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Bound for $int_0^te^-xa^2a(t-x)^-b,dx$
calculate $int_-infty^+infty cos(at) e^-bt^2 dt$Does the following Integral have a Closed Form Solution?Estimate for integral of sine to the power of $-(1+a)$ where $a>0$Help with a limit of an integral: $lim_hto inftyhint_0^infty e^-hxf(x) dx=f(0)$Is there a bound on $int_-infty^+infty int_-infty^+infty e^e^x+y+x+y dx dy$?Evaluate $intlimits_0^inftyfracsin xx, dx$Convergence of improper Integral $int_0^infty cos (x^3)mathrmdx$Integrate $int_-infty^+inftye^-2x^2 + 7dx$Compute $int_-infty^infty (x^2e^-ax^2+bx) dx$Evaluating $int_-infty^inftyfracexpleft(-a x^2right)x^2+b^2dx$
$begingroup$
I'm trying to estimate the following integral: $$int_0^te^-xa^2a(t-x)^-b,dx$$ where constants $t,a>0$ and $0<b<1/2$. I want to get a bound for this integral. The preferred bound is like:$f(t)/a$, with $f(t)rightarrow C$ as $trightarrow infty$, where C is a constant. For example, $f(t)=1-e^-ta^2$ is good enough.
I tried integration by parts but failed.
real-analysis improper-integrals
edited Mar 30 at 2:29
Saad
20.4k92352
asked Mar 30 at 2:17
MirandaMiranda
142
$endgroup$
add a comment |
$begingroup$
I'm trying to estimate the following integral: $$int_0^te^-xa^2a(t-x)^-b,dx$$ where constants $t,a>0$ and $0<b<1/2$. I want to get a bound for this integral. The preferred bound is like:$f(t)/a$, with $f(t)rightarrow C$ as $trightarrow infty$, where C is a constant. For example, $f(t)=1-e^-ta^2$ is good enough.
I tried integration by parts but failed.
real-analysis improper-integrals
edited Mar 30 at 2:29
Saad
20.4k92352
asked Mar 30 at 2:17
MirandaMiranda
142
$endgroup$
$begingroup$
Start by enforcing the substitution $xmapsto tx$ so that $$int_0^t e^-a^2 x(t-x)^-b,dx=t^1-bint_0^1 e^-ta^2 x(1-x),dx$$
$endgroup$
– Mark Viola
Mar 30 at 2:51
add a comment |
$begingroup$
I'm trying to estimate the following integral: $$int_0^te^-xa^2a(t-x)^-b,dx$$ where constants $t,a>0$ and $0<b<1/2$. I want to get a bound for this integral. The preferred bound is like:$f(t)/a$, with $f(t)rightarrow C$ as $trightarrow infty$, where C is a constant. For example, $f(t)=1-e^-ta^2$ is good enough.
I tried integration by parts but failed.
real-analysis improper-integrals
edited Mar 30 at 2:29
Saad
20.4k92352
asked Mar 30 at 2:17
MirandaMiranda
142
$endgroup$
I'm trying to estimate the following integral: $$int_0^te^-xa^2a(t-x)^-b,dx$$ where constants $t,a>0$ and $0<b<1/2$. I want to get a bound for this integral. The preferred bound is like:$f(t)/a$, with $f(t)rightarrow C$ as $trightarrow infty$, where C is a constant. For example, $f(t)=1-e^-ta^2$ is good enough.
I tried integration by parts but failed.
real-analysis improper-integrals
real-analysis improper-integrals
edited Mar 30 at 2:29
Saad
20.4k92352
asked Mar 30 at 2:17
MirandaMiranda
142
edited Mar 30 at 2:29
Saad
20.4k92352
asked Mar 30 at 2:17
MirandaMiranda
142
edited Mar 30 at 2:29
Saad
20.4k92352
edited Mar 30 at 2:29
Saad
20.4k92352
edited Mar 30 at 2:29
Saad
20.4k92352
20.4k92352
asked Mar 30 at 2:17
MirandaMiranda
142
asked Mar 30 at 2:17
MirandaMiranda
142
asked Mar 30 at 2:17
MirandaMiranda
142
142
$begingroup$
Start by enforcing the substitution $xmapsto tx$ so that $$int_0^t e^-a^2 x(t-x)^-b,dx=t^1-bint_0^1 e^-ta^2 x(1-x),dx$$
$endgroup$
– Mark Viola
Mar 30 at 2:51
add a comment |
$begingroup$
Start by enforcing the substitution $xmapsto tx$ so that $$int_0^t e^-a^2 x(t-x)^-b,dx=t^1-bint_0^1 e^-ta^2 x(1-x),dx$$
$endgroup$
– Mark Viola
Mar 30 at 2:51
$begingroup$
Start by enforcing the substitution $xmapsto tx$ so that $$int_0^t e^-a^2 x(t-x)^-b,dx=t^1-bint_0^1 e^-ta^2 x(1-x),dx$$
$endgroup$
– Mark Viola
Mar 30 at 2:51
$begingroup$
Start by enforcing the substitution $xmapsto tx$ so that $$int_0^t e^-a^2 x(t-x)^-b,dx=t^1-bint_0^1 e^-ta^2 x(1-x),dx$$
$endgroup$
– Mark Viola
Mar 30 at 2:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
(Edit: This doesn't solve the problem!)
This is a crude approximation, but it gives what you ask for.
We have $(t-x)^-b ge 1$ for $xge t-1$ and $(t-x)^-b<1$ for $x < t-1$. Therefore:
$$
beginsplit
int_0^t e^-xa^2a(t-x)^-b, dx
&le int_0^t-1 ae^-xa^2 , dx + int_t-1^ta(t-x)^-b , dx \
&= frac1aleft(1-e^(1-t)a^2right) + fraca1-b
endsplit
$$
which has limit $frac1a + fraca1-b$.
answered Mar 30 at 3:05
MiltenMilten
3746
$endgroup$
$begingroup$
I think your result should be $1/a$ times a constant of $b$.
$endgroup$
– Miranda
Mar 30 at 3:22
$begingroup$
Oops, I forgot how to integrate... I'll edit.
$endgroup$
– Milten
Mar 30 at 3:36
$begingroup$
Is it better now?
$endgroup$
– Milten
Mar 30 at 3:40
$begingroup$
Actually, not yet. Is it possible to get a bound with just $f/a$, with f being some function. Since my final goal is to see whether this integral can be bound by $1/a$ when t goes to infinity, i.e., take the infinity limit on t after integrating, like what I asked in the question.
$endgroup$
– Miranda
Mar 30 at 5:13
$begingroup$
Oh, I see now. Sorry.
$endgroup$
– Milten
Mar 30 at 9:41
add a comment |
$begingroup$
I get
$beginarray\
int_0^te^-xa^2a(t-x)^-b,dx
&=a^2b-1e^-ta^2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2gamma(-b+1, ta^2)\
&<a^2b-1e^-ta^2Gamma(-b+1)\
&le a^2b-1e^-ta^2Gamma(1/2)\
&le sqrtpia^2b-1e^-ta^2\
endarray
$
where
$gamma(,)$ is the incomplete gamma function.
Here's my steps.
$beginarray\
I(a, b, t)
&=int_0^te^-xa^2a(t-x)^-bdx\
&=aint_0^te^-(t-x)a^2x^-bdx\
&=aint_0^te^-ta^2e^xa^2x^-bdx\
&=ae^-ta^2int_0^te^xa^2x^-bdx\
&=ae^-ta^2int_0^ta^2e^y(y/a^2)^-bdy/a^2
quad y=xa^2, x=y/x^2, dx = dy/a^2\
&=ae^-ta^2a^2b-2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2gamma(-b+1, ta^2)
qquadtextincomplete gamma function\
&<a^2b-1e^-ta^2int_0^inftye^yy^-bdy\
&=a^2b-1e^-ta^2Gamma(-b+1)\
endarray
$
$0 < b < frac12
implies
frac12 < -b+1 < 1
$.
Therefore
$Gamma(-b+1)
le Gamma(1/2)
=sqrtpi
approx 1.772
$.
answered Mar 30 at 3:31
marty cohenmarty cohen
75k549130
$endgroup$
$begingroup$
But your bound goes to infinity for $ttoinfty$.
$endgroup$
– Milten
Mar 30 at 9:39
$begingroup$
(Also, it's not really the incomplete gamma function, right? Since the exponential factor has positive exponent. Not that it matters for your answer.)
$endgroup$
– Milten
Mar 30 at 9:41
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
(Edit: This doesn't solve the problem!)
This is a crude approximation, but it gives what you ask for.
We have $(t-x)^-b ge 1$ for $xge t-1$ and $(t-x)^-b<1$ for $x < t-1$. Therefore:
$$
beginsplit
int_0^t e^-xa^2a(t-x)^-b, dx
&le int_0^t-1 ae^-xa^2 , dx + int_t-1^ta(t-x)^-b , dx \
&= frac1aleft(1-e^(1-t)a^2right) + fraca1-b
endsplit
$$
which has limit $frac1a + fraca1-b$.
answered Mar 30 at 3:05
MiltenMilten
3746
$endgroup$
$begingroup$
I think your result should be $1/a$ times a constant of $b$.
$endgroup$
– Miranda
Mar 30 at 3:22
$begingroup$
Oops, I forgot how to integrate... I'll edit.
$endgroup$
– Milten
Mar 30 at 3:36
$begingroup$
Is it better now?
$endgroup$
– Milten
Mar 30 at 3:40
$begingroup$
Actually, not yet. Is it possible to get a bound with just $f/a$, with f being some function. Since my final goal is to see whether this integral can be bound by $1/a$ when t goes to infinity, i.e., take the infinity limit on t after integrating, like what I asked in the question.
$endgroup$
– Miranda
Mar 30 at 5:13
$begingroup$
Oh, I see now. Sorry.
$endgroup$
– Milten
Mar 30 at 9:41
add a comment |
$begingroup$
(Edit: This doesn't solve the problem!)
This is a crude approximation, but it gives what you ask for.
We have $(t-x)^-b ge 1$ for $xge t-1$ and $(t-x)^-b<1$ for $x < t-1$. Therefore:
$$
beginsplit
int_0^t e^-xa^2a(t-x)^-b, dx
&le int_0^t-1 ae^-xa^2 , dx + int_t-1^ta(t-x)^-b , dx \
&= frac1aleft(1-e^(1-t)a^2right) + fraca1-b
endsplit
$$
which has limit $frac1a + fraca1-b$.
answered Mar 30 at 3:05
MiltenMilten
3746
$endgroup$
$begingroup$
I think your result should be $1/a$ times a constant of $b$.
$endgroup$
– Miranda
Mar 30 at 3:22
$begingroup$
Oops, I forgot how to integrate... I'll edit.
$endgroup$
– Milten
Mar 30 at 3:36
$begingroup$
Is it better now?
$endgroup$
– Milten
Mar 30 at 3:40
$begingroup$
Actually, not yet. Is it possible to get a bound with just $f/a$, with f being some function. Since my final goal is to see whether this integral can be bound by $1/a$ when t goes to infinity, i.e., take the infinity limit on t after integrating, like what I asked in the question.
$endgroup$
– Miranda
Mar 30 at 5:13
$begingroup$
Oh, I see now. Sorry.
$endgroup$
– Milten
Mar 30 at 9:41
add a comment |
$begingroup$
(Edit: This doesn't solve the problem!)
This is a crude approximation, but it gives what you ask for.
We have $(t-x)^-b ge 1$ for $xge t-1$ and $(t-x)^-b<1$ for $x < t-1$. Therefore:
$$
beginsplit
int_0^t e^-xa^2a(t-x)^-b, dx
&le int_0^t-1 ae^-xa^2 , dx + int_t-1^ta(t-x)^-b , dx \
&= frac1aleft(1-e^(1-t)a^2right) + fraca1-b
endsplit
$$
which has limit $frac1a + fraca1-b$.
answered Mar 30 at 3:05
MiltenMilten
3746
$endgroup$
(Edit: This doesn't solve the problem!)
This is a crude approximation, but it gives what you ask for.
We have $(t-x)^-b ge 1$ for $xge t-1$ and $(t-x)^-b<1$ for $x < t-1$. Therefore:
$$
beginsplit
int_0^t e^-xa^2a(t-x)^-b, dx
&le int_0^t-1 ae^-xa^2 , dx + int_t-1^ta(t-x)^-b , dx \
&= frac1aleft(1-e^(1-t)a^2right) + fraca1-b
endsplit
$$
which has limit $frac1a + fraca1-b$.
answered Mar 30 at 3:05
MiltenMilten
3746
edited Mar 30 at 20:47
answered Mar 30 at 3:05
MiltenMilten
3746
answered Mar 30 at 3:05
MiltenMilten
3746
answered Mar 30 at 3:05
MiltenMilten
3746
3746
$begingroup$
I think your result should be $1/a$ times a constant of $b$.
$endgroup$
– Miranda
Mar 30 at 3:22
$begingroup$
Oops, I forgot how to integrate... I'll edit.
$endgroup$
– Milten
Mar 30 at 3:36
$begingroup$
Is it better now?
$endgroup$
– Milten
Mar 30 at 3:40
$begingroup$
Actually, not yet. Is it possible to get a bound with just $f/a$, with f being some function. Since my final goal is to see whether this integral can be bound by $1/a$ when t goes to infinity, i.e., take the infinity limit on t after integrating, like what I asked in the question.
$endgroup$
– Miranda
Mar 30 at 5:13
$begingroup$
Oh, I see now. Sorry.
$endgroup$
– Milten
Mar 30 at 9:41
add a comment |
$begingroup$
I think your result should be $1/a$ times a constant of $b$.
$endgroup$
– Miranda
Mar 30 at 3:22
$begingroup$
Oops, I forgot how to integrate... I'll edit.
$endgroup$
– Milten
Mar 30 at 3:36
$begingroup$
Is it better now?
$endgroup$
– Milten
Mar 30 at 3:40
$begingroup$
Actually, not yet. Is it possible to get a bound with just $f/a$, with f being some function. Since my final goal is to see whether this integral can be bound by $1/a$ when t goes to infinity, i.e., take the infinity limit on t after integrating, like what I asked in the question.
$endgroup$
– Miranda
Mar 30 at 5:13
$begingroup$
Oh, I see now. Sorry.
$endgroup$
– Milten
Mar 30 at 9:41
$begingroup$
I think your result should be $1/a$ times a constant of $b$.
$endgroup$
– Miranda
Mar 30 at 3:22
$begingroup$
I think your result should be $1/a$ times a constant of $b$.
$endgroup$
– Miranda
Mar 30 at 3:22
$begingroup$
Oops, I forgot how to integrate... I'll edit.
$endgroup$
– Milten
Mar 30 at 3:36
$begingroup$
Oops, I forgot how to integrate... I'll edit.
$endgroup$
– Milten
Mar 30 at 3:36
$begingroup$
Is it better now?
$endgroup$
– Milten
Mar 30 at 3:40
$begingroup$
Is it better now?
$endgroup$
– Milten
Mar 30 at 3:40
$begingroup$
Actually, not yet. Is it possible to get a bound with just $f/a$, with f being some function. Since my final goal is to see whether this integral can be bound by $1/a$ when t goes to infinity, i.e., take the infinity limit on t after integrating, like what I asked in the question.
$endgroup$
– Miranda
Mar 30 at 5:13
$begingroup$
Actually, not yet. Is it possible to get a bound with just $f/a$, with f being some function. Since my final goal is to see whether this integral can be bound by $1/a$ when t goes to infinity, i.e., take the infinity limit on t after integrating, like what I asked in the question.
$endgroup$
– Miranda
Mar 30 at 5:13
$begingroup$
Oh, I see now. Sorry.
$endgroup$
– Milten
Mar 30 at 9:41
$begingroup$
Oh, I see now. Sorry.
$endgroup$
– Milten
Mar 30 at 9:41
add a comment |
$begingroup$
I get
$beginarray\
int_0^te^-xa^2a(t-x)^-b,dx
&=a^2b-1e^-ta^2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2gamma(-b+1, ta^2)\
&<a^2b-1e^-ta^2Gamma(-b+1)\
&le a^2b-1e^-ta^2Gamma(1/2)\
&le sqrtpia^2b-1e^-ta^2\
endarray
$
where
$gamma(,)$ is the incomplete gamma function.
Here's my steps.
$beginarray\
I(a, b, t)
&=int_0^te^-xa^2a(t-x)^-bdx\
&=aint_0^te^-(t-x)a^2x^-bdx\
&=aint_0^te^-ta^2e^xa^2x^-bdx\
&=ae^-ta^2int_0^te^xa^2x^-bdx\
&=ae^-ta^2int_0^ta^2e^y(y/a^2)^-bdy/a^2
quad y=xa^2, x=y/x^2, dx = dy/a^2\
&=ae^-ta^2a^2b-2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2gamma(-b+1, ta^2)
qquadtextincomplete gamma function\
&<a^2b-1e^-ta^2int_0^inftye^yy^-bdy\
&=a^2b-1e^-ta^2Gamma(-b+1)\
endarray
$
$0 < b < frac12
implies
frac12 < -b+1 < 1
$.
Therefore
$Gamma(-b+1)
le Gamma(1/2)
=sqrtpi
approx 1.772
$.
answered Mar 30 at 3:31
marty cohenmarty cohen
75k549130
$endgroup$
$begingroup$
But your bound goes to infinity for $ttoinfty$.
$endgroup$
– Milten
Mar 30 at 9:39
$begingroup$
(Also, it's not really the incomplete gamma function, right? Since the exponential factor has positive exponent. Not that it matters for your answer.)
$endgroup$
– Milten
Mar 30 at 9:41
add a comment |
$begingroup$
I get
$beginarray\
int_0^te^-xa^2a(t-x)^-b,dx
&=a^2b-1e^-ta^2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2gamma(-b+1, ta^2)\
&<a^2b-1e^-ta^2Gamma(-b+1)\
&le a^2b-1e^-ta^2Gamma(1/2)\
&le sqrtpia^2b-1e^-ta^2\
endarray
$
where
$gamma(,)$ is the incomplete gamma function.
Here's my steps.
$beginarray\
I(a, b, t)
&=int_0^te^-xa^2a(t-x)^-bdx\
&=aint_0^te^-(t-x)a^2x^-bdx\
&=aint_0^te^-ta^2e^xa^2x^-bdx\
&=ae^-ta^2int_0^te^xa^2x^-bdx\
&=ae^-ta^2int_0^ta^2e^y(y/a^2)^-bdy/a^2
quad y=xa^2, x=y/x^2, dx = dy/a^2\
&=ae^-ta^2a^2b-2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2gamma(-b+1, ta^2)
qquadtextincomplete gamma function\
&<a^2b-1e^-ta^2int_0^inftye^yy^-bdy\
&=a^2b-1e^-ta^2Gamma(-b+1)\
endarray
$
$0 < b < frac12
implies
frac12 < -b+1 < 1
$.
Therefore
$Gamma(-b+1)
le Gamma(1/2)
=sqrtpi
approx 1.772
$.
answered Mar 30 at 3:31
marty cohenmarty cohen
75k549130
$endgroup$
$begingroup$
But your bound goes to infinity for $ttoinfty$.
$endgroup$
– Milten
Mar 30 at 9:39
$begingroup$
(Also, it's not really the incomplete gamma function, right? Since the exponential factor has positive exponent. Not that it matters for your answer.)
$endgroup$
– Milten
Mar 30 at 9:41
add a comment |
$begingroup$
I get
$beginarray\
int_0^te^-xa^2a(t-x)^-b,dx
&=a^2b-1e^-ta^2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2gamma(-b+1, ta^2)\
&<a^2b-1e^-ta^2Gamma(-b+1)\
&le a^2b-1e^-ta^2Gamma(1/2)\
&le sqrtpia^2b-1e^-ta^2\
endarray
$
where
$gamma(,)$ is the incomplete gamma function.
Here's my steps.
$beginarray\
I(a, b, t)
&=int_0^te^-xa^2a(t-x)^-bdx\
&=aint_0^te^-(t-x)a^2x^-bdx\
&=aint_0^te^-ta^2e^xa^2x^-bdx\
&=ae^-ta^2int_0^te^xa^2x^-bdx\
&=ae^-ta^2int_0^ta^2e^y(y/a^2)^-bdy/a^2
quad y=xa^2, x=y/x^2, dx = dy/a^2\
&=ae^-ta^2a^2b-2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2gamma(-b+1, ta^2)
qquadtextincomplete gamma function\
&<a^2b-1e^-ta^2int_0^inftye^yy^-bdy\
&=a^2b-1e^-ta^2Gamma(-b+1)\
endarray
$
$0 < b < frac12
implies
frac12 < -b+1 < 1
$.
Therefore
$Gamma(-b+1)
le Gamma(1/2)
=sqrtpi
approx 1.772
$.
answered Mar 30 at 3:31
marty cohenmarty cohen
75k549130
$endgroup$
I get
$beginarray\
int_0^te^-xa^2a(t-x)^-b,dx
&=a^2b-1e^-ta^2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2gamma(-b+1, ta^2)\
&<a^2b-1e^-ta^2Gamma(-b+1)\
&le a^2b-1e^-ta^2Gamma(1/2)\
&le sqrtpia^2b-1e^-ta^2\
endarray
$
where
$gamma(,)$ is the incomplete gamma function.
Here's my steps.
$beginarray\
I(a, b, t)
&=int_0^te^-xa^2a(t-x)^-bdx\
&=aint_0^te^-(t-x)a^2x^-bdx\
&=aint_0^te^-ta^2e^xa^2x^-bdx\
&=ae^-ta^2int_0^te^xa^2x^-bdx\
&=ae^-ta^2int_0^ta^2e^y(y/a^2)^-bdy/a^2
quad y=xa^2, x=y/x^2, dx = dy/a^2\
&=ae^-ta^2a^2b-2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2int_0^ta^2e^yy^-bdy\
&=a^2b-1e^-ta^2gamma(-b+1, ta^2)
qquadtextincomplete gamma function\
&<a^2b-1e^-ta^2int_0^inftye^yy^-bdy\
&=a^2b-1e^-ta^2Gamma(-b+1)\
endarray
$
$0 < b < frac12
implies
frac12 < -b+1 < 1
$.
Therefore
$Gamma(-b+1)
le Gamma(1/2)
=sqrtpi
approx 1.772
$.
answered Mar 30 at 3:31
marty cohenmarty cohen
75k549130
edited Mar 30 at 21:18
answered Mar 30 at 3:31
marty cohenmarty cohen
75k549130
answered Mar 30 at 3:31
marty cohenmarty cohen
75k549130
answered Mar 30 at 3:31
marty cohenmarty cohen
75k549130
75k549130
$begingroup$
But your bound goes to infinity for $ttoinfty$.
$endgroup$
– Milten
Mar 30 at 9:39
$begingroup$
(Also, it's not really the incomplete gamma function, right? Since the exponential factor has positive exponent. Not that it matters for your answer.)
$endgroup$
– Milten
Mar 30 at 9:41
add a comment |
$begingroup$
But your bound goes to infinity for $ttoinfty$.
$endgroup$
– Milten
Mar 30 at 9:39
$begingroup$
(Also, it's not really the incomplete gamma function, right? Since the exponential factor has positive exponent. Not that it matters for your answer.)
$endgroup$
– Milten
Mar 30 at 9:41
$begingroup$
But your bound goes to infinity for $ttoinfty$.
$endgroup$
– Milten
Mar 30 at 9:39
$begingroup$
But your bound goes to infinity for $ttoinfty$.
$endgroup$
– Milten
Mar 30 at 9:39
$begingroup$
(Also, it's not really the incomplete gamma function, right? Since the exponential factor has positive exponent. Not that it matters for your answer.)
$endgroup$
– Milten
Mar 30 at 9:41
$begingroup$
(Also, it's not really the incomplete gamma function, right? Since the exponential factor has positive exponent. Not that it matters for your answer.)
$endgroup$
– Milten
Mar 30 at 9:41
add a comment |
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$begingroup$
Start by enforcing the substitution $xmapsto tx$ so that $$int_0^t e^-a^2 x(t-x)^-b,dx=t^1-bint_0^1 e^-ta^2 x(1-x),dx$$
$endgroup$
– Mark Viola
Mar 30 at 2:51