Determine the value of (p,q)?
up vote
1
down vote
favorite
find all value of $(p,q)$ for which integral
$$int_0^1 x^pln^q(1/x)dx$$
converges ?
My answer is $p > q-1$
Is It correct?
Proof : $$int_0^1 x^pln^q(1/x)dxleint_0^1x^{p-q}dx= frac{1}{p-q+1}$$
real-analysis integration
asked Nov 25 at 13:56
jasmine
1,395416
add a comment |
up vote
1
down vote
favorite
find all value of $(p,q)$ for which integral
$$int_0^1 x^pln^q(1/x)dx$$
converges ?
My answer is $p > q-1$
Is It correct?
Proof : $$int_0^1 x^pln^q(1/x)dxleint_0^1x^{p-q}dx= frac{1}{p-q+1}$$
real-analysis integration
asked Nov 25 at 13:56
jasmine
1,395416
3
What about other cases? You clearly have not found all values.
– xbh
Nov 25 at 14:00
1
As $$int_{0}^{1}xlnleft(frac{1}{x}right)text{d}x=frac{1}{4}$$ your answer seems to be incomplete (it's only one of many examples which do not follow the condition you have found).
– Galc127
Nov 25 at 14:01
@xbh then what is the answer ?
– jasmine
Nov 25 at 14:06
@Galc127 okkss but u have take $p=q=1$
– jasmine
Nov 25 at 14:07
1
@jasmine, Oops, my bad... you can take $p=0,q=1$ and you have $$intlimits_{0}^{1}lnleft(frac{1}{x}right)text{d}x=1$$
– Galc127
Nov 25 at 14:11
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
find all value of $(p,q)$ for which integral
$$int_0^1 x^pln^q(1/x)dx$$
converges ?
My answer is $p > q-1$
Is It correct?
Proof : $$int_0^1 x^pln^q(1/x)dxleint_0^1x^{p-q}dx= frac{1}{p-q+1}$$
real-analysis integration
asked Nov 25 at 13:56
jasmine
1,395416
find all value of $(p,q)$ for which integral
$$int_0^1 x^pln^q(1/x)dx$$
converges ?
My answer is $p > q-1$
Is It correct?
Proof : $$int_0^1 x^pln^q(1/x)dxleint_0^1x^{p-q}dx= frac{1}{p-q+1}$$
real-analysis integration
real-analysis integration
asked Nov 25 at 13:56
jasmine
1,395416
asked Nov 25 at 13:56
jasmine
1,395416
edited Nov 26 at 15:33
asked Nov 25 at 13:56
jasmine
1,395416
asked Nov 25 at 13:56
jasmine
1,395416
asked Nov 25 at 13:56
jasmine
1,395416
1,395416
3
What about other cases? You clearly have not found all values.
– xbh
Nov 25 at 14:00
1
As $$int_{0}^{1}xlnleft(frac{1}{x}right)text{d}x=frac{1}{4}$$ your answer seems to be incomplete (it's only one of many examples which do not follow the condition you have found).
– Galc127
Nov 25 at 14:01
@xbh then what is the answer ?
– jasmine
Nov 25 at 14:06
@Galc127 okkss but u have take $p=q=1$
– jasmine
Nov 25 at 14:07
1
@jasmine, Oops, my bad... you can take $p=0,q=1$ and you have $$intlimits_{0}^{1}lnleft(frac{1}{x}right)text{d}x=1$$
– Galc127
Nov 25 at 14:11
add a comment |
3
What about other cases? You clearly have not found all values.
– xbh
Nov 25 at 14:00
1
As $$int_{0}^{1}xlnleft(frac{1}{x}right)text{d}x=frac{1}{4}$$ your answer seems to be incomplete (it's only one of many examples which do not follow the condition you have found).
– Galc127
Nov 25 at 14:01
@xbh then what is the answer ?
– jasmine
Nov 25 at 14:06
@Galc127 okkss but u have take $p=q=1$
– jasmine
Nov 25 at 14:07
1
@jasmine, Oops, my bad... you can take $p=0,q=1$ and you have $$intlimits_{0}^{1}lnleft(frac{1}{x}right)text{d}x=1$$
– Galc127
Nov 25 at 14:11
3
3
What about other cases? You clearly have not found all values.
– xbh
Nov 25 at 14:00
What about other cases? You clearly have not found all values.
– xbh
Nov 25 at 14:00
1
1
As $$int_{0}^{1}xlnleft(frac{1}{x}right)text{d}x=frac{1}{4}$$ your answer seems to be incomplete (it's only one of many examples which do not follow the condition you have found).
– Galc127
Nov 25 at 14:01
As $$int_{0}^{1}xlnleft(frac{1}{x}right)text{d}x=frac{1}{4}$$ your answer seems to be incomplete (it's only one of many examples which do not follow the condition you have found).
– Galc127
Nov 25 at 14:01
@xbh then what is the answer ?
– jasmine
Nov 25 at 14:06
@xbh then what is the answer ?
– jasmine
Nov 25 at 14:06
@Galc127 okkss but u have take $p=q=1$
– jasmine
Nov 25 at 14:07
@Galc127 okkss but u have take $p=q=1$
– jasmine
Nov 25 at 14:07
1
1
@jasmine, Oops, my bad... you can take $p=0,q=1$ and you have $$intlimits_{0}^{1}lnleft(frac{1}{x}right)text{d}x=1$$
– Galc127
Nov 25 at 14:11
@jasmine, Oops, my bad... you can take $p=0,q=1$ and you have $$intlimits_{0}^{1}lnleft(frac{1}{x}right)text{d}x=1$$
– Galc127
Nov 25 at 14:11
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Hint:
Let $lndfrac1x=u$ then
$$I=int_0^infty u^qe^{-(p+1)u} du$$
which may expressed as gamma function.
answered Nov 25 at 14:01
Nosrati
26.2k62353
sir is there any other method?? as i didn't study gamma function .
– jasmine
Nov 25 at 14:06
@jasmine Then how about directly testing the convergence for Gamma function?
– xbh
Nov 25 at 14:09
@xbh then what is the correct answer , i need answer ????
– jasmine
Nov 25 at 14:16
1
@jasmine We have $lnalpha<alpha-1$ then $$I<int_0^1x^{p-q}(1-x)^{q} dx<int_0^1(1-x)^{q} dx=int_0^1x^q dx<infty$$ if $1+q>0$.
– Nosrati
Dec 2 at 17:36
thanks u sir @Nosrati
– jasmine
Dec 3 at 0:22
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint:
Let $lndfrac1x=u$ then
$$I=int_0^infty u^qe^{-(p+1)u} du$$
which may expressed as gamma function.
answered Nov 25 at 14:01
Nosrati
26.2k62353
sir is there any other method?? as i didn't study gamma function .
– jasmine
Nov 25 at 14:06
@jasmine Then how about directly testing the convergence for Gamma function?
– xbh
Nov 25 at 14:09
@xbh then what is the correct answer , i need answer ????
– jasmine
Nov 25 at 14:16
1
@jasmine We have $lnalpha<alpha-1$ then $$I<int_0^1x^{p-q}(1-x)^{q} dx<int_0^1(1-x)^{q} dx=int_0^1x^q dx<infty$$ if $1+q>0$.
– Nosrati
Dec 2 at 17:36
thanks u sir @Nosrati
– jasmine
Dec 3 at 0:22
add a comment |
up vote
2
down vote
accepted
Hint:
Let $lndfrac1x=u$ then
$$I=int_0^infty u^qe^{-(p+1)u} du$$
which may expressed as gamma function.
answered Nov 25 at 14:01
Nosrati
26.2k62353
sir is there any other method?? as i didn't study gamma function .
– jasmine
Nov 25 at 14:06
@jasmine Then how about directly testing the convergence for Gamma function?
– xbh
Nov 25 at 14:09
@xbh then what is the correct answer , i need answer ????
– jasmine
Nov 25 at 14:16
1
@jasmine We have $lnalpha<alpha-1$ then $$I<int_0^1x^{p-q}(1-x)^{q} dx<int_0^1(1-x)^{q} dx=int_0^1x^q dx<infty$$ if $1+q>0$.
– Nosrati
Dec 2 at 17:36
thanks u sir @Nosrati
– jasmine
Dec 3 at 0:22
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint:
Let $lndfrac1x=u$ then
$$I=int_0^infty u^qe^{-(p+1)u} du$$
which may expressed as gamma function.
answered Nov 25 at 14:01
Nosrati
26.2k62353
Hint:
Let $lndfrac1x=u$ then
$$I=int_0^infty u^qe^{-(p+1)u} du$$
which may expressed as gamma function.
answered Nov 25 at 14:01
Nosrati
26.2k62353
answered Nov 25 at 14:01
Nosrati
26.2k62353
answered Nov 25 at 14:01
Nosrati
26.2k62353
answered Nov 25 at 14:01
Nosrati
26.2k62353
26.2k62353
sir is there any other method?? as i didn't study gamma function .
– jasmine
Nov 25 at 14:06
@jasmine Then how about directly testing the convergence for Gamma function?
– xbh
Nov 25 at 14:09
@xbh then what is the correct answer , i need answer ????
– jasmine
Nov 25 at 14:16
1
@jasmine We have $lnalpha<alpha-1$ then $$I<int_0^1x^{p-q}(1-x)^{q} dx<int_0^1(1-x)^{q} dx=int_0^1x^q dx<infty$$ if $1+q>0$.
– Nosrati
Dec 2 at 17:36
thanks u sir @Nosrati
– jasmine
Dec 3 at 0:22
add a comment |
sir is there any other method?? as i didn't study gamma function .
– jasmine
Nov 25 at 14:06
@jasmine Then how about directly testing the convergence for Gamma function?
– xbh
Nov 25 at 14:09
@xbh then what is the correct answer , i need answer ????
– jasmine
Nov 25 at 14:16
1
@jasmine We have $lnalpha<alpha-1$ then $$I<int_0^1x^{p-q}(1-x)^{q} dx<int_0^1(1-x)^{q} dx=int_0^1x^q dx<infty$$ if $1+q>0$.
– Nosrati
Dec 2 at 17:36
thanks u sir @Nosrati
– jasmine
Dec 3 at 0:22
sir is there any other method?? as i didn't study gamma function .
– jasmine
Nov 25 at 14:06
sir is there any other method?? as i didn't study gamma function .
– jasmine
Nov 25 at 14:06
@jasmine Then how about directly testing the convergence for Gamma function?
– xbh
Nov 25 at 14:09
@jasmine Then how about directly testing the convergence for Gamma function?
– xbh
Nov 25 at 14:09
@xbh then what is the correct answer , i need answer ????
– jasmine
Nov 25 at 14:16
@xbh then what is the correct answer , i need answer ????
– jasmine
Nov 25 at 14:16
1
1
@jasmine We have $lnalpha<alpha-1$ then $$I<int_0^1x^{p-q}(1-x)^{q} dx<int_0^1(1-x)^{q} dx=int_0^1x^q dx<infty$$ if $1+q>0$.
– Nosrati
Dec 2 at 17:36
@jasmine We have $lnalpha<alpha-1$ then $$I<int_0^1x^{p-q}(1-x)^{q} dx<int_0^1(1-x)^{q} dx=int_0^1x^q dx<infty$$ if $1+q>0$.
– Nosrati
Dec 2 at 17:36
thanks u sir @Nosrati
– jasmine
Dec 3 at 0:22
thanks u sir @Nosrati
– jasmine
Dec 3 at 0:22
add a comment |
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3
What about other cases? You clearly have not found all values.
– xbh
Nov 25 at 14:00
1
As $$int_{0}^{1}xlnleft(frac{1}{x}right)text{d}x=frac{1}{4}$$ your answer seems to be incomplete (it's only one of many examples which do not follow the condition you have found).
– Galc127
Nov 25 at 14:01
@xbh then what is the answer ?
– jasmine
Nov 25 at 14:06
@Galc127 okkss but u have take $p=q=1$
– jasmine
Nov 25 at 14:07
1
@jasmine, Oops, my bad... you can take $p=0,q=1$ and you have $$intlimits_{0}^{1}lnleft(frac{1}{x}right)text{d}x=1$$
– Galc127
Nov 25 at 14:11