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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$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







Cominou

















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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















3












$begingroup$

$intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx=
(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{x^2}e^{-(x+y)^2}e^{-(x+z)^2}dx)|_{y=0,z=0}$ $=(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{-(x+y+z)^2}e^{2yz}dx)|_{y=0,z=0}$
$=(frac{d^n}{dy^n}frac{d^n}{dz^n}sqrt{pi}e^{2yz})|_{y=0,z=0}=2^n n!sqrt{pi}$
(modulo mistakes that you will easily repair)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @MariuszIwaniuk you are right - aparently I don't know how to differentiate $exp$ :)
    $endgroup$
    – user8268
    2 days ago











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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












$begingroup$

$intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx=
(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{x^2}e^{-(x+y)^2}e^{-(x+z)^2}dx)|_{y=0,z=0}$ $=(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{-(x+y+z)^2}e^{2yz}dx)|_{y=0,z=0}$
$=(frac{d^n}{dy^n}frac{d^n}{dz^n}sqrt{pi}e^{2yz})|_{y=0,z=0}=2^n n!sqrt{pi}$
(modulo mistakes that you will easily repair)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @MariuszIwaniuk you are right - aparently I don't know how to differentiate $exp$ :)
    $endgroup$
    – user8268
    2 days ago
















3












$begingroup$

$intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx=
(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{x^2}e^{-(x+y)^2}e^{-(x+z)^2}dx)|_{y=0,z=0}$ $=(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{-(x+y+z)^2}e^{2yz}dx)|_{y=0,z=0}$
$=(frac{d^n}{dy^n}frac{d^n}{dz^n}sqrt{pi}e^{2yz})|_{y=0,z=0}=2^n n!sqrt{pi}$
(modulo mistakes that you will easily repair)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @MariuszIwaniuk you are right - aparently I don't know how to differentiate $exp$ :)
    $endgroup$
    – user8268
    2 days ago














3












3








3





$begingroup$

$intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx=
(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{x^2}e^{-(x+y)^2}e^{-(x+z)^2}dx)|_{y=0,z=0}$ $=(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{-(x+y+z)^2}e^{2yz}dx)|_{y=0,z=0}$
$=(frac{d^n}{dy^n}frac{d^n}{dz^n}sqrt{pi}e^{2yz})|_{y=0,z=0}=2^n n!sqrt{pi}$
(modulo mistakes that you will easily repair)






share|cite|improve this answer











$endgroup$



$intlimits_{-infty}^infty e^{x^2}left(frac{d^n}{dx^n}(e^{-x^2})right)^2dx=
(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{x^2}e^{-(x+y)^2}e^{-(x+z)^2}dx)|_{y=0,z=0}$ $=(frac{d^n}{dy^n}frac{d^n}{dz^n}intlimits_{-infty}^infty e^{-(x+y+z)^2}e^{2yz}dx)|_{y=0,z=0}$
$=(frac{d^n}{dy^n}frac{d^n}{dz^n}sqrt{pi}e^{2yz})|_{y=0,z=0}=2^n n!sqrt{pi}$
(modulo mistakes that you will easily repair)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























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


















  • $begingroup$
    @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


















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