Integration $ I=int_{0}^{infty} x^{c-1} ln(1+x) dx $
$begingroup$
I am trying to handle the integration that's given below.
$$ I=int_{0}^{infty} x^{c-1} ln(1+x) dx $$
where $c$ and $x$ are both positive numbers. I found a conditional solution for the above integration from the 'table of integrals' book by Gradshteyn as:
$$ int_{0}^{infty} x^{mu-1} ln(1+gamma x) dx= frac{pi}{mu gamma^{u}sin mupi} $$
with the condition $left [-1<Re mu <0, |arg gamma|<pi right]$. My expression does not satisfy the conditions of the above given solution.
Any hint/help will be really appreciated. Thanks.
calculus integration definite-integrals
edited Dec 13 '18 at 12:52
Arjang
5,60162363
asked Dec 13 '18 at 12:04
AGaniAGani
32
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add a comment |
$begingroup$
I am trying to handle the integration that's given below.
$$ I=int_{0}^{infty} x^{c-1} ln(1+x) dx $$
where $c$ and $x$ are both positive numbers. I found a conditional solution for the above integration from the 'table of integrals' book by Gradshteyn as:
$$ int_{0}^{infty} x^{mu-1} ln(1+gamma x) dx= frac{pi}{mu gamma^{u}sin mupi} $$
with the condition $left [-1<Re mu <0, |arg gamma|<pi right]$. My expression does not satisfy the conditions of the above given solution.
Any hint/help will be really appreciated. Thanks.
calculus integration definite-integrals
edited Dec 13 '18 at 12:52
Arjang
5,60162363
asked Dec 13 '18 at 12:04
AGaniAGani
32
$endgroup$
add a comment |
$begingroup$
I am trying to handle the integration that's given below.
$$ I=int_{0}^{infty} x^{c-1} ln(1+x) dx $$
where $c$ and $x$ are both positive numbers. I found a conditional solution for the above integration from the 'table of integrals' book by Gradshteyn as:
$$ int_{0}^{infty} x^{mu-1} ln(1+gamma x) dx= frac{pi}{mu gamma^{u}sin mupi} $$
with the condition $left [-1<Re mu <0, |arg gamma|<pi right]$. My expression does not satisfy the conditions of the above given solution.
Any hint/help will be really appreciated. Thanks.
calculus integration definite-integrals
edited Dec 13 '18 at 12:52
Arjang
5,60162363
asked Dec 13 '18 at 12:04
AGaniAGani
32
$endgroup$
I am trying to handle the integration that's given below.
$$ I=int_{0}^{infty} x^{c-1} ln(1+x) dx $$
where $c$ and $x$ are both positive numbers. I found a conditional solution for the above integration from the 'table of integrals' book by Gradshteyn as:
$$ int_{0}^{infty} x^{mu-1} ln(1+gamma x) dx= frac{pi}{mu gamma^{u}sin mupi} $$
with the condition $left [-1<Re mu <0, |arg gamma|<pi right]$. My expression does not satisfy the conditions of the above given solution.
Any hint/help will be really appreciated. Thanks.
calculus integration definite-integrals
calculus integration definite-integrals
edited Dec 13 '18 at 12:52
Arjang
5,60162363
asked Dec 13 '18 at 12:04
AGaniAGani
32
edited Dec 13 '18 at 12:52
Arjang
5,60162363
asked Dec 13 '18 at 12:04
AGaniAGani
32
edited Dec 13 '18 at 12:52
Arjang
5,60162363
edited Dec 13 '18 at 12:52
Arjang
5,60162363
edited Dec 13 '18 at 12:52
Arjang
5,60162363
5,60162363
asked Dec 13 '18 at 12:04
AGaniAGani
32
asked Dec 13 '18 at 12:04
AGaniAGani
32
asked Dec 13 '18 at 12:04
AGaniAGani
32
32
add a comment |
add a comment |
1 Answer
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Since $c>0$ your integral diverges, as $Igeint_1^infty x^{-1}ln xdx=[tfrac{1}{2}ln^2x]_1^infty=infty$.
answered Dec 13 '18 at 12:19
J.G.J.G.
25.7k22540
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add a comment |
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$begingroup$
Since $c>0$ your integral diverges, as $Igeint_1^infty x^{-1}ln xdx=[tfrac{1}{2}ln^2x]_1^infty=infty$.
answered Dec 13 '18 at 12:19
J.G.J.G.
25.7k22540
$endgroup$
add a comment |
$begingroup$
Since $c>0$ your integral diverges, as $Igeint_1^infty x^{-1}ln xdx=[tfrac{1}{2}ln^2x]_1^infty=infty$.
answered Dec 13 '18 at 12:19
J.G.J.G.
25.7k22540
$endgroup$
add a comment |
$begingroup$
Since $c>0$ your integral diverges, as $Igeint_1^infty x^{-1}ln xdx=[tfrac{1}{2}ln^2x]_1^infty=infty$.
answered Dec 13 '18 at 12:19
J.G.J.G.
25.7k22540
$endgroup$
Since $c>0$ your integral diverges, as $Igeint_1^infty x^{-1}ln xdx=[tfrac{1}{2}ln^2x]_1^infty=infty$.
answered Dec 13 '18 at 12:19
J.G.J.G.
25.7k22540
edited Dec 13 '18 at 12:50
answered Dec 13 '18 at 12:19
J.G.J.G.
25.7k22540
answered Dec 13 '18 at 12:19
J.G.J.G.
25.7k22540
answered Dec 13 '18 at 12:19
J.G.J.G.
25.7k22540
25.7k22540
add a comment |
add a comment |
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