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How to deal with poles at the boundaries of integration

0

$begingroup$

How can one evaluate this integral with poles at boundaries?

$$I = int_0^a fracr^nr^1/2(a-r)^1/2dr$$, $$n=0,1,2,ldots$$

complex-analysis definite-integrals

share|cite|improve this question

edited Apr 2 at 17:17

Zaxos

asked Apr 2 at 17:06

ZaxosZaxos

61

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add a comment | 

0

$begingroup$

How can one evaluate this integral with poles at boundaries?

$$I = int_0^a fracr^nr^1/2(a-r)^1/2dr$$, $$n=0,1,2,ldots$$

complex-analysis definite-integrals

share|cite|improve this question

edited Apr 2 at 17:17

Zaxos

asked Apr 2 at 17:06

ZaxosZaxos

61

$endgroup$

add a comment | 

$begingroup$

How can one evaluate this integral with poles at boundaries?

$$I = int_0^a fracr^nr^1/2(a-r)^1/2dr$$, $$n=0,1,2,ldots$$

complex-analysis definite-integrals

share|cite|improve this question

edited Apr 2 at 17:17

Zaxos

asked Apr 2 at 17:06

ZaxosZaxos

61

$endgroup$

share|cite|improve this question

edited Apr 2 at 17:17

Zaxos

asked Apr 2 at 17:06

ZaxosZaxos

61

share|cite|improve this question

edited Apr 2 at 17:17

Zaxos

asked Apr 2 at 17:06

ZaxosZaxos

61

asked Apr 2 at 17:06

ZaxosZaxos

61

asked Apr 2 at 17:06

ZaxosZaxos

61

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So generalising the integral we have
$$I_n = int_0^a fracr^nr^1/2(a-r)^1/2dr=int_0^a r^n-frac12(a-r)^-frac12dr$$
Then by using integration by parts we get
$$I_n=2(n-frac12)int_0^ar^n-frac32(a-r)^frac12dr$$
$$=(2n-1)int_0^ar^n-frac32(a-r)(a-r)^-frac12dr$$
$$=a(2n-1)int_0^ar^n-frac32(a-r)^-frac12dr-(2n-1)int_0^ar^n-frac12(a-r)^-frac12dr$$
$$=a(2n-1)I_n-1-(2n-1)I_n$$
$$therefore I_n=abigg(frac2n-12nbigg)I_n-1$$
To solve for a base case we can find $I_0$ by using the substitution $r=asin^2(theta)to dr=2asin(theta)cos(theta)dtheta$
$$I_0=int_0^ar^-frac12(a-r)^-frac12dr=int_0^fracpi2 frac2asin(theta)cos(theta)sqrtasin^2(theta)sqrta-asin^2(theta)dtheta=pi$$
for all $ainmathbbC$. Hence we can solve the recurrence relation to get
$$I_n=abigg(frac2n-12nbigg)I_n-1$$
$$=a^2bigg(frac2n-12nbigg)bigg(frac2n-32n-2bigg)I_n-2$$
By continuing this pattern we eventually reach $I_0$ such that
$$I_n=a^nfrac(2n-1)(2n-3)(2n-5)...(1)(2n)(2n-2)(2n-4)...(2)I_0$$
But $I_0=pi$ hence
$$boxedI_n=a^nfrac(2n-1)!!(2n)!!pi=Big(fraca4Big)^nfrac(2n)!(n!)^2pi$$
where $n!!$ is the double factorial.

share|cite|improve this answer

edited Apr 2 at 17:44

answered Apr 2 at 17:27

Peter ForemanPeter Foreman

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0

$begingroup$

So generalising the integral we have
$$I_n = int_0^a fracr^nr^1/2(a-r)^1/2dr=int_0^a r^n-frac12(a-r)^-frac12dr$$
Then by using integration by parts we get
$$I_n=2(n-frac12)int_0^ar^n-frac32(a-r)^frac12dr$$
$$=(2n-1)int_0^ar^n-frac32(a-r)(a-r)^-frac12dr$$
$$=a(2n-1)int_0^ar^n-frac32(a-r)^-frac12dr-(2n-1)int_0^ar^n-frac12(a-r)^-frac12dr$$
$$=a(2n-1)I_n-1-(2n-1)I_n$$
$$therefore I_n=abigg(frac2n-12nbigg)I_n-1$$
To solve for a base case we can find $I_0$ by using the substitution $r=asin^2(theta)to dr=2asin(theta)cos(theta)dtheta$
$$I_0=int_0^ar^-frac12(a-r)^-frac12dr=int_0^fracpi2 frac2asin(theta)cos(theta)sqrtasin^2(theta)sqrta-asin^2(theta)dtheta=pi$$
for all $ainmathbbC$. Hence we can solve the recurrence relation to get
$$I_n=abigg(frac2n-12nbigg)I_n-1$$
$$=a^2bigg(frac2n-12nbigg)bigg(frac2n-32n-2bigg)I_n-2$$
By continuing this pattern we eventually reach $I_0$ such that
$$I_n=a^nfrac(2n-1)(2n-3)(2n-5)...(1)(2n)(2n-2)(2n-4)...(2)I_0$$
But $I_0=pi$ hence
$$boxedI_n=a^nfrac(2n-1)!!(2n)!!pi=Big(fraca4Big)^nfrac(2n)!(n!)^2pi$$
where $n!!$ is the double factorial.

share|cite|improve this answer

edited Apr 2 at 17:44

answered Apr 2 at 17:27

Peter ForemanPeter Foreman

8,5171321

$endgroup$

add a comment | 

0

$begingroup$

So generalising the integral we have
$$I_n = int_0^a fracr^nr^1/2(a-r)^1/2dr=int_0^a r^n-frac12(a-r)^-frac12dr$$
Then by using integration by parts we get
$$I_n=2(n-frac12)int_0^ar^n-frac32(a-r)^frac12dr$$
$$=(2n-1)int_0^ar^n-frac32(a-r)(a-r)^-frac12dr$$
$$=a(2n-1)int_0^ar^n-frac32(a-r)^-frac12dr-(2n-1)int_0^ar^n-frac12(a-r)^-frac12dr$$
$$=a(2n-1)I_n-1-(2n-1)I_n$$
$$therefore I_n=abigg(frac2n-12nbigg)I_n-1$$
To solve for a base case we can find $I_0$ by using the substitution $r=asin^2(theta)to dr=2asin(theta)cos(theta)dtheta$
$$I_0=int_0^ar^-frac12(a-r)^-frac12dr=int_0^fracpi2 frac2asin(theta)cos(theta)sqrtasin^2(theta)sqrta-asin^2(theta)dtheta=pi$$
for all $ainmathbbC$. Hence we can solve the recurrence relation to get
$$I_n=abigg(frac2n-12nbigg)I_n-1$$
$$=a^2bigg(frac2n-12nbigg)bigg(frac2n-32n-2bigg)I_n-2$$
By continuing this pattern we eventually reach $I_0$ such that
$$I_n=a^nfrac(2n-1)(2n-3)(2n-5)...(1)(2n)(2n-2)(2n-4)...(2)I_0$$
But $I_0=pi$ hence
$$boxedI_n=a^nfrac(2n-1)!!(2n)!!pi=Big(fraca4Big)^nfrac(2n)!(n!)^2pi$$
where $n!!$ is the double factorial.

share|cite|improve this answer

edited Apr 2 at 17:44

answered Apr 2 at 17:27

Peter ForemanPeter Foreman

8,5171321

$endgroup$

add a comment | 

$begingroup$

So generalising the integral we have
$$I_n = int_0^a fracr^nr^1/2(a-r)^1/2dr=int_0^a r^n-frac12(a-r)^-frac12dr$$
Then by using integration by parts we get
$$I_n=2(n-frac12)int_0^ar^n-frac32(a-r)^frac12dr$$
$$=(2n-1)int_0^ar^n-frac32(a-r)(a-r)^-frac12dr$$
$$=a(2n-1)int_0^ar^n-frac32(a-r)^-frac12dr-(2n-1)int_0^ar^n-frac12(a-r)^-frac12dr$$
$$=a(2n-1)I_n-1-(2n-1)I_n$$
$$therefore I_n=abigg(frac2n-12nbigg)I_n-1$$
To solve for a base case we can find $I_0$ by using the substitution $r=asin^2(theta)to dr=2asin(theta)cos(theta)dtheta$
$$I_0=int_0^ar^-frac12(a-r)^-frac12dr=int_0^fracpi2 frac2asin(theta)cos(theta)sqrtasin^2(theta)sqrta-asin^2(theta)dtheta=pi$$
for all $ainmathbbC$. Hence we can solve the recurrence relation to get
$$I_n=abigg(frac2n-12nbigg)I_n-1$$
$$=a^2bigg(frac2n-12nbigg)bigg(frac2n-32n-2bigg)I_n-2$$
By continuing this pattern we eventually reach $I_0$ such that
$$I_n=a^nfrac(2n-1)(2n-3)(2n-5)...(1)(2n)(2n-2)(2n-4)...(2)I_0$$
But $I_0=pi$ hence
$$boxedI_n=a^nfrac(2n-1)!!(2n)!!pi=Big(fraca4Big)^nfrac(2n)!(n!)^2pi$$
where $n!!$ is the double factorial.

share|cite|improve this answer

edited Apr 2 at 17:44

answered Apr 2 at 17:27

Peter ForemanPeter Foreman

8,5171321

$endgroup$

share|cite|improve this answer

edited Apr 2 at 17:44

answered Apr 2 at 17:27

Peter ForemanPeter Foreman

8,5171321

answered Apr 2 at 17:27

Peter ForemanPeter Foreman

8,5171321

answered Apr 2 at 17:27

Peter ForemanPeter Foreman

8,5171321