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What is the identity gained after these manipulations?
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Background
Consider the following divergent integral:
$$ int_-infty^infty e^-e^x e^ax x^b dx $$
Where $a$ and $b$ are positive integers such that $b > a$. Multipling $e^-x cdot e^x$ inside the integral.
$$ int_-infty^infty e^-e^x e^x e^(a-1)x x^b dx $$
Using integration by parts and noticing $(e^-e^x)' = -e^-e^xe^x $:
$$ implies - int_-infty^infty (e^-e^x)' e^(a-1)x x^b dx = - int_-infty^infty e^-e^x (e^(a-1)x x^b)' dx $$
Again multiplying $e^-x cdot e^x$ inside the integral and repeating the process:
$$ = - int_-infty^infty e^-e^x e^x cdot e^-x(e^(a-1)x x^b)' dx $$
$$ = int_-infty^infty (e^-e^x)'cdot e^-x(e^(a-1)x x^b)' dx $$
$$ = int_-infty^infty (e^-e^x)cdot (e^-x(e^(a-1)x x^b)')' dx $$
This trick will only work $a$ times. The leading order term in $x$ will be something like: $ a!int_-infty^infty e^-e^x x^b dx $
My idea was to take:
$$ lim_n to infty fracint_-n^n e^-e^x e^ax x^b dx + (-1)^a (a-1)! int_-n^n e^-e^x x^b dxint_-n^n e^-e^x x^b-1 dx $$
Now one can use two ways two solve this:
The first by using L'hopital rule and differentiating. The other is by using integration by parts. I've tried to do this myself but I've got disagreement between my answers (I think I'm messing up the calculation).
Question
Is my idea correct? What is the identity gained after these manipulations? Furthermore, what is the identity of gained after the $x^k$ coefficient using limits and integration by parts? for:
$$ lim_n to infty fracint_-n^n e^-e^x e^ax x^b dx + (-1)^a (a-1)! int_-n^n e^-e^x x^b dxint_-n^n e^-e^x x^b-1 dx = ?$$
calculus integration limits
asked Jul 28 at 9:09
More Anonymous
1387
add a comment |Â
up vote
0
down vote
favorite
Background
Consider the following divergent integral:
$$ int_-infty^infty e^-e^x e^ax x^b dx $$
Where $a$ and $b$ are positive integers such that $b > a$. Multipling $e^-x cdot e^x$ inside the integral.
$$ int_-infty^infty e^-e^x e^x e^(a-1)x x^b dx $$
Using integration by parts and noticing $(e^-e^x)' = -e^-e^xe^x $:
$$ implies - int_-infty^infty (e^-e^x)' e^(a-1)x x^b dx = - int_-infty^infty e^-e^x (e^(a-1)x x^b)' dx $$
Again multiplying $e^-x cdot e^x$ inside the integral and repeating the process:
$$ = - int_-infty^infty e^-e^x e^x cdot e^-x(e^(a-1)x x^b)' dx $$
$$ = int_-infty^infty (e^-e^x)'cdot e^-x(e^(a-1)x x^b)' dx $$
$$ = int_-infty^infty (e^-e^x)cdot (e^-x(e^(a-1)x x^b)')' dx $$
This trick will only work $a$ times. The leading order term in $x$ will be something like: $ a!int_-infty^infty e^-e^x x^b dx $
My idea was to take:
$$ lim_n to infty fracint_-n^n e^-e^x e^ax x^b dx + (-1)^a (a-1)! int_-n^n e^-e^x x^b dxint_-n^n e^-e^x x^b-1 dx $$
Now one can use two ways two solve this:
The first by using L'hopital rule and differentiating. The other is by using integration by parts. I've tried to do this myself but I've got disagreement between my answers (I think I'm messing up the calculation).
Question
Is my idea correct? What is the identity gained after these manipulations? Furthermore, what is the identity of gained after the $x^k$ coefficient using limits and integration by parts? for:
$$ lim_n to infty fracint_-n^n e^-e^x e^ax x^b dx + (-1)^a (a-1)! int_-n^n e^-e^x x^b dxint_-n^n e^-e^x x^b-1 dx = ?$$
calculus integration limits
asked Jul 28 at 9:09
More Anonymous
1387
Why do you say the integral is divergent?
– Julián Aguirre
Jul 28 at 11:26
@JuliánAguirre after repeated integration by parts one will be left with something like: $e^-e^x x^b$ as shown in the background this diverges for $- infty$
– More Anonymous
Jul 28 at 11:38
1
The integral is convergent. When $xto+infty$ it is dominated by $e^-e^x$; when $xto-infty$ by $e^ax$. May be you did not consider the boundary terms in one of the integration by parts.
– Julián Aguirre
Jul 28 at 11:42
1
As Julián Aguirre commented, not only they are convergent but they have very nice expressions.
– Claude Leibovici
Jul 28 at 13:52
@ClaudeLeibovici can u please comment the expression?
– More Anonymous
Jul 28 at 19:17
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Background
Consider the following divergent integral:
$$ int_-infty^infty e^-e^x e^ax x^b dx $$
Where $a$ and $b$ are positive integers such that $b > a$. Multipling $e^-x cdot e^x$ inside the integral.
$$ int_-infty^infty e^-e^x e^x e^(a-1)x x^b dx $$
Using integration by parts and noticing $(e^-e^x)' = -e^-e^xe^x $:
$$ implies - int_-infty^infty (e^-e^x)' e^(a-1)x x^b dx = - int_-infty^infty e^-e^x (e^(a-1)x x^b)' dx $$
Again multiplying $e^-x cdot e^x$ inside the integral and repeating the process:
$$ = - int_-infty^infty e^-e^x e^x cdot e^-x(e^(a-1)x x^b)' dx $$
$$ = int_-infty^infty (e^-e^x)'cdot e^-x(e^(a-1)x x^b)' dx $$
$$ = int_-infty^infty (e^-e^x)cdot (e^-x(e^(a-1)x x^b)')' dx $$
This trick will only work $a$ times. The leading order term in $x$ will be something like: $ a!int_-infty^infty e^-e^x x^b dx $
My idea was to take:
$$ lim_n to infty fracint_-n^n e^-e^x e^ax x^b dx + (-1)^a (a-1)! int_-n^n e^-e^x x^b dxint_-n^n e^-e^x x^b-1 dx $$
Now one can use two ways two solve this:
The first by using L'hopital rule and differentiating. The other is by using integration by parts. I've tried to do this myself but I've got disagreement between my answers (I think I'm messing up the calculation).
Question
Is my idea correct? What is the identity gained after these manipulations? Furthermore, what is the identity of gained after the $x^k$ coefficient using limits and integration by parts? for:
$$ lim_n to infty fracint_-n^n e^-e^x e^ax x^b dx + (-1)^a (a-1)! int_-n^n e^-e^x x^b dxint_-n^n e^-e^x x^b-1 dx = ?$$
calculus integration limits
asked Jul 28 at 9:09
More Anonymous
1387
Background
Consider the following divergent integral:
$$ int_-infty^infty e^-e^x e^ax x^b dx $$
Where $a$ and $b$ are positive integers such that $b > a$. Multipling $e^-x cdot e^x$ inside the integral.
$$ int_-infty^infty e^-e^x e^x e^(a-1)x x^b dx $$
Using integration by parts and noticing $(e^-e^x)' = -e^-e^xe^x $:
$$ implies - int_-infty^infty (e^-e^x)' e^(a-1)x x^b dx = - int_-infty^infty e^-e^x (e^(a-1)x x^b)' dx $$
Again multiplying $e^-x cdot e^x$ inside the integral and repeating the process:
$$ = - int_-infty^infty e^-e^x e^x cdot e^-x(e^(a-1)x x^b)' dx $$
$$ = int_-infty^infty (e^-e^x)'cdot e^-x(e^(a-1)x x^b)' dx $$
$$ = int_-infty^infty (e^-e^x)cdot (e^-x(e^(a-1)x x^b)')' dx $$
This trick will only work $a$ times. The leading order term in $x$ will be something like: $ a!int_-infty^infty e^-e^x x^b dx $
My idea was to take:
$$ lim_n to infty fracint_-n^n e^-e^x e^ax x^b dx + (-1)^a (a-1)! int_-n^n e^-e^x x^b dxint_-n^n e^-e^x x^b-1 dx $$
Now one can use two ways two solve this:
The first by using L'hopital rule and differentiating. The other is by using integration by parts. I've tried to do this myself but I've got disagreement between my answers (I think I'm messing up the calculation).
Question
Is my idea correct? What is the identity gained after these manipulations? Furthermore, what is the identity of gained after the $x^k$ coefficient using limits and integration by parts? for:
$$ lim_n to infty fracint_-n^n e^-e^x e^ax x^b dx + (-1)^a (a-1)! int_-n^n e^-e^x x^b dxint_-n^n e^-e^x x^b-1 dx = ?$$
calculus integration limits
asked Jul 28 at 9:09
More Anonymous
1387
edited Jul 28 at 9:34
asked Jul 28 at 9:09
More Anonymous
1387
asked Jul 28 at 9:09
More Anonymous
1387
asked Jul 28 at 9:09
More Anonymous
1387
1387
Why do you say the integral is divergent?
– Julián Aguirre
Jul 28 at 11:26
@JuliánAguirre after repeated integration by parts one will be left with something like: $e^-e^x x^b$ as shown in the background this diverges for $- infty$
– More Anonymous
Jul 28 at 11:38
1
The integral is convergent. When $xto+infty$ it is dominated by $e^-e^x$; when $xto-infty$ by $e^ax$. May be you did not consider the boundary terms in one of the integration by parts.
– Julián Aguirre
Jul 28 at 11:42
1
As Julián Aguirre commented, not only they are convergent but they have very nice expressions.
– Claude Leibovici
Jul 28 at 13:52
@ClaudeLeibovici can u please comment the expression?
– More Anonymous
Jul 28 at 19:17
add a comment |Â
Why do you say the integral is divergent?
– Julián Aguirre
Jul 28 at 11:26
@JuliánAguirre after repeated integration by parts one will be left with something like: $e^-e^x x^b$ as shown in the background this diverges for $- infty$
– More Anonymous
Jul 28 at 11:38
1
The integral is convergent. When $xto+infty$ it is dominated by $e^-e^x$; when $xto-infty$ by $e^ax$. May be you did not consider the boundary terms in one of the integration by parts.
– Julián Aguirre
Jul 28 at 11:42
1
As Julián Aguirre commented, not only they are convergent but they have very nice expressions.
– Claude Leibovici
Jul 28 at 13:52
@ClaudeLeibovici can u please comment the expression?
– More Anonymous
Jul 28 at 19:17
Why do you say the integral is divergent?
– Julián Aguirre
Jul 28 at 11:26
Why do you say the integral is divergent?
– Julián Aguirre
Jul 28 at 11:26
@JuliánAguirre after repeated integration by parts one will be left with something like: $e^-e^x x^b$ as shown in the background this diverges for $- infty$
– More Anonymous
Jul 28 at 11:38
@JuliánAguirre after repeated integration by parts one will be left with something like: $e^-e^x x^b$ as shown in the background this diverges for $- infty$
– More Anonymous
Jul 28 at 11:38
1
1
The integral is convergent. When $xto+infty$ it is dominated by $e^-e^x$; when $xto-infty$ by $e^ax$. May be you did not consider the boundary terms in one of the integration by parts.
– Julián Aguirre
Jul 28 at 11:42
The integral is convergent. When $xto+infty$ it is dominated by $e^-e^x$; when $xto-infty$ by $e^ax$. May be you did not consider the boundary terms in one of the integration by parts.
– Julián Aguirre
Jul 28 at 11:42
1
1
As Julián Aguirre commented, not only they are convergent but they have very nice expressions.
– Claude Leibovici
Jul 28 at 13:52
As Julián Aguirre commented, not only they are convergent but they have very nice expressions.
– Claude Leibovici
Jul 28 at 13:52
@ClaudeLeibovici can u please comment the expression?
– More Anonymous
Jul 28 at 19:17
@ClaudeLeibovici can u please comment the expression?
– More Anonymous
Jul 28 at 19:17
add a comment |Â
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Why do you say the integral is divergent?
– Julián Aguirre
Jul 28 at 11:26
@JuliánAguirre after repeated integration by parts one will be left with something like: $e^-e^x x^b$ as shown in the background this diverges for $- infty$
– More Anonymous
Jul 28 at 11:38
1
The integral is convergent. When $xto+infty$ it is dominated by $e^-e^x$; when $xto-infty$ by $e^ax$. May be you did not consider the boundary terms in one of the integration by parts.
– Julián Aguirre
Jul 28 at 11:42
1
As Julián Aguirre commented, not only they are convergent but they have very nice expressions.
– Claude Leibovici
Jul 28 at 13:52
@ClaudeLeibovici can u please comment the expression?
– More Anonymous
Jul 28 at 19:17