How to evaluate the following integral involving a gaussian?Evaluating $intlimits_-infty^infty exp(iax)over1+ixdx$how can I evaluate this integral?Evaluating a double integral involving exponential of trigonometric functionsGeneral result of the following integralHow do I evaluate the integral $int frac1xsin xdx$?Evaluate integralHow do I evaluate the following integral $int_-infty^infty e^-sigma^2 x^2/2; mathrm dx$?Evaluating the integral $int e^ dx$What is the simplest technique to evaluate the following definite triple integral?Integral involving 2-dimensional Gaussian function
California: "For quality assurance, this phone call is being recorded"
Is there any Biblical Basis for 400 years of silence between Old and New Testament?
Will TSA allow me to carry a CPAP?
Is it legal in the UK for politicians to lie to the public for political gain?
Story about a toddler with god-like powers, dangerous tantrums
Credit card offering 0.5 miles for every cent rounded up. Too good to be true?
Is the capacitor drawn or wired wrongly?
The term for the person/group a political party aligns themselves with to appear concerned about the general public
How should I push back against my job assigning "homework"?
Comma Code - Ch. 4 Automate the Boring Stuff
What does it mean by "d-ism of Leibniz" and "dotage of Newton" in simple English?
How is it possible for this NPC to be alive during the Curse of Strahd adventure?
Can an old DSLR be upgraded to match modern smartphone image quality
How can I make 20-200 ohm variable resistor look like a 20-240 ohm resistor?
Is the decompression of compressed and encrypted data without decryption also theoretically impossible?
Word for a small burst of laughter that can't be held back
Can you please explain this joke: "I'm going bananas is what I tell my bananas before I leave the house"?
Access to all elements on the page
Will dual-learning in a glider make my airplane learning safer?
Computing the differentials in the Adams spectral sequence
Should we freeze the number of people coming in to the study for Kaplan-Meier test
Where do the UpdateInstallationWizard.aspx logs get saved in Azure PaaS?
How can a single Member of the House block a Congressional bill?
Can commodity 3d printer extrusion hardware and filament be used for injection molding?
How to evaluate the following integral involving a gaussian?
Evaluating $intlimits_-infty^infty exp(iax)over1+ixdx$how can I evaluate this integral?Evaluating a double integral involving exponential of trigonometric functionsGeneral result of the following integralHow do I evaluate the integral $int frac1xsin xdx$?Evaluate integralHow do I evaluate the following integral $int_-infty^infty e^-sigma^2 x^2/2; mathrm dx$?Evaluating the integral $int e^ dx$What is the simplest technique to evaluate the following definite triple integral?Integral involving 2-dimensional Gaussian function
$begingroup$
I want to evaluate the following integral:
$$intlimits_0 ^infty x sinpx exp(-a^2x^2) dx$$
Now I am unsure how to proceed. I know that this is an even function so I can extend the limit terms to $-infty, infty $ and then divide by 2. I have tried to evaluate this on Wolfram Alpha, but it only shows the answer while I am interested in the procedure.
integration definite-integrals
edited May 25 at 1:21
David G. Stork
12.9k41936
asked May 24 at 22:59
daljit97daljit97
333211
$endgroup$
add a comment |
$begingroup$
I want to evaluate the following integral:
$$intlimits_0 ^infty x sinpx exp(-a^2x^2) dx$$
Now I am unsure how to proceed. I know that this is an even function so I can extend the limit terms to $-infty, infty $ and then divide by 2. I have tried to evaluate this on Wolfram Alpha, but it only shows the answer while I am interested in the procedure.
integration definite-integrals
edited May 25 at 1:21
David G. Stork
12.9k41936
asked May 24 at 22:59
daljit97daljit97
333211
$endgroup$
$begingroup$
Have you tried differentiating wrt $p$ or $a$
$endgroup$
– Henry Lee
May 24 at 23:02
2
$begingroup$
i think the integral is indeed divergent???
$endgroup$
– logo
May 24 at 23:04
1
$begingroup$
Should the last term be $exp(-cx^2)$ for some constant $c$? In its current form, the integral diverges.
$endgroup$
– VHarisop
May 24 at 23:07
1
$begingroup$
@VHarishop Sorry, I made a mistake the exponential term should be $ exp(-a^2x^2)$ and not $ exp(frac-a^2x^2)$
$endgroup$
– daljit97
May 24 at 23:08
add a comment |
$begingroup$
I want to evaluate the following integral:
$$intlimits_0 ^infty x sinpx exp(-a^2x^2) dx$$
Now I am unsure how to proceed. I know that this is an even function so I can extend the limit terms to $-infty, infty $ and then divide by 2. I have tried to evaluate this on Wolfram Alpha, but it only shows the answer while I am interested in the procedure.
integration definite-integrals
edited May 25 at 1:21
David G. Stork
12.9k41936
asked May 24 at 22:59
daljit97daljit97
333211
$endgroup$
I want to evaluate the following integral:
$$intlimits_0 ^infty x sinpx exp(-a^2x^2) dx$$
Now I am unsure how to proceed. I know that this is an even function so I can extend the limit terms to $-infty, infty $ and then divide by 2. I have tried to evaluate this on Wolfram Alpha, but it only shows the answer while I am interested in the procedure.
integration definite-integrals
integration definite-integrals
edited May 25 at 1:21
David G. Stork
12.9k41936
asked May 24 at 22:59
daljit97daljit97
333211
edited May 25 at 1:21
David G. Stork
12.9k41936
asked May 24 at 22:59
daljit97daljit97
333211
edited May 25 at 1:21
David G. Stork
12.9k41936
edited May 25 at 1:21
David G. Stork
12.9k41936
edited May 25 at 1:21
David G. Stork
12.9k41936
12.9k41936
asked May 24 at 22:59
daljit97daljit97
333211
asked May 24 at 22:59
daljit97daljit97
333211
asked May 24 at 22:59
daljit97daljit97
333211
333211
$begingroup$
Have you tried differentiating wrt $p$ or $a$
$endgroup$
– Henry Lee
May 24 at 23:02
2
$begingroup$
i think the integral is indeed divergent???
$endgroup$
– logo
May 24 at 23:04
1
$begingroup$
Should the last term be $exp(-cx^2)$ for some constant $c$? In its current form, the integral diverges.
$endgroup$
– VHarisop
May 24 at 23:07
1
$begingroup$
@VHarishop Sorry, I made a mistake the exponential term should be $ exp(-a^2x^2)$ and not $ exp(frac-a^2x^2)$
$endgroup$
– daljit97
May 24 at 23:08
add a comment |
$begingroup$
Have you tried differentiating wrt $p$ or $a$
$endgroup$
– Henry Lee
May 24 at 23:02
2
$begingroup$
i think the integral is indeed divergent???
$endgroup$
– logo
May 24 at 23:04
1
$begingroup$
Should the last term be $exp(-cx^2)$ for some constant $c$? In its current form, the integral diverges.
$endgroup$
– VHarisop
May 24 at 23:07
1
$begingroup$
@VHarishop Sorry, I made a mistake the exponential term should be $ exp(-a^2x^2)$ and not $ exp(frac-a^2x^2)$
$endgroup$
– daljit97
May 24 at 23:08
$begingroup$
Have you tried differentiating wrt $p$ or $a$
$endgroup$
– Henry Lee
May 24 at 23:02
$begingroup$
Have you tried differentiating wrt $p$ or $a$
$endgroup$
– Henry Lee
May 24 at 23:02
2
2
$begingroup$
i think the integral is indeed divergent???
$endgroup$
– logo
May 24 at 23:04
$begingroup$
i think the integral is indeed divergent???
$endgroup$
– logo
May 24 at 23:04
1
1
$begingroup$
Should the last term be $exp(-cx^2)$ for some constant $c$? In its current form, the integral diverges.
$endgroup$
– VHarisop
May 24 at 23:07
$begingroup$
Should the last term be $exp(-cx^2)$ for some constant $c$? In its current form, the integral diverges.
$endgroup$
– VHarisop
May 24 at 23:07
1
1
$begingroup$
@VHarishop Sorry, I made a mistake the exponential term should be $ exp(-a^2x^2)$ and not $ exp(frac-a^2x^2)$
$endgroup$
– daljit97
May 24 at 23:08
$begingroup$
@VHarishop Sorry, I made a mistake the exponential term should be $ exp(-a^2x^2)$ and not $ exp(frac-a^2x^2)$
$endgroup$
– daljit97
May 24 at 23:08
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Start with:
$$Ileft( p right)=int_0^infty cos left( px right)exp (-a^2x^2)dx$$
We can use differentiation under the integral sign:
$$I'left( p right)=-int_0^infty xsin left( px right)exp (-a^2x^2)dx$$
Integration by parts using $u=sin left( px right)quad andquad dv=-xexp left( -a^2x^2 right)dx$
$$I'left( p right)=left. sin left( px right)fracexp left( -a^2x^2 right)2a^2 right|_0^infty -fracp2a^2int_0^infty cos left( px right)exp left( -a^2x^2 right)dx$$
The first term on the right vanishes, and we have the first-order differential equation:
$$fracI'left( p right)Ileft( p right)=-fracp2a^2Rightarrow ln left( Ileft( p right) right)=-fracp^24a^2+C$$
Using $$Ileft( 0 right)=fracsqrtpi a$$
We can find $C=ln left( fracsqrtpi a right)$
hence
$$ln left( Ileft( p right) right)=-fracp^24a^2+ln left( fracsqrtpi a right)$$
So
$$Ileft( p right)=fracsqrtpi aexp left( -fracp^24a^2 right)$$
Finally the integral in question equals
$$-I'left( p right)=-fracddpleft( fracsqrtpi aexp left( -fracp^24a^2 right) right)$$
answered May 24 at 23:42
logologo
37413
$endgroup$
$begingroup$
How is $I(0)= int_0 ^ infty cos(0) = sqrtpi/a$?
$endgroup$
– daljit97
May 25 at 0:42
1
$begingroup$
$Ileft( 0 right)=int_0^infty cos left( 0 right)exp (-a^2x^2)dx=int_0^infty exp (-a^2x^2)dx$
$endgroup$
– logo
May 25 at 1:45
add a comment |
$begingroup$
Integrating by parts we get $-frac 1 2a^2 e^-a^2x^2 sin(px)|_0^infty+ frac p 2a^2int_0^infty e^-a^2x^2 cos(px)dx$. The first term is $0$ and the second term is the real part of a Gaussian characteristic function up to a constant factor.
The answer is $frac psqrtpi 4a^3 e^-p^2/2a^2$
answered May 24 at 23:15
Kavi Rama MurthyKavi Rama Murthy
86.3k53873
$endgroup$
add a comment |
$begingroup$
$$
beginalign
int_0^infty xsinpxexp(-a^2x^2) dx
& =int_0^infty xsum_ngeq0dfrac(-1)^n(px)^2n+1(2n+1)!exp(-a^2x^2) dx \
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+1)!int_0^infty x^2n+2exp(-a^2x^2) dx \
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+1)!dfrac12a^2n+3int_0^infty u^n+frac12e^-u du ~~~;~~~a^2x^2=u\
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+2)Gamma(2n+3)dfrac12a^2n+3Gamma(n+frac32)\
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+2)2^2n+2sqrtpi^-1Gamma(n+2)Gamma(n+frac32)dfrac12a^2n+3Gamma(n+frac32)\
& =sqrtpisum_ngeq0dfrac(-1)^np^2n+12^2n+2dfrac12a^2n+3\
& =sqrtpidfracp4a^3sum_ngeq0left(dfrac-p^24a^2right)^ndfrac1n!\
& =sqrtpidfracp4a^3expleft(dfrac-p^24a^2right)
endalign
$$
answered May 25 at 1:15
NosratiNosrati
27.3k62355
$endgroup$
$begingroup$
Oh my goodness... so much work for such a simple integral (done by parts)! Obviously a technical solution with a different approach!
$endgroup$
– David G. Stork
May 25 at 1:22
$begingroup$
@DavidG.Stork This is a technical solution with different approach.
$endgroup$
– Nosrati
May 25 at 1:25
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3238744%2fhow-to-evaluate-the-following-integral-involving-a-gaussian%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Start with:
$$Ileft( p right)=int_0^infty cos left( px right)exp (-a^2x^2)dx$$
We can use differentiation under the integral sign:
$$I'left( p right)=-int_0^infty xsin left( px right)exp (-a^2x^2)dx$$
Integration by parts using $u=sin left( px right)quad andquad dv=-xexp left( -a^2x^2 right)dx$
$$I'left( p right)=left. sin left( px right)fracexp left( -a^2x^2 right)2a^2 right|_0^infty -fracp2a^2int_0^infty cos left( px right)exp left( -a^2x^2 right)dx$$
The first term on the right vanishes, and we have the first-order differential equation:
$$fracI'left( p right)Ileft( p right)=-fracp2a^2Rightarrow ln left( Ileft( p right) right)=-fracp^24a^2+C$$
Using $$Ileft( 0 right)=fracsqrtpi a$$
We can find $C=ln left( fracsqrtpi a right)$
hence
$$ln left( Ileft( p right) right)=-fracp^24a^2+ln left( fracsqrtpi a right)$$
So
$$Ileft( p right)=fracsqrtpi aexp left( -fracp^24a^2 right)$$
Finally the integral in question equals
$$-I'left( p right)=-fracddpleft( fracsqrtpi aexp left( -fracp^24a^2 right) right)$$
answered May 24 at 23:42
logologo
37413
$endgroup$
$begingroup$
How is $I(0)= int_0 ^ infty cos(0) = sqrtpi/a$?
$endgroup$
– daljit97
May 25 at 0:42
1
$begingroup$
$Ileft( 0 right)=int_0^infty cos left( 0 right)exp (-a^2x^2)dx=int_0^infty exp (-a^2x^2)dx$
$endgroup$
– logo
May 25 at 1:45
add a comment |
$begingroup$
Start with:
$$Ileft( p right)=int_0^infty cos left( px right)exp (-a^2x^2)dx$$
We can use differentiation under the integral sign:
$$I'left( p right)=-int_0^infty xsin left( px right)exp (-a^2x^2)dx$$
Integration by parts using $u=sin left( px right)quad andquad dv=-xexp left( -a^2x^2 right)dx$
$$I'left( p right)=left. sin left( px right)fracexp left( -a^2x^2 right)2a^2 right|_0^infty -fracp2a^2int_0^infty cos left( px right)exp left( -a^2x^2 right)dx$$
The first term on the right vanishes, and we have the first-order differential equation:
$$fracI'left( p right)Ileft( p right)=-fracp2a^2Rightarrow ln left( Ileft( p right) right)=-fracp^24a^2+C$$
Using $$Ileft( 0 right)=fracsqrtpi a$$
We can find $C=ln left( fracsqrtpi a right)$
hence
$$ln left( Ileft( p right) right)=-fracp^24a^2+ln left( fracsqrtpi a right)$$
So
$$Ileft( p right)=fracsqrtpi aexp left( -fracp^24a^2 right)$$
Finally the integral in question equals
$$-I'left( p right)=-fracddpleft( fracsqrtpi aexp left( -fracp^24a^2 right) right)$$
answered May 24 at 23:42
logologo
37413
$endgroup$
$begingroup$
How is $I(0)= int_0 ^ infty cos(0) = sqrtpi/a$?
$endgroup$
– daljit97
May 25 at 0:42
1
$begingroup$
$Ileft( 0 right)=int_0^infty cos left( 0 right)exp (-a^2x^2)dx=int_0^infty exp (-a^2x^2)dx$
$endgroup$
– logo
May 25 at 1:45
add a comment |
$begingroup$
Start with:
$$Ileft( p right)=int_0^infty cos left( px right)exp (-a^2x^2)dx$$
We can use differentiation under the integral sign:
$$I'left( p right)=-int_0^infty xsin left( px right)exp (-a^2x^2)dx$$
Integration by parts using $u=sin left( px right)quad andquad dv=-xexp left( -a^2x^2 right)dx$
$$I'left( p right)=left. sin left( px right)fracexp left( -a^2x^2 right)2a^2 right|_0^infty -fracp2a^2int_0^infty cos left( px right)exp left( -a^2x^2 right)dx$$
The first term on the right vanishes, and we have the first-order differential equation:
$$fracI'left( p right)Ileft( p right)=-fracp2a^2Rightarrow ln left( Ileft( p right) right)=-fracp^24a^2+C$$
Using $$Ileft( 0 right)=fracsqrtpi a$$
We can find $C=ln left( fracsqrtpi a right)$
hence
$$ln left( Ileft( p right) right)=-fracp^24a^2+ln left( fracsqrtpi a right)$$
So
$$Ileft( p right)=fracsqrtpi aexp left( -fracp^24a^2 right)$$
Finally the integral in question equals
$$-I'left( p right)=-fracddpleft( fracsqrtpi aexp left( -fracp^24a^2 right) right)$$
answered May 24 at 23:42
logologo
37413
$endgroup$
Start with:
$$Ileft( p right)=int_0^infty cos left( px right)exp (-a^2x^2)dx$$
We can use differentiation under the integral sign:
$$I'left( p right)=-int_0^infty xsin left( px right)exp (-a^2x^2)dx$$
Integration by parts using $u=sin left( px right)quad andquad dv=-xexp left( -a^2x^2 right)dx$
$$I'left( p right)=left. sin left( px right)fracexp left( -a^2x^2 right)2a^2 right|_0^infty -fracp2a^2int_0^infty cos left( px right)exp left( -a^2x^2 right)dx$$
The first term on the right vanishes, and we have the first-order differential equation:
$$fracI'left( p right)Ileft( p right)=-fracp2a^2Rightarrow ln left( Ileft( p right) right)=-fracp^24a^2+C$$
Using $$Ileft( 0 right)=fracsqrtpi a$$
We can find $C=ln left( fracsqrtpi a right)$
hence
$$ln left( Ileft( p right) right)=-fracp^24a^2+ln left( fracsqrtpi a right)$$
So
$$Ileft( p right)=fracsqrtpi aexp left( -fracp^24a^2 right)$$
Finally the integral in question equals
$$-I'left( p right)=-fracddpleft( fracsqrtpi aexp left( -fracp^24a^2 right) right)$$
answered May 24 at 23:42
logologo
37413
edited May 24 at 23:59
answered May 24 at 23:42
logologo
37413
answered May 24 at 23:42
logologo
37413
answered May 24 at 23:42
logologo
37413
37413
$begingroup$
How is $I(0)= int_0 ^ infty cos(0) = sqrtpi/a$?
$endgroup$
– daljit97
May 25 at 0:42
1
$begingroup$
$Ileft( 0 right)=int_0^infty cos left( 0 right)exp (-a^2x^2)dx=int_0^infty exp (-a^2x^2)dx$
$endgroup$
– logo
May 25 at 1:45
add a comment |
$begingroup$
How is $I(0)= int_0 ^ infty cos(0) = sqrtpi/a$?
$endgroup$
– daljit97
May 25 at 0:42
1
$begingroup$
$Ileft( 0 right)=int_0^infty cos left( 0 right)exp (-a^2x^2)dx=int_0^infty exp (-a^2x^2)dx$
$endgroup$
– logo
May 25 at 1:45
$begingroup$
How is $I(0)= int_0 ^ infty cos(0) = sqrtpi/a$?
$endgroup$
– daljit97
May 25 at 0:42
$begingroup$
How is $I(0)= int_0 ^ infty cos(0) = sqrtpi/a$?
$endgroup$
– daljit97
May 25 at 0:42
1
1
$begingroup$
$Ileft( 0 right)=int_0^infty cos left( 0 right)exp (-a^2x^2)dx=int_0^infty exp (-a^2x^2)dx$
$endgroup$
– logo
May 25 at 1:45
$begingroup$
$Ileft( 0 right)=int_0^infty cos left( 0 right)exp (-a^2x^2)dx=int_0^infty exp (-a^2x^2)dx$
$endgroup$
– logo
May 25 at 1:45
add a comment |
$begingroup$
Integrating by parts we get $-frac 1 2a^2 e^-a^2x^2 sin(px)|_0^infty+ frac p 2a^2int_0^infty e^-a^2x^2 cos(px)dx$. The first term is $0$ and the second term is the real part of a Gaussian characteristic function up to a constant factor.
The answer is $frac psqrtpi 4a^3 e^-p^2/2a^2$
answered May 24 at 23:15
Kavi Rama MurthyKavi Rama Murthy
86.3k53873
$endgroup$
add a comment |
$begingroup$
Integrating by parts we get $-frac 1 2a^2 e^-a^2x^2 sin(px)|_0^infty+ frac p 2a^2int_0^infty e^-a^2x^2 cos(px)dx$. The first term is $0$ and the second term is the real part of a Gaussian characteristic function up to a constant factor.
The answer is $frac psqrtpi 4a^3 e^-p^2/2a^2$
answered May 24 at 23:15
Kavi Rama MurthyKavi Rama Murthy
86.3k53873
$endgroup$
add a comment |
$begingroup$
Integrating by parts we get $-frac 1 2a^2 e^-a^2x^2 sin(px)|_0^infty+ frac p 2a^2int_0^infty e^-a^2x^2 cos(px)dx$. The first term is $0$ and the second term is the real part of a Gaussian characteristic function up to a constant factor.
The answer is $frac psqrtpi 4a^3 e^-p^2/2a^2$
answered May 24 at 23:15
Kavi Rama MurthyKavi Rama Murthy
86.3k53873
$endgroup$
Integrating by parts we get $-frac 1 2a^2 e^-a^2x^2 sin(px)|_0^infty+ frac p 2a^2int_0^infty e^-a^2x^2 cos(px)dx$. The first term is $0$ and the second term is the real part of a Gaussian characteristic function up to a constant factor.
The answer is $frac psqrtpi 4a^3 e^-p^2/2a^2$
answered May 24 at 23:15
Kavi Rama MurthyKavi Rama Murthy
86.3k53873
edited May 25 at 5:03
answered May 24 at 23:15
Kavi Rama MurthyKavi Rama Murthy
86.3k53873
answered May 24 at 23:15
Kavi Rama MurthyKavi Rama Murthy
86.3k53873
answered May 24 at 23:15
Kavi Rama MurthyKavi Rama Murthy
86.3k53873
86.3k53873
add a comment |
add a comment |
$begingroup$
$$
beginalign
int_0^infty xsinpxexp(-a^2x^2) dx
& =int_0^infty xsum_ngeq0dfrac(-1)^n(px)^2n+1(2n+1)!exp(-a^2x^2) dx \
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+1)!int_0^infty x^2n+2exp(-a^2x^2) dx \
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+1)!dfrac12a^2n+3int_0^infty u^n+frac12e^-u du ~~~;~~~a^2x^2=u\
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+2)Gamma(2n+3)dfrac12a^2n+3Gamma(n+frac32)\
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+2)2^2n+2sqrtpi^-1Gamma(n+2)Gamma(n+frac32)dfrac12a^2n+3Gamma(n+frac32)\
& =sqrtpisum_ngeq0dfrac(-1)^np^2n+12^2n+2dfrac12a^2n+3\
& =sqrtpidfracp4a^3sum_ngeq0left(dfrac-p^24a^2right)^ndfrac1n!\
& =sqrtpidfracp4a^3expleft(dfrac-p^24a^2right)
endalign
$$
answered May 25 at 1:15
NosratiNosrati
27.3k62355
$endgroup$
$begingroup$
Oh my goodness... so much work for such a simple integral (done by parts)! Obviously a technical solution with a different approach!
$endgroup$
– David G. Stork
May 25 at 1:22
$begingroup$
@DavidG.Stork This is a technical solution with different approach.
$endgroup$
– Nosrati
May 25 at 1:25
add a comment |
$begingroup$
$$
beginalign
int_0^infty xsinpxexp(-a^2x^2) dx
& =int_0^infty xsum_ngeq0dfrac(-1)^n(px)^2n+1(2n+1)!exp(-a^2x^2) dx \
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+1)!int_0^infty x^2n+2exp(-a^2x^2) dx \
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+1)!dfrac12a^2n+3int_0^infty u^n+frac12e^-u du ~~~;~~~a^2x^2=u\
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+2)Gamma(2n+3)dfrac12a^2n+3Gamma(n+frac32)\
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+2)2^2n+2sqrtpi^-1Gamma(n+2)Gamma(n+frac32)dfrac12a^2n+3Gamma(n+frac32)\
& =sqrtpisum_ngeq0dfrac(-1)^np^2n+12^2n+2dfrac12a^2n+3\
& =sqrtpidfracp4a^3sum_ngeq0left(dfrac-p^24a^2right)^ndfrac1n!\
& =sqrtpidfracp4a^3expleft(dfrac-p^24a^2right)
endalign
$$
answered May 25 at 1:15
NosratiNosrati
27.3k62355
$endgroup$
$begingroup$
Oh my goodness... so much work for such a simple integral (done by parts)! Obviously a technical solution with a different approach!
$endgroup$
– David G. Stork
May 25 at 1:22
$begingroup$
@DavidG.Stork This is a technical solution with different approach.
$endgroup$
– Nosrati
May 25 at 1:25
add a comment |
$begingroup$
$$
beginalign
int_0^infty xsinpxexp(-a^2x^2) dx
& =int_0^infty xsum_ngeq0dfrac(-1)^n(px)^2n+1(2n+1)!exp(-a^2x^2) dx \
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+1)!int_0^infty x^2n+2exp(-a^2x^2) dx \
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+1)!dfrac12a^2n+3int_0^infty u^n+frac12e^-u du ~~~;~~~a^2x^2=u\
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+2)Gamma(2n+3)dfrac12a^2n+3Gamma(n+frac32)\
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+2)2^2n+2sqrtpi^-1Gamma(n+2)Gamma(n+frac32)dfrac12a^2n+3Gamma(n+frac32)\
& =sqrtpisum_ngeq0dfrac(-1)^np^2n+12^2n+2dfrac12a^2n+3\
& =sqrtpidfracp4a^3sum_ngeq0left(dfrac-p^24a^2right)^ndfrac1n!\
& =sqrtpidfracp4a^3expleft(dfrac-p^24a^2right)
endalign
$$
answered May 25 at 1:15
NosratiNosrati
27.3k62355
$endgroup$
$$
beginalign
int_0^infty xsinpxexp(-a^2x^2) dx
& =int_0^infty xsum_ngeq0dfrac(-1)^n(px)^2n+1(2n+1)!exp(-a^2x^2) dx \
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+1)!int_0^infty x^2n+2exp(-a^2x^2) dx \
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+1)!dfrac12a^2n+3int_0^infty u^n+frac12e^-u du ~~~;~~~a^2x^2=u\
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+2)Gamma(2n+3)dfrac12a^2n+3Gamma(n+frac32)\
& =sum_ngeq0dfrac(-1)^np^2n+1(2n+2)2^2n+2sqrtpi^-1Gamma(n+2)Gamma(n+frac32)dfrac12a^2n+3Gamma(n+frac32)\
& =sqrtpisum_ngeq0dfrac(-1)^np^2n+12^2n+2dfrac12a^2n+3\
& =sqrtpidfracp4a^3sum_ngeq0left(dfrac-p^24a^2right)^ndfrac1n!\
& =sqrtpidfracp4a^3expleft(dfrac-p^24a^2right)
endalign
$$
answered May 25 at 1:15
NosratiNosrati
27.3k62355
edited May 25 at 1:21
answered May 25 at 1:15
NosratiNosrati
27.3k62355
answered May 25 at 1:15
NosratiNosrati
27.3k62355
answered May 25 at 1:15
NosratiNosrati
27.3k62355
27.3k62355
$begingroup$
Oh my goodness... so much work for such a simple integral (done by parts)! Obviously a technical solution with a different approach!
$endgroup$
– David G. Stork
May 25 at 1:22
$begingroup$
@DavidG.Stork This is a technical solution with different approach.
$endgroup$
– Nosrati
May 25 at 1:25
add a comment |
$begingroup$
Oh my goodness... so much work for such a simple integral (done by parts)! Obviously a technical solution with a different approach!
$endgroup$
– David G. Stork
May 25 at 1:22
$begingroup$
@DavidG.Stork This is a technical solution with different approach.
$endgroup$
– Nosrati
May 25 at 1:25
$begingroup$
Oh my goodness... so much work for such a simple integral (done by parts)! Obviously a technical solution with a different approach!
$endgroup$
– David G. Stork
May 25 at 1:22
$begingroup$
Oh my goodness... so much work for such a simple integral (done by parts)! Obviously a technical solution with a different approach!
$endgroup$
– David G. Stork
May 25 at 1:22
$begingroup$
@DavidG.Stork This is a technical solution with different approach.
$endgroup$
– Nosrati
May 25 at 1:25
$begingroup$
@DavidG.Stork This is a technical solution with different approach.
$endgroup$
– Nosrati
May 25 at 1:25
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3238744%2fhow-to-evaluate-the-following-integral-involving-a-gaussian%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Have you tried differentiating wrt $p$ or $a$
$endgroup$
– Henry Lee
May 24 at 23:02
2
$begingroup$
i think the integral is indeed divergent???
$endgroup$
– logo
May 24 at 23:04
1
$begingroup$
Should the last term be $exp(-cx^2)$ for some constant $c$? In its current form, the integral diverges.
$endgroup$
– VHarisop
May 24 at 23:07
1
$begingroup$
@VHarishop Sorry, I made a mistake the exponential term should be $ exp(-a^2x^2)$ and not $ exp(frac-a^2x^2)$
$endgroup$
– daljit97
May 24 at 23:08