Cauchy principle value-improper integral
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!
calculus complex-analysis multivariable-calculus
|
show 3 more comments
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!
calculus complex-analysis multivariable-calculus
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
|
show 3 more comments
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!
calculus complex-analysis multivariable-calculus
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
calculus complex-analysis multivariable-calculus
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
|
show 3 more comments
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
|
show 3 more comments
1 Answer
1
active
oldest
votes
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?
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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?
add a comment |
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?
add a comment |
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?
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?
edited yesterday
answered yesterday
Robert Z
89.4k1056128
89.4k1056128
add a comment |
add a comment |
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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