How to make a multiple integration by parts rigourous
$begingroup$
Sometimes, in order to prove an identity about the integral of a function we can apply multiple time integration by parts until we integrate an easy function.
For example it's a nice technique with polynomial. Let's say I want to calculate the following :
$$int_{-1}^1 left(frac{1}{2^j j!}right)^2 left(frac{d^j}{dx^j} (x^2-1)^jright)^2 mathrm{d}x$$
Then clearly I can do integration by part multiple times until I get an easy integral. So here by doing one integration by part the problem boils down to calculating :
begin{align*}
&left(frac{1}{2^jj!}right)^2 Bigg( left[left(frac{d^{j-1}}{dx^{j-1}} (x^2-1)^jright)frac{d^j}{dx^j} (x^2-1)^jright]_{-1}^1 \
&hspace{5em} -int_{-1}^1 left(frac{d^{j-1}}{dx^{j-1}} (x^2-1)^jright)left(frac{d^{j+1}}{dx^{j+1}} (x^2-1)^jright) mathrm{d}xBigg) \
&= left(frac{1}{2^jj!}right)^2 int_{-1}^1 left(frac{d^{j-1}}{dx^{j-1}} (x^2-1)^jright)left(frac{d^{j+1}}{dx^{j+1}} (x^2-1)^jright) mathrm{d}x
end{align*}
as you can see I am going to continue integrating by part until one of the two polynomial cancel out.
The problem is how to prove rigourously all the steps. Normally I just put dots : "$ldots$" just to say : we continue doing the same thing until we get to an easy integral.
But this is not rigourous at all. An idea to make it rigourous is to do an induction, but the problem is that induction is long and a bit cumbersome, that's why is there a way to make the above argument rigourous without using induction ?
For example, if we can express multiple integration by parts as a sum it would clearly make the above argument rigourous.
Thank you !
real-analysis calculus integration summation
edited Jan 23 at 12:26
Sangchul Lee
95.4k12171278
asked Jan 23 at 12:16
dghkgfzyukzdghkgfzyukz
16112
$endgroup$
add a comment |
$begingroup$
Sometimes, in order to prove an identity about the integral of a function we can apply multiple time integration by parts until we integrate an easy function.
For example it's a nice technique with polynomial. Let's say I want to calculate the following :
$$int_{-1}^1 left(frac{1}{2^j j!}right)^2 left(frac{d^j}{dx^j} (x^2-1)^jright)^2 mathrm{d}x$$
Then clearly I can do integration by part multiple times until I get an easy integral. So here by doing one integration by part the problem boils down to calculating :
begin{align*}
&left(frac{1}{2^jj!}right)^2 Bigg( left[left(frac{d^{j-1}}{dx^{j-1}} (x^2-1)^jright)frac{d^j}{dx^j} (x^2-1)^jright]_{-1}^1 \
&hspace{5em} -int_{-1}^1 left(frac{d^{j-1}}{dx^{j-1}} (x^2-1)^jright)left(frac{d^{j+1}}{dx^{j+1}} (x^2-1)^jright) mathrm{d}xBigg) \
&= left(frac{1}{2^jj!}right)^2 int_{-1}^1 left(frac{d^{j-1}}{dx^{j-1}} (x^2-1)^jright)left(frac{d^{j+1}}{dx^{j+1}} (x^2-1)^jright) mathrm{d}x
end{align*}
as you can see I am going to continue integrating by part until one of the two polynomial cancel out.
The problem is how to prove rigourously all the steps. Normally I just put dots : "$ldots$" just to say : we continue doing the same thing until we get to an easy integral.
But this is not rigourous at all. An idea to make it rigourous is to do an induction, but the problem is that induction is long and a bit cumbersome, that's why is there a way to make the above argument rigourous without using induction ?
For example, if we can express multiple integration by parts as a sum it would clearly make the above argument rigourous.
Thank you !
real-analysis calculus integration summation
edited Jan 23 at 12:26
Sangchul Lee
95.4k12171278
asked Jan 23 at 12:16
dghkgfzyukzdghkgfzyukz
16112
$endgroup$
$begingroup$
How is $X$ related to $x$?
$endgroup$
– J.G.
Jan 23 at 12:24
1
$begingroup$
For OP and @J.G., I edited the question based on common-sense, hope you do not mind it.
$endgroup$
– Sangchul Lee
Jan 23 at 12:28
$begingroup$
Ever heard of the tabular method? It's exactly what you're describing. I suspect induction is lurking in the proof of that, unfortunately.
$endgroup$
– Cameron Williams
Jan 23 at 12:31
$begingroup$
@SangchulLee This is perfect, thank you for the edit !
$endgroup$
– dghkgfzyukz
Jan 23 at 13:34
$begingroup$
@CameronWilliams Never heard of this before. I looked on the internet to see what it was and I must say it's a nice technique to not mess up duriong the calculations. Thank you for talking about it. So maybe yes, inudction is forced. But I find it quite surprising that there isn't any general formula for multiple integration by parts... (for example for a multiple derivation of $uv$ we have Leibniz formula), but nothing for IPP ? :(
$endgroup$
– dghkgfzyukz
Jan 23 at 13:37
add a comment |
$begingroup$
Sometimes, in order to prove an identity about the integral of a function we can apply multiple time integration by parts until we integrate an easy function.
For example it's a nice technique with polynomial. Let's say I want to calculate the following :
$$int_{-1}^1 left(frac{1}{2^j j!}right)^2 left(frac{d^j}{dx^j} (x^2-1)^jright)^2 mathrm{d}x$$
Then clearly I can do integration by part multiple times until I get an easy integral. So here by doing one integration by part the problem boils down to calculating :
begin{align*}
&left(frac{1}{2^jj!}right)^2 Bigg( left[left(frac{d^{j-1}}{dx^{j-1}} (x^2-1)^jright)frac{d^j}{dx^j} (x^2-1)^jright]_{-1}^1 \
&hspace{5em} -int_{-1}^1 left(frac{d^{j-1}}{dx^{j-1}} (x^2-1)^jright)left(frac{d^{j+1}}{dx^{j+1}} (x^2-1)^jright) mathrm{d}xBigg) \
&= left(frac{1}{2^jj!}right)^2 int_{-1}^1 left(frac{d^{j-1}}{dx^{j-1}} (x^2-1)^jright)left(frac{d^{j+1}}{dx^{j+1}} (x^2-1)^jright) mathrm{d}x
end{align*}
as you can see I am going to continue integrating by part until one of the two polynomial cancel out.
The problem is how to prove rigourously all the steps. Normally I just put dots : "$ldots$" just to say : we continue doing the same thing until we get to an easy integral.
But this is not rigourous at all. An idea to make it rigourous is to do an induction, but the problem is that induction is long and a bit cumbersome, that's why is there a way to make the above argument rigourous without using induction ?
For example, if we can express multiple integration by parts as a sum it would clearly make the above argument rigourous.
Thank you !
real-analysis calculus integration summation
edited Jan 23 at 12:26
Sangchul Lee
95.4k12171278
asked Jan 23 at 12:16
dghkgfzyukzdghkgfzyukz
16112
$endgroup$
Sometimes, in order to prove an identity about the integral of a function we can apply multiple time integration by parts until we integrate an easy function.
For example it's a nice technique with polynomial. Let's say I want to calculate the following :
$$int_{-1}^1 left(frac{1}{2^j j!}right)^2 left(frac{d^j}{dx^j} (x^2-1)^jright)^2 mathrm{d}x$$
Then clearly I can do integration by part multiple times until I get an easy integral. So here by doing one integration by part the problem boils down to calculating :
begin{align*}
&left(frac{1}{2^jj!}right)^2 Bigg( left[left(frac{d^{j-1}}{dx^{j-1}} (x^2-1)^jright)frac{d^j}{dx^j} (x^2-1)^jright]_{-1}^1 \
&hspace{5em} -int_{-1}^1 left(frac{d^{j-1}}{dx^{j-1}} (x^2-1)^jright)left(frac{d^{j+1}}{dx^{j+1}} (x^2-1)^jright) mathrm{d}xBigg) \
&= left(frac{1}{2^jj!}right)^2 int_{-1}^1 left(frac{d^{j-1}}{dx^{j-1}} (x^2-1)^jright)left(frac{d^{j+1}}{dx^{j+1}} (x^2-1)^jright) mathrm{d}x
end{align*}
as you can see I am going to continue integrating by part until one of the two polynomial cancel out.
The problem is how to prove rigourously all the steps. Normally I just put dots : "$ldots$" just to say : we continue doing the same thing until we get to an easy integral.
But this is not rigourous at all. An idea to make it rigourous is to do an induction, but the problem is that induction is long and a bit cumbersome, that's why is there a way to make the above argument rigourous without using induction ?
For example, if we can express multiple integration by parts as a sum it would clearly make the above argument rigourous.
Thank you !
real-analysis calculus integration summation
real-analysis calculus integration summation
edited Jan 23 at 12:26
Sangchul Lee
95.4k12171278
asked Jan 23 at 12:16
dghkgfzyukzdghkgfzyukz
16112
edited Jan 23 at 12:26
Sangchul Lee
95.4k12171278
asked Jan 23 at 12:16
dghkgfzyukzdghkgfzyukz
16112
edited Jan 23 at 12:26
Sangchul Lee
95.4k12171278
edited Jan 23 at 12:26
Sangchul Lee
95.4k12171278
edited Jan 23 at 12:26
Sangchul Lee
95.4k12171278
95.4k12171278
asked Jan 23 at 12:16
dghkgfzyukzdghkgfzyukz
16112
asked Jan 23 at 12:16
dghkgfzyukzdghkgfzyukz
16112
asked Jan 23 at 12:16
dghkgfzyukzdghkgfzyukz
16112
16112
$begingroup$
How is $X$ related to $x$?
$endgroup$
– J.G.
Jan 23 at 12:24
1
$begingroup$
For OP and @J.G., I edited the question based on common-sense, hope you do not mind it.
$endgroup$
– Sangchul Lee
Jan 23 at 12:28
$begingroup$
Ever heard of the tabular method? It's exactly what you're describing. I suspect induction is lurking in the proof of that, unfortunately.
$endgroup$
– Cameron Williams
Jan 23 at 12:31
$begingroup$
@SangchulLee This is perfect, thank you for the edit !
$endgroup$
– dghkgfzyukz
Jan 23 at 13:34
$begingroup$
@CameronWilliams Never heard of this before. I looked on the internet to see what it was and I must say it's a nice technique to not mess up duriong the calculations. Thank you for talking about it. So maybe yes, inudction is forced. But I find it quite surprising that there isn't any general formula for multiple integration by parts... (for example for a multiple derivation of $uv$ we have Leibniz formula), but nothing for IPP ? :(
$endgroup$
– dghkgfzyukz
Jan 23 at 13:37
add a comment |
$begingroup$
How is $X$ related to $x$?
$endgroup$
– J.G.
Jan 23 at 12:24
1
$begingroup$
For OP and @J.G., I edited the question based on common-sense, hope you do not mind it.
$endgroup$
– Sangchul Lee
Jan 23 at 12:28
$begingroup$
Ever heard of the tabular method? It's exactly what you're describing. I suspect induction is lurking in the proof of that, unfortunately.
$endgroup$
– Cameron Williams
Jan 23 at 12:31
$begingroup$
@SangchulLee This is perfect, thank you for the edit !
$endgroup$
– dghkgfzyukz
Jan 23 at 13:34
$begingroup$
@CameronWilliams Never heard of this before. I looked on the internet to see what it was and I must say it's a nice technique to not mess up duriong the calculations. Thank you for talking about it. So maybe yes, inudction is forced. But I find it quite surprising that there isn't any general formula for multiple integration by parts... (for example for a multiple derivation of $uv$ we have Leibniz formula), but nothing for IPP ? :(
$endgroup$
– dghkgfzyukz
Jan 23 at 13:37
$begingroup$
How is $X$ related to $x$?
$endgroup$
– J.G.
Jan 23 at 12:24
$begingroup$
How is $X$ related to $x$?
$endgroup$
– J.G.
Jan 23 at 12:24
1
1
$begingroup$
For OP and @J.G., I edited the question based on common-sense, hope you do not mind it.
$endgroup$
– Sangchul Lee
Jan 23 at 12:28
$begingroup$
For OP and @J.G., I edited the question based on common-sense, hope you do not mind it.
$endgroup$
– Sangchul Lee
Jan 23 at 12:28
$begingroup$
Ever heard of the tabular method? It's exactly what you're describing. I suspect induction is lurking in the proof of that, unfortunately.
$endgroup$
– Cameron Williams
Jan 23 at 12:31
$begingroup$
Ever heard of the tabular method? It's exactly what you're describing. I suspect induction is lurking in the proof of that, unfortunately.
$endgroup$
– Cameron Williams
Jan 23 at 12:31
$begingroup$
@SangchulLee This is perfect, thank you for the edit !
$endgroup$
– dghkgfzyukz
Jan 23 at 13:34
$begingroup$
@SangchulLee This is perfect, thank you for the edit !
$endgroup$
– dghkgfzyukz
Jan 23 at 13:34
$begingroup$
@CameronWilliams Never heard of this before. I looked on the internet to see what it was and I must say it's a nice technique to not mess up duriong the calculations. Thank you for talking about it. So maybe yes, inudction is forced. But I find it quite surprising that there isn't any general formula for multiple integration by parts... (for example for a multiple derivation of $uv$ we have Leibniz formula), but nothing for IPP ? :(
$endgroup$
– dghkgfzyukz
Jan 23 at 13:37
$begingroup$
@CameronWilliams Never heard of this before. I looked on the internet to see what it was and I must say it's a nice technique to not mess up duriong the calculations. Thank you for talking about it. So maybe yes, inudction is forced. But I find it quite surprising that there isn't any general formula for multiple integration by parts... (for example for a multiple derivation of $uv$ we have Leibniz formula), but nothing for IPP ? :(
$endgroup$
– dghkgfzyukz
Jan 23 at 13:37
add a comment |
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$begingroup$
How is $X$ related to $x$?
$endgroup$
– J.G.
Jan 23 at 12:24
1
$begingroup$
For OP and @J.G., I edited the question based on common-sense, hope you do not mind it.
$endgroup$
– Sangchul Lee
Jan 23 at 12:28
$begingroup$
Ever heard of the tabular method? It's exactly what you're describing. I suspect induction is lurking in the proof of that, unfortunately.
$endgroup$
– Cameron Williams
Jan 23 at 12:31
$begingroup$
@SangchulLee This is perfect, thank you for the edit !
$endgroup$
– dghkgfzyukz
Jan 23 at 13:34
$begingroup$
@CameronWilliams Never heard of this before. I looked on the internet to see what it was and I must say it's a nice technique to not mess up duriong the calculations. Thank you for talking about it. So maybe yes, inudction is forced. But I find it quite surprising that there isn't any general formula for multiple integration by parts... (for example for a multiple derivation of $uv$ we have Leibniz formula), but nothing for IPP ? :(
$endgroup$
– dghkgfzyukz
Jan 23 at 13:37