How to determine if an improper integral converges or diverges?












1












$begingroup$


Let's say we have



$int_{0}^{infty}frac{e^{-x}}{sqrt{x}}dx$.



I'd like to find the antiderivative and then analyze the integral going from 0 to N as N goes to infinity. However, I don't know how to find this particular primitive, and the online calculators do something that we haven't gone over in class (they use the Gauss error function), which leads me to suspect that there's another way of doing this.



Does anyone have an idea? Any help is appreciated, thank you.






integration riemann-integration







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    a) Do you really want to mix $x$ and $t$? b) If you just want to know some estimates, then estimate this is reasonably close to $int_0^1 t^{-1/2} , dt$.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 23:38






  • 1




    $begingroup$
    You have an $e^{-x}$ in your integrand. Is that $x$ meant to be a $t$? If not, the integrand is much more trivial to deal with since it'd be a constant w.r.t. $t$.
    $endgroup$
    – Eevee Trainer
    Dec 4 '18 at 23:38










  • $begingroup$
    @T.Bongers Apologies, had a brain fart. Yes, I meant to have everything be the same variable; fixed now.
    $endgroup$
    – user494405
    Dec 4 '18 at 23:40
















1












$begingroup$


Let's say we have



$int_{0}^{infty}frac{e^{-x}}{sqrt{x}}dx$.



I'd like to find the antiderivative and then analyze the integral going from 0 to N as N goes to infinity. However, I don't know how to find this particular primitive, and the online calculators do something that we haven't gone over in class (they use the Gauss error function), which leads me to suspect that there's another way of doing this.



Does anyone have an idea? Any help is appreciated, thank you.






integration riemann-integration







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    a) Do you really want to mix $x$ and $t$? b) If you just want to know some estimates, then estimate this is reasonably close to $int_0^1 t^{-1/2} , dt$.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 23:38






  • 1




    $begingroup$
    You have an $e^{-x}$ in your integrand. Is that $x$ meant to be a $t$? If not, the integrand is much more trivial to deal with since it'd be a constant w.r.t. $t$.
    $endgroup$
    – Eevee Trainer
    Dec 4 '18 at 23:38










  • $begingroup$
    @T.Bongers Apologies, had a brain fart. Yes, I meant to have everything be the same variable; fixed now.
    $endgroup$
    – user494405
    Dec 4 '18 at 23:40














1












1








1





$begingroup$


Let's say we have



$int_{0}^{infty}frac{e^{-x}}{sqrt{x}}dx$.



I'd like to find the antiderivative and then analyze the integral going from 0 to N as N goes to infinity. However, I don't know how to find this particular primitive, and the online calculators do something that we haven't gone over in class (they use the Gauss error function), which leads me to suspect that there's another way of doing this.



Does anyone have an idea? Any help is appreciated, thank you.






integration riemann-integration







share|cite|improve this question











$endgroup$




Let's say we have



$int_{0}^{infty}frac{e^{-x}}{sqrt{x}}dx$.



I'd like to find the antiderivative and then analyze the integral going from 0 to N as N goes to infinity. However, I don't know how to find this particular primitive, and the online calculators do something that we haven't gone over in class (they use the Gauss error function), which leads me to suspect that there's another way of doing this.



Does anyone have an idea? Any help is appreciated, thank you.






integration riemann-integration




integration riemann-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 23:39







user494405

















asked Dec 4 '18 at 23:37









user494405user494405

37919




37919








  • 1




    $begingroup$
    a) Do you really want to mix $x$ and $t$? b) If you just want to know some estimates, then estimate this is reasonably close to $int_0^1 t^{-1/2} , dt$.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 23:38






  • 1




    $begingroup$
    You have an $e^{-x}$ in your integrand. Is that $x$ meant to be a $t$? If not, the integrand is much more trivial to deal with since it'd be a constant w.r.t. $t$.
    $endgroup$
    – Eevee Trainer
    Dec 4 '18 at 23:38










  • $begingroup$
    @T.Bongers Apologies, had a brain fart. Yes, I meant to have everything be the same variable; fixed now.
    $endgroup$
    – user494405
    Dec 4 '18 at 23:40














  • 1




    $begingroup$
    a) Do you really want to mix $x$ and $t$? b) If you just want to know some estimates, then estimate this is reasonably close to $int_0^1 t^{-1/2} , dt$.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 23:38






  • 1




    $begingroup$
    You have an $e^{-x}$ in your integrand. Is that $x$ meant to be a $t$? If not, the integrand is much more trivial to deal with since it'd be a constant w.r.t. $t$.
    $endgroup$
    – Eevee Trainer
    Dec 4 '18 at 23:38










  • $begingroup$
    @T.Bongers Apologies, had a brain fart. Yes, I meant to have everything be the same variable; fixed now.
    $endgroup$
    – user494405
    Dec 4 '18 at 23:40








1




1




$begingroup$
a) Do you really want to mix $x$ and $t$? b) If you just want to know some estimates, then estimate this is reasonably close to $int_0^1 t^{-1/2} , dt$.
$endgroup$
– T. Bongers
Dec 4 '18 at 23:38




$begingroup$
a) Do you really want to mix $x$ and $t$? b) If you just want to know some estimates, then estimate this is reasonably close to $int_0^1 t^{-1/2} , dt$.
$endgroup$
– T. Bongers
Dec 4 '18 at 23:38




1




1




$begingroup$
You have an $e^{-x}$ in your integrand. Is that $x$ meant to be a $t$? If not, the integrand is much more trivial to deal with since it'd be a constant w.r.t. $t$.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 23:38




$begingroup$
You have an $e^{-x}$ in your integrand. Is that $x$ meant to be a $t$? If not, the integrand is much more trivial to deal with since it'd be a constant w.r.t. $t$.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 23:38












$begingroup$
@T.Bongers Apologies, had a brain fart. Yes, I meant to have everything be the same variable; fixed now.
$endgroup$
– user494405
Dec 4 '18 at 23:40




$begingroup$
@T.Bongers Apologies, had a brain fart. Yes, I meant to have everything be the same variable; fixed now.
$endgroup$
– user494405
Dec 4 '18 at 23:40










1 Answer
1






active

oldest

votes


















3












$begingroup$

Finding an antiderivative is going to be difficult. Seeing if it converges will not,
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
int_0^1 frac{e^{-x}}{sqrt{x}}mathrm dx+int_1^infty frac{e^{-x}}{sqrt{x}}mathrm dx
$$
Now, for $0leq xleq 1$, we have
$$
frac{e^{-x}}{sqrt{x}}leq frac{1}{sqrt{x}}
$$
and for $1leq x$, we have
$$
frac{e^{-x}}{sqrt{x}}leq e^{-x}
$$



Alternatively, enforce the substitution $u=sqrt{x}$, then your integral is
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
2int_0^infty e^{-u^2}mathrm du=sqrt{pi}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
    $endgroup$
    – user494405
    Dec 4 '18 at 23:51










  • $begingroup$
    Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
    $endgroup$
    – user494405
    Dec 4 '18 at 23:54










  • $begingroup$
    @user494405 that's exactly right
    $endgroup$
    – qbert
    Dec 5 '18 at 0:09










  • $begingroup$
    Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
    $endgroup$
    – user494405
    Dec 5 '18 at 0:13










  • $begingroup$
    there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
    $endgroup$
    – qbert
    Dec 5 '18 at 0:16













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Finding an antiderivative is going to be difficult. Seeing if it converges will not,
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
int_0^1 frac{e^{-x}}{sqrt{x}}mathrm dx+int_1^infty frac{e^{-x}}{sqrt{x}}mathrm dx
$$
Now, for $0leq xleq 1$, we have
$$
frac{e^{-x}}{sqrt{x}}leq frac{1}{sqrt{x}}
$$
and for $1leq x$, we have
$$
frac{e^{-x}}{sqrt{x}}leq e^{-x}
$$



Alternatively, enforce the substitution $u=sqrt{x}$, then your integral is
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
2int_0^infty e^{-u^2}mathrm du=sqrt{pi}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
    $endgroup$
    – user494405
    Dec 4 '18 at 23:51










  • $begingroup$
    Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
    $endgroup$
    – user494405
    Dec 4 '18 at 23:54










  • $begingroup$
    @user494405 that's exactly right
    $endgroup$
    – qbert
    Dec 5 '18 at 0:09










  • $begingroup$
    Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
    $endgroup$
    – user494405
    Dec 5 '18 at 0:13










  • $begingroup$
    there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
    $endgroup$
    – qbert
    Dec 5 '18 at 0:16


















3












$begingroup$

Finding an antiderivative is going to be difficult. Seeing if it converges will not,
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
int_0^1 frac{e^{-x}}{sqrt{x}}mathrm dx+int_1^infty frac{e^{-x}}{sqrt{x}}mathrm dx
$$
Now, for $0leq xleq 1$, we have
$$
frac{e^{-x}}{sqrt{x}}leq frac{1}{sqrt{x}}
$$
and for $1leq x$, we have
$$
frac{e^{-x}}{sqrt{x}}leq e^{-x}
$$



Alternatively, enforce the substitution $u=sqrt{x}$, then your integral is
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
2int_0^infty e^{-u^2}mathrm du=sqrt{pi}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
    $endgroup$
    – user494405
    Dec 4 '18 at 23:51










  • $begingroup$
    Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
    $endgroup$
    – user494405
    Dec 4 '18 at 23:54










  • $begingroup$
    @user494405 that's exactly right
    $endgroup$
    – qbert
    Dec 5 '18 at 0:09










  • $begingroup$
    Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
    $endgroup$
    – user494405
    Dec 5 '18 at 0:13










  • $begingroup$
    there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
    $endgroup$
    – qbert
    Dec 5 '18 at 0:16
















3












3








3





$begingroup$

Finding an antiderivative is going to be difficult. Seeing if it converges will not,
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
int_0^1 frac{e^{-x}}{sqrt{x}}mathrm dx+int_1^infty frac{e^{-x}}{sqrt{x}}mathrm dx
$$
Now, for $0leq xleq 1$, we have
$$
frac{e^{-x}}{sqrt{x}}leq frac{1}{sqrt{x}}
$$
and for $1leq x$, we have
$$
frac{e^{-x}}{sqrt{x}}leq e^{-x}
$$



Alternatively, enforce the substitution $u=sqrt{x}$, then your integral is
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
2int_0^infty e^{-u^2}mathrm du=sqrt{pi}
$$






share|cite|improve this answer









$endgroup$



Finding an antiderivative is going to be difficult. Seeing if it converges will not,
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
int_0^1 frac{e^{-x}}{sqrt{x}}mathrm dx+int_1^infty frac{e^{-x}}{sqrt{x}}mathrm dx
$$
Now, for $0leq xleq 1$, we have
$$
frac{e^{-x}}{sqrt{x}}leq frac{1}{sqrt{x}}
$$
and for $1leq x$, we have
$$
frac{e^{-x}}{sqrt{x}}leq e^{-x}
$$



Alternatively, enforce the substitution $u=sqrt{x}$, then your integral is
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
2int_0^infty e^{-u^2}mathrm du=sqrt{pi}
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 23:42









qbertqbert

22.1k32560




22.1k32560












  • $begingroup$
    Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
    $endgroup$
    – user494405
    Dec 4 '18 at 23:51










  • $begingroup$
    Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
    $endgroup$
    – user494405
    Dec 4 '18 at 23:54










  • $begingroup$
    @user494405 that's exactly right
    $endgroup$
    – qbert
    Dec 5 '18 at 0:09










  • $begingroup$
    Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
    $endgroup$
    – user494405
    Dec 5 '18 at 0:13










  • $begingroup$
    there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
    $endgroup$
    – qbert
    Dec 5 '18 at 0:16




















  • $begingroup$
    Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
    $endgroup$
    – user494405
    Dec 4 '18 at 23:51










  • $begingroup$
    Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
    $endgroup$
    – user494405
    Dec 4 '18 at 23:54










  • $begingroup$
    @user494405 that's exactly right
    $endgroup$
    – qbert
    Dec 5 '18 at 0:09










  • $begingroup$
    Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
    $endgroup$
    – user494405
    Dec 5 '18 at 0:13










  • $begingroup$
    there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
    $endgroup$
    – qbert
    Dec 5 '18 at 0:16


















$begingroup$
Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
$endgroup$
– user494405
Dec 4 '18 at 23:51




$begingroup$
Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
$endgroup$
– user494405
Dec 4 '18 at 23:51












$begingroup$
Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
$endgroup$
– user494405
Dec 4 '18 at 23:54




$begingroup$
Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
$endgroup$
– user494405
Dec 4 '18 at 23:54












$begingroup$
@user494405 that's exactly right
$endgroup$
– qbert
Dec 5 '18 at 0:09




$begingroup$
@user494405 that's exactly right
$endgroup$
– qbert
Dec 5 '18 at 0:09












$begingroup$
Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
$endgroup$
– user494405
Dec 5 '18 at 0:13




$begingroup$
Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
$endgroup$
– user494405
Dec 5 '18 at 0:13












$begingroup$
there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
$endgroup$
– qbert
Dec 5 '18 at 0:16






$begingroup$
there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
$endgroup$
– qbert
Dec 5 '18 at 0:16




















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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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