Calculating improper integral $int limits_{0}^{infty}frac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x$












1












$begingroup$



I want to calculate the improper integral $displaystyle int
limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x$
$DeclareMathOperatorerf{erf}$




Therefore
begin{align}
I(b)&=limlimits_{bto0}left(displaystyle int limits_{b}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}xright) qquad forall binmathbb{R}:0<b<infty\
&=limlimits_{bto0}left(sqrt{pi} erf(sqrt{b}) right)=sqrt{pi}erf(sqrt{0})=sqrt{pi}
end{align}



This looks way to easy. Is this correct or am I missing something? Do you know a better way while using the following equation from our lectures?: $$displaystyle intlimits_0^infty e^{-x^2},mathrm{d}x=frac{1}{2}sqrt{pi}$$






integration definite-integrals improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    I corrected that, thank you!
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:36
















1












$begingroup$



I want to calculate the improper integral $displaystyle int
limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x$
$DeclareMathOperatorerf{erf}$




Therefore
begin{align}
I(b)&=limlimits_{bto0}left(displaystyle int limits_{b}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}xright) qquad forall binmathbb{R}:0<b<infty\
&=limlimits_{bto0}left(sqrt{pi} erf(sqrt{b}) right)=sqrt{pi}erf(sqrt{0})=sqrt{pi}
end{align}



This looks way to easy. Is this correct or am I missing something? Do you know a better way while using the following equation from our lectures?: $$displaystyle intlimits_0^infty e^{-x^2},mathrm{d}x=frac{1}{2}sqrt{pi}$$






integration definite-integrals improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    I corrected that, thank you!
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:36














1












1








1





$begingroup$



I want to calculate the improper integral $displaystyle int
limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x$
$DeclareMathOperatorerf{erf}$




Therefore
begin{align}
I(b)&=limlimits_{bto0}left(displaystyle int limits_{b}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}xright) qquad forall binmathbb{R}:0<b<infty\
&=limlimits_{bto0}left(sqrt{pi} erf(sqrt{b}) right)=sqrt{pi}erf(sqrt{0})=sqrt{pi}
end{align}



This looks way to easy. Is this correct or am I missing something? Do you know a better way while using the following equation from our lectures?: $$displaystyle intlimits_0^infty e^{-x^2},mathrm{d}x=frac{1}{2}sqrt{pi}$$






integration definite-integrals improper-integrals







share|cite|improve this question











$endgroup$





I want to calculate the improper integral $displaystyle int
limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x$
$DeclareMathOperatorerf{erf}$




Therefore
begin{align}
I(b)&=limlimits_{bto0}left(displaystyle int limits_{b}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}xright) qquad forall binmathbb{R}:0<b<infty\
&=limlimits_{bto0}left(sqrt{pi} erf(sqrt{b}) right)=sqrt{pi}erf(sqrt{0})=sqrt{pi}
end{align}



This looks way to easy. Is this correct or am I missing something? Do you know a better way while using the following equation from our lectures?: $$displaystyle intlimits_0^infty e^{-x^2},mathrm{d}x=frac{1}{2}sqrt{pi}$$






integration definite-integrals improper-integrals




integration definite-integrals improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 12:23









Martin Sleziak

44.8k10119273




44.8k10119273










asked Jan 24 at 11:26









DoesbaddelDoesbaddel

34211




34211












  • $begingroup$
    I corrected that, thank you!
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:36


















  • $begingroup$
    I corrected that, thank you!
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:36
















$begingroup$
I corrected that, thank you!
$endgroup$
– Doesbaddel
Jan 24 at 11:36




$begingroup$
I corrected that, thank you!
$endgroup$
– Doesbaddel
Jan 24 at 11:36










2 Answers
2






active

oldest

votes


















5












$begingroup$

Hint:



Just substitute $x= u^2$. So, you get
$$int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:44












  • $begingroup$
    Yes, it is the famous Gaussian integral.
    $endgroup$
    – Larry
    Jan 24 at 11:46










  • $begingroup$
    @Doesbaddel : So ist es. :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:46










  • $begingroup$
    You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
    $endgroup$
    – Yauhen Mardan
    Jan 24 at 11:47






  • 1




    $begingroup$
    @YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:49





















3












$begingroup$

It's Gamma function: $int_0^infty x^{1/2-1}e^{-x}dx=Г(1/2)=sqrt{pi}$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
    $endgroup$
    – J.G.
    Jan 24 at 12:32











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085767%2fcalculating-improper-integral-int-limits-0-infty-frac-mathrme-x%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

Hint:



Just substitute $x= u^2$. So, you get
$$int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:44












  • $begingroup$
    Yes, it is the famous Gaussian integral.
    $endgroup$
    – Larry
    Jan 24 at 11:46










  • $begingroup$
    @Doesbaddel : So ist es. :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:46










  • $begingroup$
    You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
    $endgroup$
    – Yauhen Mardan
    Jan 24 at 11:47






  • 1




    $begingroup$
    @YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:49


















5












$begingroup$

Hint:



Just substitute $x= u^2$. So, you get
$$int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:44












  • $begingroup$
    Yes, it is the famous Gaussian integral.
    $endgroup$
    – Larry
    Jan 24 at 11:46










  • $begingroup$
    @Doesbaddel : So ist es. :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:46










  • $begingroup$
    You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
    $endgroup$
    – Yauhen Mardan
    Jan 24 at 11:47






  • 1




    $begingroup$
    @YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:49
















5












5








5





$begingroup$

Hint:



Just substitute $x= u^2$. So, you get
$$int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du$$






share|cite|improve this answer









$endgroup$



Hint:



Just substitute $x= u^2$. So, you get
$$int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 at 11:31









trancelocationtrancelocation

12.6k1826




12.6k1826












  • $begingroup$
    Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:44












  • $begingroup$
    Yes, it is the famous Gaussian integral.
    $endgroup$
    – Larry
    Jan 24 at 11:46










  • $begingroup$
    @Doesbaddel : So ist es. :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:46










  • $begingroup$
    You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
    $endgroup$
    – Yauhen Mardan
    Jan 24 at 11:47






  • 1




    $begingroup$
    @YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:49




















  • $begingroup$
    Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:44












  • $begingroup$
    Yes, it is the famous Gaussian integral.
    $endgroup$
    – Larry
    Jan 24 at 11:46










  • $begingroup$
    @Doesbaddel : So ist es. :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:46










  • $begingroup$
    You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
    $endgroup$
    – Yauhen Mardan
    Jan 24 at 11:47






  • 1




    $begingroup$
    @YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:49


















$begingroup$
Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
$endgroup$
– Doesbaddel
Jan 24 at 11:44






$begingroup$
Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
$endgroup$
– Doesbaddel
Jan 24 at 11:44














$begingroup$
Yes, it is the famous Gaussian integral.
$endgroup$
– Larry
Jan 24 at 11:46




$begingroup$
Yes, it is the famous Gaussian integral.
$endgroup$
– Larry
Jan 24 at 11:46












$begingroup$
@Doesbaddel : So ist es. :-)
$endgroup$
– trancelocation
Jan 24 at 11:46




$begingroup$
@Doesbaddel : So ist es. :-)
$endgroup$
– trancelocation
Jan 24 at 11:46












$begingroup$
You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
$endgroup$
– Yauhen Mardan
Jan 24 at 11:47




$begingroup$
You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
$endgroup$
– Yauhen Mardan
Jan 24 at 11:47




1




1




$begingroup$
@YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
$endgroup$
– trancelocation
Jan 24 at 11:49






$begingroup$
@YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
$endgroup$
– trancelocation
Jan 24 at 11:49













3












$begingroup$

It's Gamma function: $int_0^infty x^{1/2-1}e^{-x}dx=Г(1/2)=sqrt{pi}$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
    $endgroup$
    – J.G.
    Jan 24 at 12:32
















3












$begingroup$

It's Gamma function: $int_0^infty x^{1/2-1}e^{-x}dx=Г(1/2)=sqrt{pi}$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
    $endgroup$
    – J.G.
    Jan 24 at 12:32














3












3








3





$begingroup$

It's Gamma function: $int_0^infty x^{1/2-1}e^{-x}dx=Г(1/2)=sqrt{pi}$






share|cite|improve this answer









$endgroup$



It's Gamma function: $int_0^infty x^{1/2-1}e^{-x}dx=Г(1/2)=sqrt{pi}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 at 11:42









Yauhen MardanYauhen Mardan

936




936








  • 1




    $begingroup$
    In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
    $endgroup$
    – J.G.
    Jan 24 at 12:32














  • 1




    $begingroup$
    In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
    $endgroup$
    – J.G.
    Jan 24 at 12:32








1




1




$begingroup$
In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
$endgroup$
– J.G.
Jan 24 at 12:32




$begingroup$
In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
$endgroup$
– J.G.
Jan 24 at 12:32


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085767%2fcalculating-improper-integral-int-limits-0-infty-frac-mathrme-x%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
Read more

Antonio Litta Visconti Arese

Image
Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
Read more