How to solve this following integral?

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$displaystyle dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx =$ $ ?$

I tried putting $x=sin(a)$ but I could do nothing about the 100 and 101 powers, they made the integration not solvable for me.

How should I do or approach this question?

edited 36 mins ago

asked 42 mins ago

Football Life

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up vote
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down vote

favorite

$displaystyle dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx =$ $ ?$

I tried putting $x=sin(a)$ but I could do nothing about the 100 and 101 powers, they made the integration not solvable for me.

How should I do or approach this question?

edited 36 mins ago

asked 42 mins ago

Football Life

526

New contributor

add a commentÂ |Â

up vote
1
down vote

favorite

$displaystyle dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx =$ $ ?$

I tried putting $x=sin(a)$ but I could do nothing about the 100 and 101 powers, they made the integration not solvable for me.

How should I do or approach this question?

edited 36 mins ago

asked 42 mins ago

Football Life

526

New contributor

$displaystyle dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx =$ $ ?$

I tried putting $x=sin(a)$ but I could do nothing about the 100 and 101 powers, they made the integration not solvable for me.

How should I do or approach this question?

integration

edited 36 mins ago

asked 42 mins ago

Football Life

526

New contributor

edited 36 mins ago

asked 42 mins ago

Football Life

526

New contributor

edited 36 mins ago

asked 42 mins ago

Football Life

526

New contributor

asked 42 mins ago

Football Life

526

asked 42 mins ago

Football Life

526

New contributor

Football Life is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
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$ dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx$

Let $displaystyle I_n = int_0^1 (1-x^m)^n dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)^n+1dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)(1-x^m)^n dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)^n dx- int_0^1x^m(1-x^m)^n dx$ ...[*]

$implies displaystyle I_n+1 = I_n- int_0^1x^m(1-x^m)^n dx$

We will integrate $displaystyle int_0^1x^m(1-x^m)^n dx$ via integration by parts. Watch closely, this is a little tricky.

$displaystyle int_0^1x^m(1-x^m)^n dx=int_0^1xcdot x^m-1(1-x^m)^ndx$

Let $u = x implies du = dx$

And $dv = x^m-1(1-x^m)^n dx$

Let $y = (1-x^m) implies dy = -mx^m-1 dx implies x^m-1 dx = -dfracdym$

$displaystyle v = intx^m-1(1-x^m)^n dx = int-dfracy^nmdy = dfrac-y^n+1m(n+1)=dfrac-(1-x^m)^n+1m(n+1)$

$displaystyle int_0^1x^m(1-x^m)^n dx$

$=displaystyle left[-dfracx(1-x^m)^n+1m(n+1)right]_0^1+dfrac1m(n+1)int_0^1(1-x^m)^n+1dx$

$= dfracI_n+1m(n+1)$

Substituting this result into [*]

$I_n+1 = I_n - dfracI_n+1m(n+1)$

$implies left[1+dfrac1m(n+1)right]=dfracI_nI_n+1$

$implies dfracI_nI_n+1 = dfracm(n+1)+1m(n+1)$

Putting $m = 50$ and $n = 100$, we have

$displaystyle dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx=dfrac50times101+150times 101=dfrac50515050$

answered 37 mins ago

Pradyuman Dixit

60610

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Let $a$, $b>0$. Then, substituting $t=x^a$,
$$I_a,b=
int_0^1(1-x^a)^b,dx=frac1aint_0^1(1-t)^bt^1/a-1,dt=fracB(b+1,1/a)a$$
where $B$ denotes the Beta function. But the beta function is expressible
in terms of the Gamma function so that
$$fracB(b+1,1/a)a=fracGamma(1/a)Gamma(b+1)aGamma(1/a+b+1).$$

Therefore
$$fracI_a,bI_a,b+1=fracGamma(b+1)Gamma(1/a+b+2)Gamma(b+2)
Gamma(1/a+b+1)=frac1/a+b+1b+1.$$
Now let $a=50$ and $b=100$.

answered 24 mins ago

Lord Shark the Unknown

91.9k955118

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2 Answers
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2 Answers
2

active

oldest

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up vote
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down vote

$ dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx$

Let $displaystyle I_n = int_0^1 (1-x^m)^n dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)^n+1dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)(1-x^m)^n dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)^n dx- int_0^1x^m(1-x^m)^n dx$ ...[*]

$implies displaystyle I_n+1 = I_n- int_0^1x^m(1-x^m)^n dx$

We will integrate $displaystyle int_0^1x^m(1-x^m)^n dx$ via integration by parts. Watch closely, this is a little tricky.

$displaystyle int_0^1x^m(1-x^m)^n dx=int_0^1xcdot x^m-1(1-x^m)^ndx$

Let $u = x implies du = dx$

And $dv = x^m-1(1-x^m)^n dx$

Let $y = (1-x^m) implies dy = -mx^m-1 dx implies x^m-1 dx = -dfracdym$

$displaystyle v = intx^m-1(1-x^m)^n dx = int-dfracy^nmdy = dfrac-y^n+1m(n+1)=dfrac-(1-x^m)^n+1m(n+1)$

$displaystyle int_0^1x^m(1-x^m)^n dx$

$=displaystyle left[-dfracx(1-x^m)^n+1m(n+1)right]_0^1+dfrac1m(n+1)int_0^1(1-x^m)^n+1dx$

$= dfracI_n+1m(n+1)$

Substituting this result into [*]

$I_n+1 = I_n - dfracI_n+1m(n+1)$

$implies left[1+dfrac1m(n+1)right]=dfracI_nI_n+1$

$implies dfracI_nI_n+1 = dfracm(n+1)+1m(n+1)$

Putting $m = 50$ and $n = 100$, we have

$displaystyle dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx=dfrac50times101+150times 101=dfrac50515050$

answered 37 mins ago

Pradyuman Dixit

60610

add a commentÂ |Â

up vote
8
down vote

$ dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx$

Let $displaystyle I_n = int_0^1 (1-x^m)^n dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)^n+1dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)(1-x^m)^n dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)^n dx- int_0^1x^m(1-x^m)^n dx$ ...[*]

$implies displaystyle I_n+1 = I_n- int_0^1x^m(1-x^m)^n dx$

We will integrate $displaystyle int_0^1x^m(1-x^m)^n dx$ via integration by parts. Watch closely, this is a little tricky.

$displaystyle int_0^1x^m(1-x^m)^n dx=int_0^1xcdot x^m-1(1-x^m)^ndx$

Let $u = x implies du = dx$

And $dv = x^m-1(1-x^m)^n dx$

Let $y = (1-x^m) implies dy = -mx^m-1 dx implies x^m-1 dx = -dfracdym$

$displaystyle v = intx^m-1(1-x^m)^n dx = int-dfracy^nmdy = dfrac-y^n+1m(n+1)=dfrac-(1-x^m)^n+1m(n+1)$

$displaystyle int_0^1x^m(1-x^m)^n dx$

$=displaystyle left[-dfracx(1-x^m)^n+1m(n+1)right]_0^1+dfrac1m(n+1)int_0^1(1-x^m)^n+1dx$

$= dfracI_n+1m(n+1)$

Substituting this result into [*]

$I_n+1 = I_n - dfracI_n+1m(n+1)$

$implies left[1+dfrac1m(n+1)right]=dfracI_nI_n+1$

$implies dfracI_nI_n+1 = dfracm(n+1)+1m(n+1)$

Putting $m = 50$ and $n = 100$, we have

$displaystyle dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx=dfrac50times101+150times 101=dfrac50515050$

answered 37 mins ago

Pradyuman Dixit

60610

add a commentÂ |Â

up vote
8
down vote

$ dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx$

Let $displaystyle I_n = int_0^1 (1-x^m)^n dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)^n+1dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)(1-x^m)^n dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)^n dx- int_0^1x^m(1-x^m)^n dx$ ...[*]

$implies displaystyle I_n+1 = I_n- int_0^1x^m(1-x^m)^n dx$

We will integrate $displaystyle int_0^1x^m(1-x^m)^n dx$ via integration by parts. Watch closely, this is a little tricky.

$displaystyle int_0^1x^m(1-x^m)^n dx=int_0^1xcdot x^m-1(1-x^m)^ndx$

Let $u = x implies du = dx$

And $dv = x^m-1(1-x^m)^n dx$

Let $y = (1-x^m) implies dy = -mx^m-1 dx implies x^m-1 dx = -dfracdym$

$displaystyle v = intx^m-1(1-x^m)^n dx = int-dfracy^nmdy = dfrac-y^n+1m(n+1)=dfrac-(1-x^m)^n+1m(n+1)$

$displaystyle int_0^1x^m(1-x^m)^n dx$

$=displaystyle left[-dfracx(1-x^m)^n+1m(n+1)right]_0^1+dfrac1m(n+1)int_0^1(1-x^m)^n+1dx$

$= dfracI_n+1m(n+1)$

Substituting this result into [*]

$I_n+1 = I_n - dfracI_n+1m(n+1)$

$implies left[1+dfrac1m(n+1)right]=dfracI_nI_n+1$

$implies dfracI_nI_n+1 = dfracm(n+1)+1m(n+1)$

Putting $m = 50$ and $n = 100$, we have

$displaystyle dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx=dfrac50times101+150times 101=dfrac50515050$

answered 37 mins ago

Pradyuman Dixit

60610

$ dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx$

Let $displaystyle I_n = int_0^1 (1-x^m)^n dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)^n+1dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)(1-x^m)^n dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)^n dx- int_0^1x^m(1-x^m)^n dx$ ...[*]

$implies displaystyle I_n+1 = I_n- int_0^1x^m(1-x^m)^n dx$

We will integrate $displaystyle int_0^1x^m(1-x^m)^n dx$ via integration by parts. Watch closely, this is a little tricky.

$displaystyle int_0^1x^m(1-x^m)^n dx=int_0^1xcdot x^m-1(1-x^m)^ndx$

Let $u = x implies du = dx$

And $dv = x^m-1(1-x^m)^n dx$

Let $y = (1-x^m) implies dy = -mx^m-1 dx implies x^m-1 dx = -dfracdym$

$displaystyle v = intx^m-1(1-x^m)^n dx = int-dfracy^nmdy = dfrac-y^n+1m(n+1)=dfrac-(1-x^m)^n+1m(n+1)$

$displaystyle int_0^1x^m(1-x^m)^n dx$

$=displaystyle left[-dfracx(1-x^m)^n+1m(n+1)right]_0^1+dfrac1m(n+1)int_0^1(1-x^m)^n+1dx$

$= dfracI_n+1m(n+1)$

Substituting this result into [*]

$I_n+1 = I_n - dfracI_n+1m(n+1)$

$implies left[1+dfrac1m(n+1)right]=dfracI_nI_n+1$

$implies dfracI_nI_n+1 = dfracm(n+1)+1m(n+1)$

Putting $m = 50$ and $n = 100$, we have

$displaystyle dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx=dfrac50times101+150times 101=dfrac50515050$

answered 37 mins ago

Pradyuman Dixit

60610

answered 37 mins ago

Pradyuman Dixit

60610

answered 37 mins ago

Pradyuman Dixit

60610

answered 37 mins ago

Pradyuman Dixit

60610

add a commentÂ |Â

up vote
3
down vote

Therefore
$$fracI_a,bI_a,b+1=fracGamma(b+1)Gamma(1/a+b+2)Gamma(b+2)
Gamma(1/a+b+1)=frac1/a+b+1b+1.$$
Now let $a=50$ and $b=100$.

answered 24 mins ago

Lord Shark the Unknown

91.9k955118

add a commentÂ |Â

up vote
3
down vote

Therefore
$$fracI_a,bI_a,b+1=fracGamma(b+1)Gamma(1/a+b+2)Gamma(b+2)
Gamma(1/a+b+1)=frac1/a+b+1b+1.$$
Now let $a=50$ and $b=100$.

answered 24 mins ago

Lord Shark the Unknown

91.9k955118

add a commentÂ |Â

up vote
3
down vote

Therefore
$$fracI_a,bI_a,b+1=fracGamma(b+1)Gamma(1/a+b+2)Gamma(b+2)
Gamma(1/a+b+1)=frac1/a+b+1b+1.$$
Now let $a=50$ and $b=100$.

answered 24 mins ago

Lord Shark the Unknown

91.9k955118

Therefore
$$fracI_a,bI_a,b+1=fracGamma(b+1)Gamma(1/a+b+2)Gamma(b+2)
Gamma(1/a+b+1)=frac1/a+b+1b+1.$$
Now let $a=50$ and $b=100$.

answered 24 mins ago

Lord Shark the Unknown

91.9k955118

answered 24 mins ago

Lord Shark the Unknown

91.9k955118

answered 24 mins ago

Lord Shark the Unknown

91.9k955118

answered 24 mins ago

Lord Shark the Unknown

91.9k955118

add a commentÂ |Â

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