Cauchy principle value-improper integral











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I'd like to calculate
$int_0^infty frac{ln(1+x)}{x^{1+a}}dx$ for $a in (0,1).$



I don't know how to start. Would you give me any hint for this problem? Thanks in advance!










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  • 1




    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    – robjohn
    yesterday










  • @robjohn Okay. I will do it next time. In this case, I just want a hint for the beginning, so that I can use residue theorem for calculating the integral. Thanks for your comment.
    – 0706
    yesterday












  • This doesn't seem like an integral that uses a Cauchy principal value. That usually indicates a singularity around which one removes a symmetric region and limits that region to a point.
    – robjohn
    yesterday










  • I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula.
    – robjohn
    yesterday










  • If you want to use contour integration to compute the integral after integration by parts, see this answer.
    – robjohn
    yesterday















up vote
3
down vote

favorite












I'd like to calculate
$int_0^infty frac{ln(1+x)}{x^{1+a}}dx$ for $a in (0,1).$



I don't know how to start. Would you give me any hint for this problem? Thanks in advance!










share|cite|improve this question


















  • 1




    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    – robjohn
    yesterday










  • @robjohn Okay. I will do it next time. In this case, I just want a hint for the beginning, so that I can use residue theorem for calculating the integral. Thanks for your comment.
    – 0706
    yesterday












  • This doesn't seem like an integral that uses a Cauchy principal value. That usually indicates a singularity around which one removes a symmetric region and limits that region to a point.
    – robjohn
    yesterday










  • I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula.
    – robjohn
    yesterday










  • If you want to use contour integration to compute the integral after integration by parts, see this answer.
    – robjohn
    yesterday













up vote
3
down vote

favorite









up vote
3
down vote

favorite











I'd like to calculate
$int_0^infty frac{ln(1+x)}{x^{1+a}}dx$ for $a in (0,1).$



I don't know how to start. Would you give me any hint for this problem? Thanks in advance!










share|cite|improve this question













I'd like to calculate
$int_0^infty frac{ln(1+x)}{x^{1+a}}dx$ for $a in (0,1).$



I don't know how to start. Would you give me any hint for this problem? Thanks in advance!







calculus complex-analysis multivariable-calculus






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share|cite|improve this question











share|cite|improve this question




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asked yesterday









0706

370110




370110








  • 1




    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    – robjohn
    yesterday










  • @robjohn Okay. I will do it next time. In this case, I just want a hint for the beginning, so that I can use residue theorem for calculating the integral. Thanks for your comment.
    – 0706
    yesterday












  • This doesn't seem like an integral that uses a Cauchy principal value. That usually indicates a singularity around which one removes a symmetric region and limits that region to a point.
    – robjohn
    yesterday










  • I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula.
    – robjohn
    yesterday










  • If you want to use contour integration to compute the integral after integration by parts, see this answer.
    – robjohn
    yesterday














  • 1




    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    – robjohn
    yesterday










  • @robjohn Okay. I will do it next time. In this case, I just want a hint for the beginning, so that I can use residue theorem for calculating the integral. Thanks for your comment.
    – 0706
    yesterday












  • This doesn't seem like an integral that uses a Cauchy principal value. That usually indicates a singularity around which one removes a symmetric region and limits that region to a point.
    – robjohn
    yesterday










  • I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula.
    – robjohn
    yesterday










  • If you want to use contour integration to compute the integral after integration by parts, see this answer.
    – robjohn
    yesterday








1




1




Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
– robjohn
yesterday




Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
– robjohn
yesterday












@robjohn Okay. I will do it next time. In this case, I just want a hint for the beginning, so that I can use residue theorem for calculating the integral. Thanks for your comment.
– 0706
yesterday






@robjohn Okay. I will do it next time. In this case, I just want a hint for the beginning, so that I can use residue theorem for calculating the integral. Thanks for your comment.
– 0706
yesterday














This doesn't seem like an integral that uses a Cauchy principal value. That usually indicates a singularity around which one removes a symmetric region and limits that region to a point.
– robjohn
yesterday




This doesn't seem like an integral that uses a Cauchy principal value. That usually indicates a singularity around which one removes a symmetric region and limits that region to a point.
– robjohn
yesterday












I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula.
– robjohn
yesterday




I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula.
– robjohn
yesterday












If you want to use contour integration to compute the integral after integration by parts, see this answer.
– robjohn
yesterday




If you want to use contour integration to compute the integral after integration by parts, see this answer.
– robjohn
yesterday










1 Answer
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Hint. Note that the integrand is positive and the improper integral is convergent. Moreover
$$begin{align}int_0^{infty}frac{ln(1+x)}{x^{1+a}},dx&=-frac{1}{a}int_0^{infty}ln(1+x)cdot d(x^{-a})\&=-frac{1}{a}left[frac{ln(1+x)}{x^a}right]_{0^+}^{+infty}+frac{1}{a}int_0^{infty} frac{dx}{x^a(1+x)}.end{align}$$
For the integral on the right side take a look at Cauchy Theorem application.
What do you obtain for the first term? What is the final result?






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    up vote
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    accepted










    Hint. Note that the integrand is positive and the improper integral is convergent. Moreover
    $$begin{align}int_0^{infty}frac{ln(1+x)}{x^{1+a}},dx&=-frac{1}{a}int_0^{infty}ln(1+x)cdot d(x^{-a})\&=-frac{1}{a}left[frac{ln(1+x)}{x^a}right]_{0^+}^{+infty}+frac{1}{a}int_0^{infty} frac{dx}{x^a(1+x)}.end{align}$$
    For the integral on the right side take a look at Cauchy Theorem application.
    What do you obtain for the first term? What is the final result?






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      Hint. Note that the integrand is positive and the improper integral is convergent. Moreover
      $$begin{align}int_0^{infty}frac{ln(1+x)}{x^{1+a}},dx&=-frac{1}{a}int_0^{infty}ln(1+x)cdot d(x^{-a})\&=-frac{1}{a}left[frac{ln(1+x)}{x^a}right]_{0^+}^{+infty}+frac{1}{a}int_0^{infty} frac{dx}{x^a(1+x)}.end{align}$$
      For the integral on the right side take a look at Cauchy Theorem application.
      What do you obtain for the first term? What is the final result?






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Hint. Note that the integrand is positive and the improper integral is convergent. Moreover
        $$begin{align}int_0^{infty}frac{ln(1+x)}{x^{1+a}},dx&=-frac{1}{a}int_0^{infty}ln(1+x)cdot d(x^{-a})\&=-frac{1}{a}left[frac{ln(1+x)}{x^a}right]_{0^+}^{+infty}+frac{1}{a}int_0^{infty} frac{dx}{x^a(1+x)}.end{align}$$
        For the integral on the right side take a look at Cauchy Theorem application.
        What do you obtain for the first term? What is the final result?






        share|cite|improve this answer














        Hint. Note that the integrand is positive and the improper integral is convergent. Moreover
        $$begin{align}int_0^{infty}frac{ln(1+x)}{x^{1+a}},dx&=-frac{1}{a}int_0^{infty}ln(1+x)cdot d(x^{-a})\&=-frac{1}{a}left[frac{ln(1+x)}{x^a}right]_{0^+}^{+infty}+frac{1}{a}int_0^{infty} frac{dx}{x^a(1+x)}.end{align}$$
        For the integral on the right side take a look at Cauchy Theorem application.
        What do you obtain for the first term? What is the final result?







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered yesterday









        Robert Z

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