Value of $intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx$Show that...
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Value of $intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx$
Show that $int^{infty}_{0}left(frac{sin(x)}{x}right)^2 < 2$Evaluate $int limits _{0}^{infty}lnleft({x+frac{1}{x}}right)cdotfrac{mathrm dx}{1+x^2}$Convergence of the integral $intlimits_{1}^{infty} left( frac{1}{sqrt{x}} - frac{1}{sqrt{x+3}} right) , dx$Does $int _1^{infty }left(sin left(x^2right)right)dx$ converge or diverge?Integrate $intfrac{dx}{xsqrt{x^2+1}}$Does $int _0^{infty }left(fleft(xright)+gleft(xright)right)^2dx$ converge or diverge?Evaluating $int x!left(frac1x+frac{1}{x-1}+…+frac11right)dx$does the integral $int _{1}^{infty }!{frac {sin left( cos left( x right) +sin left( xsqrt {3} right) right) }{x}}{dx}$ converges?Evaluating the integral: $intlimits_{0}^{infty}left(frac{sin(ax)}{x}right)^2 dx , a neq 0 $Is it true that $left(int |xf(x)|^2dxright)^{1/2}left(int |f(x)|^2dxright)^{1/2} leq C left(int (x^2+1) |f(x)|^2dxright)^{1/2}$?
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I need to find the value of this integral : $intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx$ .
I tried to integrate by parts $intlimits_{-infty}^infty e^{x^2}left(frac{d^{n+1}}{dx^{n+1}}(e^{-x^2})right)^2dx$ but I'm stuck for the expression of the antiderivative. Any help would be greatly appreciated.
(n is a positive integer)
calculus
asked 2 days ago
CominouCominou
364
$endgroup$
add a comment |
$begingroup$
I need to find the value of this integral : $intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx$ .
I tried to integrate by parts $intlimits_{-infty}^infty e^{x^2}left(frac{d^{n+1}}{dx^{n+1}}(e^{-x^2})right)^2dx$ but I'm stuck for the expression of the antiderivative. Any help would be greatly appreciated.
(n is a positive integer)
calculus
asked 2 days ago
CominouCominou
364
$endgroup$
$begingroup$
Have you tried brute force - taking the derivative (which will be a polynomial times $e^{-x^2}$) After squaring the integrand will be another polynomial times $e^{-x^2}$, which can be integrated explicitly.
$endgroup$
– herb steinberg
2 days ago
$begingroup$
What is the value of $n$, or are you trying to get a general expression for all values of $n$?
$endgroup$
– John Omielan
2 days ago
$begingroup$
@JohnOmielan n is a positive integer, so I'm trying to get a general expression
$endgroup$
– Cominou
2 days ago
add a comment |
$begingroup$
I need to find the value of this integral : $intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx$ .
I tried to integrate by parts $intlimits_{-infty}^infty e^{x^2}left(frac{d^{n+1}}{dx^{n+1}}(e^{-x^2})right)^2dx$ but I'm stuck for the expression of the antiderivative. Any help would be greatly appreciated.
(n is a positive integer)
calculus
asked 2 days ago
CominouCominou
364
$endgroup$
I need to find the value of this integral : $intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx$ .
I tried to integrate by parts $intlimits_{-infty}^infty e^{x^2}left(frac{d^{n+1}}{dx^{n+1}}(e^{-x^2})right)^2dx$ but I'm stuck for the expression of the antiderivative. Any help would be greatly appreciated.
(n is a positive integer)
calculus
calculus
asked 2 days ago
CominouCominou
364
asked 2 days ago
CominouCominou
364
edited 2 days ago
Cominou
asked 2 days ago
CominouCominou
364
asked 2 days ago
CominouCominou
364
asked 2 days ago
CominouCominou
364
364
$begingroup$
Have you tried brute force - taking the derivative (which will be a polynomial times $e^{-x^2}$) After squaring the integrand will be another polynomial times $e^{-x^2}$, which can be integrated explicitly.
$endgroup$
– herb steinberg
2 days ago
$begingroup$
What is the value of $n$, or are you trying to get a general expression for all values of $n$?
$endgroup$
– John Omielan
2 days ago
$begingroup$
@JohnOmielan n is a positive integer, so I'm trying to get a general expression
$endgroup$
– Cominou
2 days ago
add a comment |
$begingroup$
Have you tried brute force - taking the derivative (which will be a polynomial times $e^{-x^2}$) After squaring the integrand will be another polynomial times $e^{-x^2}$, which can be integrated explicitly.
$endgroup$
– herb steinberg
2 days ago
$begingroup$
What is the value of $n$, or are you trying to get a general expression for all values of $n$?
$endgroup$
– John Omielan
2 days ago
$begingroup$
@JohnOmielan n is a positive integer, so I'm trying to get a general expression
$endgroup$
– Cominou
2 days ago
$begingroup$
Have you tried brute force - taking the derivative (which will be a polynomial times $e^{-x^2}$) After squaring the integrand will be another polynomial times $e^{-x^2}$, which can be integrated explicitly.
$endgroup$
– herb steinberg
2 days ago
$begingroup$
Have you tried brute force - taking the derivative (which will be a polynomial times $e^{-x^2}$) After squaring the integrand will be another polynomial times $e^{-x^2}$, which can be integrated explicitly.
$endgroup$
– herb steinberg
2 days ago
$begingroup$
What is the value of $n$, or are you trying to get a general expression for all values of $n$?
$endgroup$
– John Omielan
2 days ago
$begingroup$
What is the value of $n$, or are you trying to get a general expression for all values of $n$?
$endgroup$
– John Omielan
2 days ago
$begingroup$
@JohnOmielan n is a positive integer, so I'm trying to get a general expression
$endgroup$
– Cominou
2 days ago
$begingroup$
@JohnOmielan n is a positive integer, so I'm trying to get a general expression
$endgroup$
– Cominou
2 days ago
add a comment |
1 Answer
1
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$begingroup$
$intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx=
(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{x^2}e^{-(x+y)^2}e^{-(x+z)^2}dx)|_{y=0,z=0}$ $=(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{-(x+y+z)^2}e^{2yz}dx)|_{y=0,z=0}$
$=(frac{d^n}{dy^n}frac{d^n}{dz^n}sqrt{pi}e^{2yz})|_{y=0,z=0}=2^n n!sqrt{pi}$
(modulo mistakes that you will easily repair)
answered 2 days ago
user8268user8268
17k12747
$endgroup$
$begingroup$
@MariuszIwaniuk you are right - aparently I don't know how to differentiate $exp$ :)
$endgroup$
– user8268
2 days ago
add a comment |
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1 Answer
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$begingroup$
$intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx=
(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{x^2}e^{-(x+y)^2}e^{-(x+z)^2}dx)|_{y=0,z=0}$ $=(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{-(x+y+z)^2}e^{2yz}dx)|_{y=0,z=0}$
$=(frac{d^n}{dy^n}frac{d^n}{dz^n}sqrt{pi}e^{2yz})|_{y=0,z=0}=2^n n!sqrt{pi}$
(modulo mistakes that you will easily repair)
answered 2 days ago
user8268user8268
17k12747
$endgroup$
$begingroup$
@MariuszIwaniuk you are right - aparently I don't know how to differentiate $exp$ :)
$endgroup$
– user8268
2 days ago
add a comment |
$begingroup$
$intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx=
(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{x^2}e^{-(x+y)^2}e^{-(x+z)^2}dx)|_{y=0,z=0}$ $=(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{-(x+y+z)^2}e^{2yz}dx)|_{y=0,z=0}$
$=(frac{d^n}{dy^n}frac{d^n}{dz^n}sqrt{pi}e^{2yz})|_{y=0,z=0}=2^n n!sqrt{pi}$
(modulo mistakes that you will easily repair)
answered 2 days ago
user8268user8268
17k12747
$endgroup$
$begingroup$
@MariuszIwaniuk you are right - aparently I don't know how to differentiate $exp$ :)
$endgroup$
– user8268
2 days ago
add a comment |
$begingroup$
$intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx=
(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{x^2}e^{-(x+y)^2}e^{-(x+z)^2}dx)|_{y=0,z=0}$ $=(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{-(x+y+z)^2}e^{2yz}dx)|_{y=0,z=0}$
$=(frac{d^n}{dy^n}frac{d^n}{dz^n}sqrt{pi}e^{2yz})|_{y=0,z=0}=2^n n!sqrt{pi}$
(modulo mistakes that you will easily repair)
answered 2 days ago
user8268user8268
17k12747
$endgroup$
$intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx=
(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{x^2}e^{-(x+y)^2}e^{-(x+z)^2}dx)|_{y=0,z=0}$ $=(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{-(x+y+z)^2}e^{2yz}dx)|_{y=0,z=0}$
$=(frac{d^n}{dy^n}frac{d^n}{dz^n}sqrt{pi}e^{2yz})|_{y=0,z=0}=2^n n!sqrt{pi}$
(modulo mistakes that you will easily repair)
answered 2 days ago
user8268user8268
17k12747
edited 2 days ago
answered 2 days ago
user8268user8268
17k12747
answered 2 days ago
user8268user8268
17k12747
answered 2 days ago
user8268user8268
17k12747
17k12747
$begingroup$
@MariuszIwaniuk you are right - aparently I don't know how to differentiate $exp$ :)
$endgroup$
– user8268
2 days ago
add a comment |
$begingroup$
@MariuszIwaniuk you are right - aparently I don't know how to differentiate $exp$ :)
$endgroup$
– user8268
2 days ago
$begingroup$
@MariuszIwaniuk you are right - aparently I don't know how to differentiate $exp$ :)
$endgroup$
– user8268
2 days ago
$begingroup$
@MariuszIwaniuk you are right - aparently I don't know how to differentiate $exp$ :)
$endgroup$
– user8268
2 days ago
add a comment |
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$begingroup$
Have you tried brute force - taking the derivative (which will be a polynomial times $e^{-x^2}$) After squaring the integrand will be another polynomial times $e^{-x^2}$, which can be integrated explicitly.
$endgroup$
– herb steinberg
2 days ago
$begingroup$
What is the value of $n$, or are you trying to get a general expression for all values of $n$?
$endgroup$
– John Omielan
2 days ago
$begingroup$
@JohnOmielan n is a positive integer, so I'm trying to get a general expression
$endgroup$
– Cominou
2 days ago