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
asked 13 hours ago
0706
370110
|
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
asked 13 hours ago
0706
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♦
10 hours ago
@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
10 hours ago
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♦
10 hours ago
I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula.
– robjohn♦
10 hours ago
If you want to use contour integration to compute the integral after integration by parts, see this answer.
– robjohn♦
10 hours ago
|
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
asked 13 hours ago
0706
370110
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 13 hours ago
0706
370110
asked 13 hours ago
0706
370110
asked 13 hours ago
0706
370110
asked 13 hours ago
0706
370110
asked 13 hours ago
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♦
10 hours ago
@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
10 hours ago
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♦
10 hours ago
I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula.
– robjohn♦
10 hours ago
If you want to use contour integration to compute the integral after integration by parts, see this answer.
– robjohn♦
10 hours ago
|
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♦
10 hours ago
@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
10 hours ago
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♦
10 hours ago
I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula.
– robjohn♦
10 hours ago
If you want to use contour integration to compute the integral after integration by parts, see this answer.
– robjohn♦
10 hours ago
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♦
10 hours ago
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♦
10 hours ago
@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
10 hours ago
@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
10 hours ago
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♦
10 hours ago
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♦
10 hours ago
I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula.
– robjohn♦
10 hours ago
I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula.
– robjohn♦
10 hours ago
If you want to use contour integration to compute the integral after integration by parts, see this answer.
– robjohn♦
10 hours ago
If you want to use contour integration to compute the integral after integration by parts, see this answer.
– robjohn♦
10 hours ago
|
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?
answered 13 hours ago
Robert Z
89.3k1056128
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?
answered 13 hours ago
Robert Z
89.3k1056128
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?
answered 13 hours ago
Robert Z
89.3k1056128
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?
answered 13 hours ago
Robert Z
89.3k1056128
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?
answered 13 hours ago
Robert Z
89.3k1056128
edited 10 hours ago
answered 13 hours ago
Robert Z
89.3k1056128
answered 13 hours ago
Robert Z
89.3k1056128
answered 13 hours ago
Robert Z
89.3k1056128
89.3k1056128
add a comment |
add a comment |
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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♦
10 hours ago
@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
10 hours ago
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♦
10 hours ago
I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula.
– robjohn♦
10 hours ago
If you want to use contour integration to compute the integral after integration by parts, see this answer.
– robjohn♦
10 hours ago