim integrals but on second line you consider a 2-dim integral. Use parenthesis and let the $dx,dy$ at the end for 2-dim integrals. Thus what you are asking is not clear.
$endgroup$
– zwim
Jan 23 at 12:01










1 Answer
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$begingroup$

Trying to calculate the iterated integrals is not the way here - one of those integrals will work in an elementary way, but the other won't.



No, this is a case of Fubini's theorem - the two integrals are the same because they're the same double integral over a triangle, integrated in the two possible orders. We wish to show that
$$int_0^1 int_0^x f(x,y),dy,dx = int_0^1int_y^1 f(x,y),dx,dy$$
for nice enough $f$. The function we're trying to integrate is continuous and bounded on this bounded set, and that's certainly nice enough.



Draw the picture:



Triangle



Integrating over $y$ first, our condition is that $0le yle x$. Integrating over $x$ first, our condition is that $yle xle 1$. Then, in both cases, the outer variable runs from $0$ to $1$.






share|cite|improve this answer









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  • $begingroup$
    thank you for your exposure ! @jmerry
    $endgroup$
    – andrew
    Jan 23 at 12:11













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

Trying to calculate the iterated integrals is not the way here - one of those integrals will work in an elementary way, but the other won't.



No, this is a case of Fubini's theorem - the two integrals are the same because they're the same double integral over a triangle, integrated in the two possible orders. We wish to show that
$$int_0^1 int_0^x f(x,y),dy,dx = int_0^1int_y^1 f(x,y),dx,dy$$
for nice enough $f$. The function we're trying to integrate is continuous and bounded on this bounded set, and that's certainly nice enough.



Draw the picture:



Triangle



Integrating over $y$ first, our condition is that $0le yle x$. Integrating over $x$ first, our condition is that $yle xle 1$. Then, in both cases, the outer variable runs from $0$ to $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you for your exposure ! @jmerry
    $endgroup$
    – andrew
    Jan 23 at 12:11


















3












$begingroup$

Trying to calculate the iterated integrals is not the way here - one of those integrals will work in an elementary way, but the other won't.



No, this is a case of Fubini's theorem - the two integrals are the same because they're the same double integral over a triangle, integrated in the two possible orders. We wish to show that
$$int_0^1 int_0^x f(x,y),dy,dx = int_0^1int_y^1 f(x,y),dx,dy$$
for nice enough $f$. The function we're trying to integrate is continuous and bounded on this bounded set, and that's certainly nice enough.



Draw the picture:



Triangle



Integrating over $y$ first, our condition is that $0le yle x$. Integrating over $x$ first, our condition is that $yle xle 1$. Then, in both cases, the outer variable runs from $0$ to $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you for your exposure ! @jmerry
    $endgroup$
    – andrew
    Jan 23 at 12:11
















3












3








3





$begingroup$

Trying to calculate the iterated integrals is not the way here - one of those integrals will work in an elementary way, but the other won't.



No, this is a case of Fubini's theorem - the two integrals are the same because they're the same double integral over a triangle, integrated in the two possible orders. We wish to show that
$$int_0^1 int_0^x f(x,y),dy,dx = int_0^1int_y^1 f(x,y),dx,dy$$
for nice enough $f$. The function we're trying to integrate is continuous and bounded on this bounded set, and that's certainly nice enough.



Draw the picture:



Triangle



Integrating over $y$ first, our condition is that $0le yle x$. Integrating over $x$ first, our condition is that $yle xle 1$. Then, in both cases, the outer variable runs from $0$ to $1$.






share|cite|improve this answer









$endgroup$



Trying to calculate the iterated integrals is not the way here - one of those integrals will work in an elementary way, but the other won't.



No, this is a case of Fubini's theorem - the two integrals are the same because they're the same double integral over a triangle, integrated in the two possible orders. We wish to show that
$$int_0^1 int_0^x f(x,y),dy,dx = int_0^1int_y^1 f(x,y),dx,dy$$
for nice enough $f$. The function we're trying to integrate is continuous and bounded on this bounded set, and that's certainly nice enough.



Draw the picture:



Triangle



Integrating over $y$ first, our condition is that $0le yle x$. Integrating over $x$ first, our condition is that $yle xle 1$. Then, in both cases, the outer variable runs from $0$ to $1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 23 at 11:52









jmerryjmerry

11.8k1527




11.8k1527












  • $begingroup$
    thank you for your exposure ! @jmerry
    $endgroup$
    – andrew
    Jan 23 at 12:11




















  • $begingroup$
    thank you for your exposure ! @jmerry
    $endgroup$
    – andrew
    Jan 23 at 12:11


















$begingroup$
thank you for your exposure ! @jmerry
$endgroup$
– andrew
Jan 23 at 12:11






$begingroup$
thank you for your exposure ! @jmerry
$endgroup$
– andrew
Jan 23 at 12:11




















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