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Determine whether the integral $ int^+infty_0frace^-t sqrt t , dt$ converges or diverges?
Clash Royale CLAN TAG#URR8PPP
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$$ int^+infty_0frace^-t sqrt t , dt$$
I have computed it getting $$ frac2e3 therefore text it converges.$$ using integration by parts letting $$u = frac 1 sqrt t $$ and $$ dV= e^-t , dt$$
Is it the right way to do it ?
calculus integration convergence improper-integrals
edited Aug 6 at 7:46
Nosrati
20.1k41644
asked Aug 5 at 14:45
Mayar
385
add a comment |Â
up vote
6
down vote
favorite
$$ int^+infty_0frace^-t sqrt t , dt$$
I have computed it getting $$ frac2e3 therefore text it converges.$$ using integration by parts letting $$u = frac 1 sqrt t $$ and $$ dV= e^-t , dt$$
Is it the right way to do it ?
calculus integration convergence improper-integrals
edited Aug 6 at 7:46
Nosrati
20.1k41644
asked Aug 5 at 14:45
Mayar
385
Do you know of the gamma function?
– TheSimpliFire
Aug 5 at 14:48
1
@TheSimpliFire no I don't
– Mayar
Aug 5 at 14:49
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
$$ int^+infty_0frace^-t sqrt t , dt$$
I have computed it getting $$ frac2e3 therefore text it converges.$$ using integration by parts letting $$u = frac 1 sqrt t $$ and $$ dV= e^-t , dt$$
Is it the right way to do it ?
calculus integration convergence improper-integrals
edited Aug 6 at 7:46
Nosrati
20.1k41644
asked Aug 5 at 14:45
Mayar
385
$$ int^+infty_0frace^-t sqrt t , dt$$
I have computed it getting $$ frac2e3 therefore text it converges.$$ using integration by parts letting $$u = frac 1 sqrt t $$ and $$ dV= e^-t , dt$$
Is it the right way to do it ?
calculus integration convergence improper-integrals
edited Aug 6 at 7:46
Nosrati
20.1k41644
asked Aug 5 at 14:45
Mayar
385
edited Aug 6 at 7:46
Nosrati
20.1k41644
edited Aug 6 at 7:46
Nosrati
20.1k41644
edited Aug 6 at 7:46
Nosrati
20.1k41644
20.1k41644
asked Aug 5 at 14:45
Mayar
385
asked Aug 5 at 14:45
Mayar
385
asked Aug 5 at 14:45
Mayar
385
385
Do you know of the gamma function?
– TheSimpliFire
Aug 5 at 14:48
1
@TheSimpliFire no I don't
– Mayar
Aug 5 at 14:49
add a comment |Â
Do you know of the gamma function?
– TheSimpliFire
Aug 5 at 14:48
1
@TheSimpliFire no I don't
– Mayar
Aug 5 at 14:49
Do you know of the gamma function?
– TheSimpliFire
Aug 5 at 14:48
Do you know of the gamma function?
– TheSimpliFire
Aug 5 at 14:48
1
1
@TheSimpliFire no I don't
– Mayar
Aug 5 at 14:49
@TheSimpliFire no I don't
– Mayar
Aug 5 at 14:49
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
15
down vote
If you simply want to decide convergence and you don't notice the trick in Henry Lee's answer, you can look at the following. Near zero,
$$
int_0^1 frace^-tsqrt t,dtleq int_0^1frac1sqrt t,dt,
$$
which is convergent.
At infinity,
$$
int_1^infty frace^-tsqrt t,dtleqint_1^infty e^-t,dt<infty.
$$
So the integral converges.
answered Aug 5 at 15:03
Martin Argerami
116k1071164
It is exactly the same idea I had. I let also my answer since I've suggested e different criteria for convergence.
– gimusi
Aug 6 at 8:12
add a comment |Â
up vote
7
down vote
$$I=int_0^inftyfrace^-tsqrtt,dt$$
$u=sqrtt,,dt=2sqrttdu$
$$I=2int_0^infty e^-u^2,du=sqrtpi$$ As it is a standard integral
EDIT:
$$I=2int_0^infty e^-x^2,dx$$
then
$$I^2=4left(int_0^infty e^-x^2,dxright)^2=4left(int_0^infty e^-x^2right)left(int_0^infty e^-y^2 , dyright) = 4int_0^infty int_0^infty e^-(x^2+y^2) , dx , dy$$
now we can use polar coordinates to simplify this.
$x^2+y^2=r^2,$ and $dA=dx,dy=r,dr,dtheta$ so our integral becomes:
$$I^2=4int_0^fracpi2 int_0^infty e^-r^2r,dr,dtheta$$
now $u=-r^2$ so $fracdudr=-2r, therefore,dr=fracdu-2r$ and the integral becomes:
$$I^2=-2int_0^fracpi2 int_0^-inftye^u , du , dtheta = 2int_0^fracpi2 int_-infty^0 e^u,du,dtheta = 2int_0^fracpi2 left[e^uright]_-infty^0 , dtheta = 2int_0^fracpi2 , dtheta =2cdotfracpi2=pi$$
so if $I^2=pi$ then $I=sqrtpi$
answered Aug 5 at 14:49
Henry Lee
49210
This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
– Henry Lee
Aug 5 at 14:51
Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
– Mayar
Aug 5 at 19:28
I don't really understand your notation, but I will add to my answer to show why this is the case
– Henry Lee
Aug 5 at 19:30
Okay thank you .
– Mayar
Aug 5 at 19:31
I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
– Henry Lee
Aug 5 at 19:44
 |Â
show 2 more comments
up vote
4
down vote
$$int_0^+inftyfrace^-tsqrttdt=int_0^1frace^-tsqrtt , dt + int_1^+inftyfrace^-tsqrtt , dt=A+B$$
for $A$ when $e^-tsim1$ then
$$A=int_0^1 frace^-tsqrtt,dtsimint_0^1 frac1sqrtt , dt < infty$$
for $B$ when $tgeq1$ then
$$B=int_1^+inftyfrace^-tsqrtt,dtleqint_1^+inftye^-t , dt < infty$$
edited Aug 5 at 15:15
Michael Hardy
204k23186463
answered Aug 5 at 15:01
Nosrati
20.1k41644
That's also exactly my idea to proceed. Maybe for the integral A you should recall that we need to refer to limit comparison test to prove convergence.
– gimusi
Aug 6 at 8:33
add a comment |Â
up vote
3
down vote
As Henry Lee's answer mentions, you get $sqrtpi,$ so $2e/3$ is wrong. But if the function is everywhere positive and you computed it correctly and get a finite number rather than $+infty,$ then it converges. If it's not everywhere positive, then the meaning of convergence may become problematic. For example,
$$
lim_a,to,+inftyint_0^a fracsin x x , dx = frac pi 2
$$
and that is a sort of convergence, but
$$
intlimits_,x ,:, (sin x)/x , ge,0 fracsin x x,dx = +infty quadtext and quad intlimits_,x ,:, (sin x)/x , < ,0 fracsin x x,dx = -infty
$$
and some questions arise about when one should consider the thing convergent.
But one thing to bear in mind is that the question of whether something converges or not is often simpler than the question of what it converges to. Thus
$$
0 < int_0^1 frace^-tsqrt t , dt le int_0^1 frac 1 sqrt t , dt < +infty
$$
and
$$
0 < int_1^infty frace^-tsqrt t,dt le int_1^infty e^-t , dx = e^-1 < +infty
$$
so what you have converges.
answered Aug 5 at 15:14
Michael Hardy
204k23186463
add a comment |Â
up vote
0
down vote
If we can calculate a finite value for the integral of course the integral converges but if our goal is simply to check for the convergence we don’t need to calculate it explicitly but we can proceed for example by some quick convergence test.
For example, in this case we have that
$$ int^+infty_0frace^-t sqrt t , dt=int^1_0frace^-t sqrt t , dt+
int^+infty_1frace^-t sqrt t , dt$$
and as $tto 0^+$
$$frace^-t sqrt t sim
frac1 sqrt t$$
and as $tto infty$
$$frace^-t sqrt t sim
frac1 e^t$$
therefore both integral converge by limit comparison test and therefore the given integral converges.
answered Aug 6 at 7:55
gimusi
65.5k73684
you are late ;)
– Nosrati
Aug 6 at 8:07
@user108128 Sometimes it is good, so I can think to a better answer ;)
– gimusi
Aug 6 at 8:08
@user108128 I see now that my answer is very similar to that given by Martin, I let mine since I gave a suggestion by limit comparison test which avoid to use inequalities.
– gimusi
Aug 6 at 8:11
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
If you simply want to decide convergence and you don't notice the trick in Henry Lee's answer, you can look at the following. Near zero,
$$
int_0^1 frace^-tsqrt t,dtleq int_0^1frac1sqrt t,dt,
$$
which is convergent.
At infinity,
$$
int_1^infty frace^-tsqrt t,dtleqint_1^infty e^-t,dt<infty.
$$
So the integral converges.
answered Aug 5 at 15:03
Martin Argerami
116k1071164
It is exactly the same idea I had. I let also my answer since I've suggested e different criteria for convergence.
– gimusi
Aug 6 at 8:12
add a comment |Â
up vote
15
down vote
If you simply want to decide convergence and you don't notice the trick in Henry Lee's answer, you can look at the following. Near zero,
$$
int_0^1 frace^-tsqrt t,dtleq int_0^1frac1sqrt t,dt,
$$
which is convergent.
At infinity,
$$
int_1^infty frace^-tsqrt t,dtleqint_1^infty e^-t,dt<infty.
$$
So the integral converges.
answered Aug 5 at 15:03
Martin Argerami
116k1071164
It is exactly the same idea I had. I let also my answer since I've suggested e different criteria for convergence.
– gimusi
Aug 6 at 8:12
add a comment |Â
up vote
15
down vote
up vote
15
down vote
If you simply want to decide convergence and you don't notice the trick in Henry Lee's answer, you can look at the following. Near zero,
$$
int_0^1 frace^-tsqrt t,dtleq int_0^1frac1sqrt t,dt,
$$
which is convergent.
At infinity,
$$
int_1^infty frace^-tsqrt t,dtleqint_1^infty e^-t,dt<infty.
$$
So the integral converges.
answered Aug 5 at 15:03
Martin Argerami
116k1071164
If you simply want to decide convergence and you don't notice the trick in Henry Lee's answer, you can look at the following. Near zero,
$$
int_0^1 frace^-tsqrt t,dtleq int_0^1frac1sqrt t,dt,
$$
which is convergent.
At infinity,
$$
int_1^infty frace^-tsqrt t,dtleqint_1^infty e^-t,dt<infty.
$$
So the integral converges.
answered Aug 5 at 15:03
Martin Argerami
116k1071164
answered Aug 5 at 15:03
Martin Argerami
116k1071164
answered Aug 5 at 15:03
Martin Argerami
116k1071164
answered Aug 5 at 15:03
Martin Argerami
116k1071164
116k1071164
It is exactly the same idea I had. I let also my answer since I've suggested e different criteria for convergence.
– gimusi
Aug 6 at 8:12
add a comment |Â
It is exactly the same idea I had. I let also my answer since I've suggested e different criteria for convergence.
– gimusi
Aug 6 at 8:12
It is exactly the same idea I had. I let also my answer since I've suggested e different criteria for convergence.
– gimusi
Aug 6 at 8:12
It is exactly the same idea I had. I let also my answer since I've suggested e different criteria for convergence.
– gimusi
Aug 6 at 8:12
add a comment |Â
up vote
7
down vote
$$I=int_0^inftyfrace^-tsqrtt,dt$$
$u=sqrtt,,dt=2sqrttdu$
$$I=2int_0^infty e^-u^2,du=sqrtpi$$ As it is a standard integral
EDIT:
$$I=2int_0^infty e^-x^2,dx$$
then
$$I^2=4left(int_0^infty e^-x^2,dxright)^2=4left(int_0^infty e^-x^2right)left(int_0^infty e^-y^2 , dyright) = 4int_0^infty int_0^infty e^-(x^2+y^2) , dx , dy$$
now we can use polar coordinates to simplify this.
$x^2+y^2=r^2,$ and $dA=dx,dy=r,dr,dtheta$ so our integral becomes:
$$I^2=4int_0^fracpi2 int_0^infty e^-r^2r,dr,dtheta$$
now $u=-r^2$ so $fracdudr=-2r, therefore,dr=fracdu-2r$ and the integral becomes:
$$I^2=-2int_0^fracpi2 int_0^-inftye^u , du , dtheta = 2int_0^fracpi2 int_-infty^0 e^u,du,dtheta = 2int_0^fracpi2 left[e^uright]_-infty^0 , dtheta = 2int_0^fracpi2 , dtheta =2cdotfracpi2=pi$$
so if $I^2=pi$ then $I=sqrtpi$
answered Aug 5 at 14:49
Henry Lee
49210
This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
– Henry Lee
Aug 5 at 14:51
Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
– Mayar
Aug 5 at 19:28
I don't really understand your notation, but I will add to my answer to show why this is the case
– Henry Lee
Aug 5 at 19:30
Okay thank you .
– Mayar
Aug 5 at 19:31
I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
– Henry Lee
Aug 5 at 19:44
 |Â
show 2 more comments
up vote
7
down vote
$$I=int_0^inftyfrace^-tsqrtt,dt$$
$u=sqrtt,,dt=2sqrttdu$
$$I=2int_0^infty e^-u^2,du=sqrtpi$$ As it is a standard integral
EDIT:
$$I=2int_0^infty e^-x^2,dx$$
then
$$I^2=4left(int_0^infty e^-x^2,dxright)^2=4left(int_0^infty e^-x^2right)left(int_0^infty e^-y^2 , dyright) = 4int_0^infty int_0^infty e^-(x^2+y^2) , dx , dy$$
now we can use polar coordinates to simplify this.
$x^2+y^2=r^2,$ and $dA=dx,dy=r,dr,dtheta$ so our integral becomes:
$$I^2=4int_0^fracpi2 int_0^infty e^-r^2r,dr,dtheta$$
now $u=-r^2$ so $fracdudr=-2r, therefore,dr=fracdu-2r$ and the integral becomes:
$$I^2=-2int_0^fracpi2 int_0^-inftye^u , du , dtheta = 2int_0^fracpi2 int_-infty^0 e^u,du,dtheta = 2int_0^fracpi2 left[e^uright]_-infty^0 , dtheta = 2int_0^fracpi2 , dtheta =2cdotfracpi2=pi$$
so if $I^2=pi$ then $I=sqrtpi$
answered Aug 5 at 14:49
Henry Lee
49210
This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
– Henry Lee
Aug 5 at 14:51
Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
– Mayar
Aug 5 at 19:28
I don't really understand your notation, but I will add to my answer to show why this is the case
– Henry Lee
Aug 5 at 19:30
Okay thank you .
– Mayar
Aug 5 at 19:31
I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
– Henry Lee
Aug 5 at 19:44
 |Â
show 2 more comments
up vote
7
down vote
up vote
7
down vote
$$I=int_0^inftyfrace^-tsqrtt,dt$$
$u=sqrtt,,dt=2sqrttdu$
$$I=2int_0^infty e^-u^2,du=sqrtpi$$ As it is a standard integral
EDIT:
$$I=2int_0^infty e^-x^2,dx$$
then
$$I^2=4left(int_0^infty e^-x^2,dxright)^2=4left(int_0^infty e^-x^2right)left(int_0^infty e^-y^2 , dyright) = 4int_0^infty int_0^infty e^-(x^2+y^2) , dx , dy$$
now we can use polar coordinates to simplify this.
$x^2+y^2=r^2,$ and $dA=dx,dy=r,dr,dtheta$ so our integral becomes:
$$I^2=4int_0^fracpi2 int_0^infty e^-r^2r,dr,dtheta$$
now $u=-r^2$ so $fracdudr=-2r, therefore,dr=fracdu-2r$ and the integral becomes:
$$I^2=-2int_0^fracpi2 int_0^-inftye^u , du , dtheta = 2int_0^fracpi2 int_-infty^0 e^u,du,dtheta = 2int_0^fracpi2 left[e^uright]_-infty^0 , dtheta = 2int_0^fracpi2 , dtheta =2cdotfracpi2=pi$$
so if $I^2=pi$ then $I=sqrtpi$
answered Aug 5 at 14:49
Henry Lee
49210
$$I=int_0^inftyfrace^-tsqrtt,dt$$
$u=sqrtt,,dt=2sqrttdu$
$$I=2int_0^infty e^-u^2,du=sqrtpi$$ As it is a standard integral
EDIT:
$$I=2int_0^infty e^-x^2,dx$$
then
$$I^2=4left(int_0^infty e^-x^2,dxright)^2=4left(int_0^infty e^-x^2right)left(int_0^infty e^-y^2 , dyright) = 4int_0^infty int_0^infty e^-(x^2+y^2) , dx , dy$$
now we can use polar coordinates to simplify this.
$x^2+y^2=r^2,$ and $dA=dx,dy=r,dr,dtheta$ so our integral becomes:
$$I^2=4int_0^fracpi2 int_0^infty e^-r^2r,dr,dtheta$$
now $u=-r^2$ so $fracdudr=-2r, therefore,dr=fracdu-2r$ and the integral becomes:
$$I^2=-2int_0^fracpi2 int_0^-inftye^u , du , dtheta = 2int_0^fracpi2 int_-infty^0 e^u,du,dtheta = 2int_0^fracpi2 left[e^uright]_-infty^0 , dtheta = 2int_0^fracpi2 , dtheta =2cdotfracpi2=pi$$
so if $I^2=pi$ then $I=sqrtpi$
answered Aug 5 at 14:49
Henry Lee
49210
edited Aug 6 at 14:05
answered Aug 5 at 14:49
Henry Lee
49210
answered Aug 5 at 14:49
Henry Lee
49210
answered Aug 5 at 14:49
Henry Lee
49210
49210
This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
– Henry Lee
Aug 5 at 14:51
Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
– Mayar
Aug 5 at 19:28
I don't really understand your notation, but I will add to my answer to show why this is the case
– Henry Lee
Aug 5 at 19:30
Okay thank you .
– Mayar
Aug 5 at 19:31
I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
– Henry Lee
Aug 5 at 19:44
 |Â
show 2 more comments
This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
– Henry Lee
Aug 5 at 14:51
Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
– Mayar
Aug 5 at 19:28
I don't really understand your notation, but I will add to my answer to show why this is the case
– Henry Lee
Aug 5 at 19:30
Okay thank you .
– Mayar
Aug 5 at 19:31
I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
– Henry Lee
Aug 5 at 19:44
This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
– Henry Lee
Aug 5 at 14:51
This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
– Henry Lee
Aug 5 at 14:51
Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
– Mayar
Aug 5 at 19:28
Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
– Mayar
Aug 5 at 19:28
I don't really understand your notation, but I will add to my answer to show why this is the case
– Henry Lee
Aug 5 at 19:30
I don't really understand your notation, but I will add to my answer to show why this is the case
– Henry Lee
Aug 5 at 19:30
Okay thank you .
– Mayar
Aug 5 at 19:31
Okay thank you .
– Mayar
Aug 5 at 19:31
I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
– Henry Lee
Aug 5 at 19:44
I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
– Henry Lee
Aug 5 at 19:44
 |Â
show 2 more comments
up vote
4
down vote
$$int_0^+inftyfrace^-tsqrttdt=int_0^1frace^-tsqrtt , dt + int_1^+inftyfrace^-tsqrtt , dt=A+B$$
for $A$ when $e^-tsim1$ then
$$A=int_0^1 frace^-tsqrtt,dtsimint_0^1 frac1sqrtt , dt < infty$$
for $B$ when $tgeq1$ then
$$B=int_1^+inftyfrace^-tsqrtt,dtleqint_1^+inftye^-t , dt < infty$$
edited Aug 5 at 15:15
Michael Hardy
204k23186463
answered Aug 5 at 15:01
Nosrati
20.1k41644
That's also exactly my idea to proceed. Maybe for the integral A you should recall that we need to refer to limit comparison test to prove convergence.
– gimusi
Aug 6 at 8:33
add a comment |Â
up vote
4
down vote
$$int_0^+inftyfrace^-tsqrttdt=int_0^1frace^-tsqrtt , dt + int_1^+inftyfrace^-tsqrtt , dt=A+B$$
for $A$ when $e^-tsim1$ then
$$A=int_0^1 frace^-tsqrtt,dtsimint_0^1 frac1sqrtt , dt < infty$$
for $B$ when $tgeq1$ then
$$B=int_1^+inftyfrace^-tsqrtt,dtleqint_1^+inftye^-t , dt < infty$$
edited Aug 5 at 15:15
Michael Hardy
204k23186463
answered Aug 5 at 15:01
Nosrati
20.1k41644
That's also exactly my idea to proceed. Maybe for the integral A you should recall that we need to refer to limit comparison test to prove convergence.
– gimusi
Aug 6 at 8:33
add a comment |Â
up vote
4
down vote
up vote
4
down vote
$$int_0^+inftyfrace^-tsqrttdt=int_0^1frace^-tsqrtt , dt + int_1^+inftyfrace^-tsqrtt , dt=A+B$$
for $A$ when $e^-tsim1$ then
$$A=int_0^1 frace^-tsqrtt,dtsimint_0^1 frac1sqrtt , dt < infty$$
for $B$ when $tgeq1$ then
$$B=int_1^+inftyfrace^-tsqrtt,dtleqint_1^+inftye^-t , dt < infty$$
edited Aug 5 at 15:15
Michael Hardy
204k23186463
answered Aug 5 at 15:01
Nosrati
20.1k41644
$$int_0^+inftyfrace^-tsqrttdt=int_0^1frace^-tsqrtt , dt + int_1^+inftyfrace^-tsqrtt , dt=A+B$$
for $A$ when $e^-tsim1$ then
$$A=int_0^1 frace^-tsqrtt,dtsimint_0^1 frac1sqrtt , dt < infty$$
for $B$ when $tgeq1$ then
$$B=int_1^+inftyfrace^-tsqrtt,dtleqint_1^+inftye^-t , dt < infty$$
edited Aug 5 at 15:15
Michael Hardy
204k23186463
answered Aug 5 at 15:01
Nosrati
20.1k41644
edited Aug 5 at 15:15
Michael Hardy
204k23186463
edited Aug 5 at 15:15
Michael Hardy
204k23186463
edited Aug 5 at 15:15
Michael Hardy
204k23186463
204k23186463
answered Aug 5 at 15:01
Nosrati
20.1k41644
answered Aug 5 at 15:01
Nosrati
20.1k41644
answered Aug 5 at 15:01
Nosrati
20.1k41644
20.1k41644
That's also exactly my idea to proceed. Maybe for the integral A you should recall that we need to refer to limit comparison test to prove convergence.
– gimusi
Aug 6 at 8:33
add a comment |Â
That's also exactly my idea to proceed. Maybe for the integral A you should recall that we need to refer to limit comparison test to prove convergence.
– gimusi
Aug 6 at 8:33
That's also exactly my idea to proceed. Maybe for the integral A you should recall that we need to refer to limit comparison test to prove convergence.
– gimusi
Aug 6 at 8:33
That's also exactly my idea to proceed. Maybe for the integral A you should recall that we need to refer to limit comparison test to prove convergence.
– gimusi
Aug 6 at 8:33
add a comment |Â
up vote
3
down vote
As Henry Lee's answer mentions, you get $sqrtpi,$ so $2e/3$ is wrong. But if the function is everywhere positive and you computed it correctly and get a finite number rather than $+infty,$ then it converges. If it's not everywhere positive, then the meaning of convergence may become problematic. For example,
$$
lim_a,to,+inftyint_0^a fracsin x x , dx = frac pi 2
$$
and that is a sort of convergence, but
$$
intlimits_,x ,:, (sin x)/x , ge,0 fracsin x x,dx = +infty quadtext and quad intlimits_,x ,:, (sin x)/x , < ,0 fracsin x x,dx = -infty
$$
and some questions arise about when one should consider the thing convergent.
But one thing to bear in mind is that the question of whether something converges or not is often simpler than the question of what it converges to. Thus
$$
0 < int_0^1 frace^-tsqrt t , dt le int_0^1 frac 1 sqrt t , dt < +infty
$$
and
$$
0 < int_1^infty frace^-tsqrt t,dt le int_1^infty e^-t , dx = e^-1 < +infty
$$
so what you have converges.
answered Aug 5 at 15:14
Michael Hardy
204k23186463
add a comment |Â
up vote
3
down vote
As Henry Lee's answer mentions, you get $sqrtpi,$ so $2e/3$ is wrong. But if the function is everywhere positive and you computed it correctly and get a finite number rather than $+infty,$ then it converges. If it's not everywhere positive, then the meaning of convergence may become problematic. For example,
$$
lim_a,to,+inftyint_0^a fracsin x x , dx = frac pi 2
$$
and that is a sort of convergence, but
$$
intlimits_,x ,:, (sin x)/x , ge,0 fracsin x x,dx = +infty quadtext and quad intlimits_,x ,:, (sin x)/x , < ,0 fracsin x x,dx = -infty
$$
and some questions arise about when one should consider the thing convergent.
But one thing to bear in mind is that the question of whether something converges or not is often simpler than the question of what it converges to. Thus
$$
0 < int_0^1 frace^-tsqrt t , dt le int_0^1 frac 1 sqrt t , dt < +infty
$$
and
$$
0 < int_1^infty frace^-tsqrt t,dt le int_1^infty e^-t , dx = e^-1 < +infty
$$
so what you have converges.
answered Aug 5 at 15:14
Michael Hardy
204k23186463
add a comment |Â
up vote
3
down vote
up vote
3
down vote
As Henry Lee's answer mentions, you get $sqrtpi,$ so $2e/3$ is wrong. But if the function is everywhere positive and you computed it correctly and get a finite number rather than $+infty,$ then it converges. If it's not everywhere positive, then the meaning of convergence may become problematic. For example,
$$
lim_a,to,+inftyint_0^a fracsin x x , dx = frac pi 2
$$
and that is a sort of convergence, but
$$
intlimits_,x ,:, (sin x)/x , ge,0 fracsin x x,dx = +infty quadtext and quad intlimits_,x ,:, (sin x)/x , < ,0 fracsin x x,dx = -infty
$$
and some questions arise about when one should consider the thing convergent.
But one thing to bear in mind is that the question of whether something converges or not is often simpler than the question of what it converges to. Thus
$$
0 < int_0^1 frace^-tsqrt t , dt le int_0^1 frac 1 sqrt t , dt < +infty
$$
and
$$
0 < int_1^infty frace^-tsqrt t,dt le int_1^infty e^-t , dx = e^-1 < +infty
$$
so what you have converges.
answered Aug 5 at 15:14
Michael Hardy
204k23186463
As Henry Lee's answer mentions, you get $sqrtpi,$ so $2e/3$ is wrong. But if the function is everywhere positive and you computed it correctly and get a finite number rather than $+infty,$ then it converges. If it's not everywhere positive, then the meaning of convergence may become problematic. For example,
$$
lim_a,to,+inftyint_0^a fracsin x x , dx = frac pi 2
$$
and that is a sort of convergence, but
$$
intlimits_,x ,:, (sin x)/x , ge,0 fracsin x x,dx = +infty quadtext and quad intlimits_,x ,:, (sin x)/x , < ,0 fracsin x x,dx = -infty
$$
and some questions arise about when one should consider the thing convergent.
But one thing to bear in mind is that the question of whether something converges or not is often simpler than the question of what it converges to. Thus
$$
0 < int_0^1 frace^-tsqrt t , dt le int_0^1 frac 1 sqrt t , dt < +infty
$$
and
$$
0 < int_1^infty frace^-tsqrt t,dt le int_1^infty e^-t , dx = e^-1 < +infty
$$
so what you have converges.
answered Aug 5 at 15:14
Michael Hardy
204k23186463
edited Aug 6 at 2:15
answered Aug 5 at 15:14
Michael Hardy
204k23186463
answered Aug 5 at 15:14
Michael Hardy
204k23186463
answered Aug 5 at 15:14
Michael Hardy
204k23186463
204k23186463
add a comment |Â
add a comment |Â
up vote
0
down vote
If we can calculate a finite value for the integral of course the integral converges but if our goal is simply to check for the convergence we don’t need to calculate it explicitly but we can proceed for example by some quick convergence test.
For example, in this case we have that
$$ int^+infty_0frace^-t sqrt t , dt=int^1_0frace^-t sqrt t , dt+
int^+infty_1frace^-t sqrt t , dt$$
and as $tto 0^+$
$$frace^-t sqrt t sim
frac1 sqrt t$$
and as $tto infty$
$$frace^-t sqrt t sim
frac1 e^t$$
therefore both integral converge by limit comparison test and therefore the given integral converges.
answered Aug 6 at 7:55
gimusi
65.5k73684
you are late ;)
– Nosrati
Aug 6 at 8:07
@user108128 Sometimes it is good, so I can think to a better answer ;)
– gimusi
Aug 6 at 8:08
@user108128 I see now that my answer is very similar to that given by Martin, I let mine since I gave a suggestion by limit comparison test which avoid to use inequalities.
– gimusi
Aug 6 at 8:11
add a comment |Â
up vote
0
down vote
If we can calculate a finite value for the integral of course the integral converges but if our goal is simply to check for the convergence we don’t need to calculate it explicitly but we can proceed for example by some quick convergence test.
For example, in this case we have that
$$ int^+infty_0frace^-t sqrt t , dt=int^1_0frace^-t sqrt t , dt+
int^+infty_1frace^-t sqrt t , dt$$
and as $tto 0^+$
$$frace^-t sqrt t sim
frac1 sqrt t$$
and as $tto infty$
$$frace^-t sqrt t sim
frac1 e^t$$
therefore both integral converge by limit comparison test and therefore the given integral converges.
answered Aug 6 at 7:55
gimusi
65.5k73684
you are late ;)
– Nosrati
Aug 6 at 8:07
@user108128 Sometimes it is good, so I can think to a better answer ;)
– gimusi
Aug 6 at 8:08
@user108128 I see now that my answer is very similar to that given by Martin, I let mine since I gave a suggestion by limit comparison test which avoid to use inequalities.
– gimusi
Aug 6 at 8:11
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If we can calculate a finite value for the integral of course the integral converges but if our goal is simply to check for the convergence we don’t need to calculate it explicitly but we can proceed for example by some quick convergence test.
For example, in this case we have that
$$ int^+infty_0frace^-t sqrt t , dt=int^1_0frace^-t sqrt t , dt+
int^+infty_1frace^-t sqrt t , dt$$
and as $tto 0^+$
$$frace^-t sqrt t sim
frac1 sqrt t$$
and as $tto infty$
$$frace^-t sqrt t sim
frac1 e^t$$
therefore both integral converge by limit comparison test and therefore the given integral converges.
answered Aug 6 at 7:55
gimusi
65.5k73684
If we can calculate a finite value for the integral of course the integral converges but if our goal is simply to check for the convergence we don’t need to calculate it explicitly but we can proceed for example by some quick convergence test.
For example, in this case we have that
$$ int^+infty_0frace^-t sqrt t , dt=int^1_0frace^-t sqrt t , dt+
int^+infty_1frace^-t sqrt t , dt$$
and as $tto 0^+$
$$frace^-t sqrt t sim
frac1 sqrt t$$
and as $tto infty$
$$frace^-t sqrt t sim
frac1 e^t$$
therefore both integral converge by limit comparison test and therefore the given integral converges.
answered Aug 6 at 7:55
gimusi
65.5k73684
edited Aug 6 at 8:07
answered Aug 6 at 7:55
gimusi
65.5k73684
answered Aug 6 at 7:55
gimusi
65.5k73684
answered Aug 6 at 7:55
gimusi
65.5k73684
65.5k73684
you are late ;)
– Nosrati
Aug 6 at 8:07
@user108128 Sometimes it is good, so I can think to a better answer ;)
– gimusi
Aug 6 at 8:08
@user108128 I see now that my answer is very similar to that given by Martin, I let mine since I gave a suggestion by limit comparison test which avoid to use inequalities.
– gimusi
Aug 6 at 8:11
add a comment |Â
you are late ;)
– Nosrati
Aug 6 at 8:07
@user108128 Sometimes it is good, so I can think to a better answer ;)
– gimusi
Aug 6 at 8:08
@user108128 I see now that my answer is very similar to that given by Martin, I let mine since I gave a suggestion by limit comparison test which avoid to use inequalities.
– gimusi
Aug 6 at 8:11
you are late ;)
– Nosrati
Aug 6 at 8:07
you are late ;)
– Nosrati
Aug 6 at 8:07
@user108128 Sometimes it is good, so I can think to a better answer ;)
– gimusi
Aug 6 at 8:08
@user108128 Sometimes it is good, so I can think to a better answer ;)
– gimusi
Aug 6 at 8:08
@user108128 I see now that my answer is very similar to that given by Martin, I let mine since I gave a suggestion by limit comparison test which avoid to use inequalities.
– gimusi
Aug 6 at 8:11
@user108128 I see now that my answer is very similar to that given by Martin, I let mine since I gave a suggestion by limit comparison test which avoid to use inequalities.
– gimusi
Aug 6 at 8:11
add a comment |Â
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Do you know of the gamma function?
– TheSimpliFire
Aug 5 at 14:48
1
@TheSimpliFire no I don't
– Mayar
Aug 5 at 14:49