Evaluate the integral of $frac1(x^2+a^2)^2$ on $[0, infty]$ [on hold]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
-4
down vote

favorite












find the integral where $a>0$, $xinmathbbR$




$$int_0^infty frac1(x^2+a^2)^2dx$$




it's from a test in Complex Analysis, so Please use tools accordingly.




integration complex-analysis




share|cite|improve this question













put on hold as off-topic by José Carlos Santos, Claude Leibovici, Tyrone, Dylan, amWhy 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Claude Leibovici, Tyrone, Dylan, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 3




    What tools from Complex Analysis do you know? Which have you tried?
    – Bruce
    2 days ago










  • Just out of interest you might want to see how it can be done using Leibniz's formula.
    – Bruce
    2 days ago










  • As the limits are not parameters you don't actually use the full blown Leibniz formula ... just differentiate the integrand under the integral sign with respect to the parameter, $a$.
    – Bruce
    2 days ago














up vote
-4
down vote

favorite












find the integral where $a>0$, $xinmathbbR$




$$int_0^infty frac1(x^2+a^2)^2dx$$




it's from a test in Complex Analysis, so Please use tools accordingly.




integration complex-analysis




share|cite|improve this question













put on hold as off-topic by José Carlos Santos, Claude Leibovici, Tyrone, Dylan, amWhy 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Claude Leibovici, Tyrone, Dylan, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 3




    What tools from Complex Analysis do you know? Which have you tried?
    – Bruce
    2 days ago










  • Just out of interest you might want to see how it can be done using Leibniz's formula.
    – Bruce
    2 days ago










  • As the limits are not parameters you don't actually use the full blown Leibniz formula ... just differentiate the integrand under the integral sign with respect to the parameter, $a$.
    – Bruce
    2 days ago












up vote
-4
down vote

favorite









up vote
-4
down vote

favorite











find the integral where $a>0$, $xinmathbbR$




$$int_0^infty frac1(x^2+a^2)^2dx$$




it's from a test in Complex Analysis, so Please use tools accordingly.




integration complex-analysis




share|cite|improve this question













find the integral where $a>0$, $xinmathbbR$




$$int_0^infty frac1(x^2+a^2)^2dx$$




it's from a test in Complex Analysis, so Please use tools accordingly.





integration complex-analysis





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago









user 108128

18.6k41544




18.6k41544









asked 2 days ago









SlyxBrd

275




275




put on hold as off-topic by José Carlos Santos, Claude Leibovici, Tyrone, Dylan, amWhy 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Claude Leibovici, Tyrone, Dylan, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by José Carlos Santos, Claude Leibovici, Tyrone, Dylan, amWhy 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Claude Leibovici, Tyrone, Dylan, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 3




    What tools from Complex Analysis do you know? Which have you tried?
    – Bruce
    2 days ago










  • Just out of interest you might want to see how it can be done using Leibniz's formula.
    – Bruce
    2 days ago










  • As the limits are not parameters you don't actually use the full blown Leibniz formula ... just differentiate the integrand under the integral sign with respect to the parameter, $a$.
    – Bruce
    2 days ago












  • 3




    What tools from Complex Analysis do you know? Which have you tried?
    – Bruce
    2 days ago










  • Just out of interest you might want to see how it can be done using Leibniz's formula.
    – Bruce
    2 days ago










  • As the limits are not parameters you don't actually use the full blown Leibniz formula ... just differentiate the integrand under the integral sign with respect to the parameter, $a$.
    – Bruce
    2 days ago







3




3




What tools from Complex Analysis do you know? Which have you tried?
– Bruce
2 days ago




What tools from Complex Analysis do you know? Which have you tried?
– Bruce
2 days ago












Just out of interest you might want to see how it can be done using Leibniz's formula.
– Bruce
2 days ago




Just out of interest you might want to see how it can be done using Leibniz's formula.
– Bruce
2 days ago












As the limits are not parameters you don't actually use the full blown Leibniz formula ... just differentiate the integrand under the integral sign with respect to the parameter, $a$.
– Bruce
2 days ago




As the limits are not parameters you don't actually use the full blown Leibniz formula ... just differentiate the integrand under the integral sign with respect to the parameter, $a$.
– Bruce
2 days ago










2 Answers
2






active

oldest

votes

















up vote
2
down vote













Let $I=int_0^infty frac1x^2+a^2~dx=left[frac1atan^-1(fracxa)right]_0^infty=fracpi2a$



Treat $a$ as the parameter



$$fracdIda=int_0^infty frac-2a(x^2+a^2)^2dx=frac-2pi4a^2$$



yielding



$$int_0^infty frac1(x^2+a^2)^2dx=fracpi4a^3$$






share|cite|improve this answer























  • OP wanted a solution with Complex Analysis!
    – user 108128
    2 days ago










  • If you look above the solution I said it was "just for interest".... I love seeing various different ways of doing things and as OP hadn't shown any working I thought it may be fun.
    – Bruce
    2 days ago










  • huh! for fun? do you think people take their time for fun here? :(
    – user 108128
    2 days ago











  • @user108128 Certainly some do, and it's worth noting that this site is built around creating a repertory of knowledge; answers can be written with an audience different from the OP in mind.
    – Simply Beautiful Art
    2 days ago










  • @SimplyBeautifulArt Regards you. This site does not work fairly, and in this case I encountered contradictions many many times. Sometimes this is the case, but most of the time, justice is not implemented and it is customary.
    – user 108128
    2 days ago

















up vote
0
down vote













$$I=frac12int_-infty^infty frac1(x+ia)^2(x-ia)^2dx$$ Let us consider as a contour ($C$) a semicircle with boundary diameter on the real, going from $−R$ to $R$. Our function $f(z)=frac1(z+ia)^2(z-ia)^2$ has singularities at $z_1= ia$ and $z_2=-ia$, but we should only look at $z_1$ because of the contour chosen. Let also denote with $digamma$ the arc of the semicircle.
$$int_C f(z)dz=int_-R^Rf(z)dz+int_digamma f(z)dz$$ The integral over $digamma$ tends to zero as $R rightarrow infty$ and since $f(z)$ has poles of order $2$ we need to take the derivate once when we use either Residue theorem or Cauchy's formula. $$I=frac12 left(2pi i lim_zto z_1fracddzleft((z-z_1)^2
frac1(z-z_1)^2(z-z_2)^2right)right)$$






share|cite|improve this answer





















  • thank you. How can you tell that the integration over the arc goes to zero as R goes to infinity?
    – SlyxBrd
    2 days ago










  • Because I use this: en.wikipedia.org/wiki/Estimation_lemma
    – Zacky
    2 days ago

















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













Let $I=int_0^infty frac1x^2+a^2~dx=left[frac1atan^-1(fracxa)right]_0^infty=fracpi2a$



Treat $a$ as the parameter



$$fracdIda=int_0^infty frac-2a(x^2+a^2)^2dx=frac-2pi4a^2$$



yielding



$$int_0^infty frac1(x^2+a^2)^2dx=fracpi4a^3$$






share|cite|improve this answer























  • OP wanted a solution with Complex Analysis!
    – user 108128
    2 days ago










  • If you look above the solution I said it was "just for interest".... I love seeing various different ways of doing things and as OP hadn't shown any working I thought it may be fun.
    – Bruce
    2 days ago










  • huh! for fun? do you think people take their time for fun here? :(
    – user 108128
    2 days ago











  • @user108128 Certainly some do, and it's worth noting that this site is built around creating a repertory of knowledge; answers can be written with an audience different from the OP in mind.
    – Simply Beautiful Art
    2 days ago










  • @SimplyBeautifulArt Regards you. This site does not work fairly, and in this case I encountered contradictions many many times. Sometimes this is the case, but most of the time, justice is not implemented and it is customary.
    – user 108128
    2 days ago














up vote
2
down vote













Let $I=int_0^infty frac1x^2+a^2~dx=left[frac1atan^-1(fracxa)right]_0^infty=fracpi2a$



Treat $a$ as the parameter



$$fracdIda=int_0^infty frac-2a(x^2+a^2)^2dx=frac-2pi4a^2$$



yielding



$$int_0^infty frac1(x^2+a^2)^2dx=fracpi4a^3$$






share|cite|improve this answer























  • OP wanted a solution with Complex Analysis!
    – user 108128
    2 days ago










  • If you look above the solution I said it was "just for interest".... I love seeing various different ways of doing things and as OP hadn't shown any working I thought it may be fun.
    – Bruce
    2 days ago










  • huh! for fun? do you think people take their time for fun here? :(
    – user 108128
    2 days ago











  • @user108128 Certainly some do, and it's worth noting that this site is built around creating a repertory of knowledge; answers can be written with an audience different from the OP in mind.
    – Simply Beautiful Art
    2 days ago










  • @SimplyBeautifulArt Regards you. This site does not work fairly, and in this case I encountered contradictions many many times. Sometimes this is the case, but most of the time, justice is not implemented and it is customary.
    – user 108128
    2 days ago












up vote
2
down vote










up vote
2
down vote









Let $I=int_0^infty frac1x^2+a^2~dx=left[frac1atan^-1(fracxa)right]_0^infty=fracpi2a$



Treat $a$ as the parameter



$$fracdIda=int_0^infty frac-2a(x^2+a^2)^2dx=frac-2pi4a^2$$



yielding



$$int_0^infty frac1(x^2+a^2)^2dx=fracpi4a^3$$






share|cite|improve this answer















Let $I=int_0^infty frac1x^2+a^2~dx=left[frac1atan^-1(fracxa)right]_0^infty=fracpi2a$



Treat $a$ as the parameter



$$fracdIda=int_0^infty frac-2a(x^2+a^2)^2dx=frac-2pi4a^2$$



yielding



$$int_0^infty frac1(x^2+a^2)^2dx=fracpi4a^3$$







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago









pointguard0

520315




520315











answered 2 days ago









Bruce

13511




13511











  • OP wanted a solution with Complex Analysis!
    – user 108128
    2 days ago










  • If you look above the solution I said it was "just for interest".... I love seeing various different ways of doing things and as OP hadn't shown any working I thought it may be fun.
    – Bruce
    2 days ago










  • huh! for fun? do you think people take their time for fun here? :(
    – user 108128
    2 days ago











  • @user108128 Certainly some do, and it's worth noting that this site is built around creating a repertory of knowledge; answers can be written with an audience different from the OP in mind.
    – Simply Beautiful Art
    2 days ago










  • @SimplyBeautifulArt Regards you. This site does not work fairly, and in this case I encountered contradictions many many times. Sometimes this is the case, but most of the time, justice is not implemented and it is customary.
    – user 108128
    2 days ago
















  • OP wanted a solution with Complex Analysis!
    – user 108128
    2 days ago










  • If you look above the solution I said it was "just for interest".... I love seeing various different ways of doing things and as OP hadn't shown any working I thought it may be fun.
    – Bruce
    2 days ago










  • huh! for fun? do you think people take their time for fun here? :(
    – user 108128
    2 days ago











  • @user108128 Certainly some do, and it's worth noting that this site is built around creating a repertory of knowledge; answers can be written with an audience different from the OP in mind.
    – Simply Beautiful Art
    2 days ago










  • @SimplyBeautifulArt Regards you. This site does not work fairly, and in this case I encountered contradictions many many times. Sometimes this is the case, but most of the time, justice is not implemented and it is customary.
    – user 108128
    2 days ago















OP wanted a solution with Complex Analysis!
– user 108128
2 days ago




OP wanted a solution with Complex Analysis!
– user 108128
2 days ago












If you look above the solution I said it was "just for interest".... I love seeing various different ways of doing things and as OP hadn't shown any working I thought it may be fun.
– Bruce
2 days ago




If you look above the solution I said it was "just for interest".... I love seeing various different ways of doing things and as OP hadn't shown any working I thought it may be fun.
– Bruce
2 days ago












huh! for fun? do you think people take their time for fun here? :(
– user 108128
2 days ago





huh! for fun? do you think people take their time for fun here? :(
– user 108128
2 days ago













@user108128 Certainly some do, and it's worth noting that this site is built around creating a repertory of knowledge; answers can be written with an audience different from the OP in mind.
– Simply Beautiful Art
2 days ago




@user108128 Certainly some do, and it's worth noting that this site is built around creating a repertory of knowledge; answers can be written with an audience different from the OP in mind.
– Simply Beautiful Art
2 days ago












@SimplyBeautifulArt Regards you. This site does not work fairly, and in this case I encountered contradictions many many times. Sometimes this is the case, but most of the time, justice is not implemented and it is customary.
– user 108128
2 days ago




@SimplyBeautifulArt Regards you. This site does not work fairly, and in this case I encountered contradictions many many times. Sometimes this is the case, but most of the time, justice is not implemented and it is customary.
– user 108128
2 days ago










up vote
0
down vote













$$I=frac12int_-infty^infty frac1(x+ia)^2(x-ia)^2dx$$ Let us consider as a contour ($C$) a semicircle with boundary diameter on the real, going from $−R$ to $R$. Our function $f(z)=frac1(z+ia)^2(z-ia)^2$ has singularities at $z_1= ia$ and $z_2=-ia$, but we should only look at $z_1$ because of the contour chosen. Let also denote with $digamma$ the arc of the semicircle.
$$int_C f(z)dz=int_-R^Rf(z)dz+int_digamma f(z)dz$$ The integral over $digamma$ tends to zero as $R rightarrow infty$ and since $f(z)$ has poles of order $2$ we need to take the derivate once when we use either Residue theorem or Cauchy's formula. $$I=frac12 left(2pi i lim_zto z_1fracddzleft((z-z_1)^2
frac1(z-z_1)^2(z-z_2)^2right)right)$$






share|cite|improve this answer





















  • thank you. How can you tell that the integration over the arc goes to zero as R goes to infinity?
    – SlyxBrd
    2 days ago










  • Because I use this: en.wikipedia.org/wiki/Estimation_lemma
    – Zacky
    2 days ago














up vote
0
down vote













$$I=frac12int_-infty^infty frac1(x+ia)^2(x-ia)^2dx$$ Let us consider as a contour ($C$) a semicircle with boundary diameter on the real, going from $−R$ to $R$. Our function $f(z)=frac1(z+ia)^2(z-ia)^2$ has singularities at $z_1= ia$ and $z_2=-ia$, but we should only look at $z_1$ because of the contour chosen. Let also denote with $digamma$ the arc of the semicircle.
$$int_C f(z)dz=int_-R^Rf(z)dz+int_digamma f(z)dz$$ The integral over $digamma$ tends to zero as $R rightarrow infty$ and since $f(z)$ has poles of order $2$ we need to take the derivate once when we use either Residue theorem or Cauchy's formula. $$I=frac12 left(2pi i lim_zto z_1fracddzleft((z-z_1)^2
frac1(z-z_1)^2(z-z_2)^2right)right)$$






share|cite|improve this answer





















  • thank you. How can you tell that the integration over the arc goes to zero as R goes to infinity?
    – SlyxBrd
    2 days ago










  • Because I use this: en.wikipedia.org/wiki/Estimation_lemma
    – Zacky
    2 days ago












up vote
0
down vote










up vote
0
down vote









$$I=frac12int_-infty^infty frac1(x+ia)^2(x-ia)^2dx$$ Let us consider as a contour ($C$) a semicircle with boundary diameter on the real, going from $−R$ to $R$. Our function $f(z)=frac1(z+ia)^2(z-ia)^2$ has singularities at $z_1= ia$ and $z_2=-ia$, but we should only look at $z_1$ because of the contour chosen. Let also denote with $digamma$ the arc of the semicircle.
$$int_C f(z)dz=int_-R^Rf(z)dz+int_digamma f(z)dz$$ The integral over $digamma$ tends to zero as $R rightarrow infty$ and since $f(z)$ has poles of order $2$ we need to take the derivate once when we use either Residue theorem or Cauchy's formula. $$I=frac12 left(2pi i lim_zto z_1fracddzleft((z-z_1)^2
frac1(z-z_1)^2(z-z_2)^2right)right)$$






share|cite|improve this answer













$$I=frac12int_-infty^infty frac1(x+ia)^2(x-ia)^2dx$$ Let us consider as a contour ($C$) a semicircle with boundary diameter on the real, going from $−R$ to $R$. Our function $f(z)=frac1(z+ia)^2(z-ia)^2$ has singularities at $z_1= ia$ and $z_2=-ia$, but we should only look at $z_1$ because of the contour chosen. Let also denote with $digamma$ the arc of the semicircle.
$$int_C f(z)dz=int_-R^Rf(z)dz+int_digamma f(z)dz$$ The integral over $digamma$ tends to zero as $R rightarrow infty$ and since $f(z)$ has poles of order $2$ we need to take the derivate once when we use either Residue theorem or Cauchy's formula. $$I=frac12 left(2pi i lim_zto z_1fracddzleft((z-z_1)^2
frac1(z-z_1)^2(z-z_2)^2right)right)$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered 2 days ago









Zacky

1,9711325




1,9711325











  • thank you. How can you tell that the integration over the arc goes to zero as R goes to infinity?
    – SlyxBrd
    2 days ago










  • Because I use this: en.wikipedia.org/wiki/Estimation_lemma
    – Zacky
    2 days ago
















  • thank you. How can you tell that the integration over the arc goes to zero as R goes to infinity?
    – SlyxBrd
    2 days ago










  • Because I use this: en.wikipedia.org/wiki/Estimation_lemma
    – Zacky
    2 days ago















thank you. How can you tell that the integration over the arc goes to zero as R goes to infinity?
– SlyxBrd
2 days ago




thank you. How can you tell that the integration over the arc goes to zero as R goes to infinity?
– SlyxBrd
2 days ago












Because I use this: en.wikipedia.org/wiki/Estimation_lemma
– Zacky
2 days ago




Because I use this: en.wikipedia.org/wiki/Estimation_lemma
– Zacky
2 days ago


Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
Read more