Integration $ I=int_{0}^{infty} x^{c-1} ln(1+x) dx $












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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







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    0












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















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      0








      0


      0



      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






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      edited Dec 13 '18 at 12:52









      Arjang

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      5,59162363










      asked Dec 13 '18 at 12:04









      AGaniAGani

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      32






















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          Since $c>0$ your integral diverges, as $Igeint_1^infty x^{-1}ln xdx=[tfrac{1}{2}ln^2x]_1^infty=infty$.






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            $begingroup$

            Since $c>0$ your integral diverges, as $Igeint_1^infty x^{-1}ln xdx=[tfrac{1}{2}ln^2x]_1^infty=infty$.






            share|cite|improve this answer











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              $begingroup$

              Since $c>0$ your integral diverges, as $Igeint_1^infty x^{-1}ln xdx=[tfrac{1}{2}ln^2x]_1^infty=infty$.






              share|cite|improve this answer











              $endgroup$
















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                3





                $begingroup$

                Since $c>0$ your integral diverges, as $Igeint_1^infty x^{-1}ln xdx=[tfrac{1}{2}ln^2x]_1^infty=infty$.






                share|cite|improve this answer











                $endgroup$



                Since $c>0$ your integral diverges, as $Igeint_1^infty x^{-1}ln xdx=[tfrac{1}{2}ln^2x]_1^infty=infty$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 13 '18 at 12:50

























                answered Dec 13 '18 at 12:19









                J.G.J.G.

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