# Use Mathematical Induction and Integration by Parts to prove integral

• Oct 8th 2013, 08:27 PM
vidomagru
Use Mathematical Induction and Integration by Parts to prove integral
Prove that

$\int\limits_0^1{x^n(1-x)^ydt} = \frac{n!}{(y+1)(y+2)...(y+n+1)}$

where $n$ is a nonnegative integer and $y$ is any real number, which is not a negative integer.

[Hint: Use the method of mathematical induction and integration by parts.]

I do not understand how to proceed with induction?
• Oct 8th 2013, 08:38 PM
SlipEternal
Re: Use Mathematical Induction and Integration by Parts to prove integral
Is $x$ a variable that depends on $t$? What is the relationship?[/tex]
• Oct 8th 2013, 08:40 PM
vidomagru
Re: Use Mathematical Induction and Integration by Parts to prove integral
Quote:

Originally Posted by SlipEternal
Is $x$ a variable that depends on $t$? What is the relationship?[/tex]

It is not clear in the problem, but I believe so.
• Oct 8th 2013, 08:43 PM
SlipEternal
Re: Use Mathematical Induction and Integration by Parts to prove integral
Quote:

Originally Posted by vidomagru
It is not clear in the problem, but I believe so.

Without a relationship, I must assume that $x$ is a constant, as well. So $\int x^n(1-x)^ydt = x^n(1-x)^yt + C$. I think something is missing...
• Oct 8th 2013, 08:53 PM
ibdutt
Re: Use Mathematical Induction and Integration by Parts to prove integral
That way we just cannot get what appears in the denominator on the RHS. it looks as if there is something missing.
• Oct 9th 2013, 01:22 AM
topsquark
Re: Use Mathematical Induction and Integration by Parts to prove integral
Quote:

Originally Posted by vidomagru
Prove that

$\int\limits_0^1{x^n(1-x)^ydt} = \frac{n!}{(y+1)(y+2)...(y+n+1)}$

where $n$ is a nonnegative integer and $y$ is any real number, which is not a negative integer.

[Hint: Use the method of mathematical induction and integration by parts.]

I do not understand how to proceed with induction?

I can't do it by elementary means as I can't seem to manage the induction either, but I can verify that the dt is really a dx. The integral is the definition of the beta function.

$\int_0^1 x^n (1 - x)^y~dx = \frac{n!}{(y + 1)(y + 2)...(y + n + 1)}$

-Dan
• Oct 9th 2013, 07:17 AM
SlipEternal
Re: Use Mathematical Induction and Integration by Parts to prove integral
Quote:

Originally Posted by topsquark
I can't do it by elementary means as I can't seem to manage the induction either, but I can verify that the dt is really a dx. The integral is the definition of the beta function.

$\int_0^1 x^n (1 - x)^y~dx = \frac{n!}{(y + 1)(y + 2)...(y + n + 1)}$

-Dan

Ok, I get it now. The dt was throwing me. With dx, it actually seems fairly trivial, so I may be missing something.

Base case: n = 0

$\int_0^1(1-x)^y~dx = \left. -\dfrac{(1-x)^{y+1}}{y+1} \right]_0^1 = \dfrac{1}{y+1} = \dfrac{0!}{y+0+1}$

So, it satisfies the base case. Suppose it is true for all nonnegative integers less than $n>0$. I want to show it is true for $n$.

The integration by parts is very straightforward: $u = x^n, dv = (1-x)^y~dx \Rightarrow du = nx^{n-1}~dx, v = -\dfrac{(1-x)^{y+1}}{y+1}$.

Now, we have:

$\int_0^1 x^n(1-x)^y~dx = \left[ -\dfrac{x^n(1-x)^{y+1}}{y+1} \right]_0^1 + \dfrac{n}{y+1}\int_0^1 x^{n-1}(1-x)^{y+1}~dx$

The first term is zero at both 1 and 0. For the second term, since $y+1 \in \mathbb{R}$ and $n-1$ is a nonnegative integer less than $n>0$, so by the induction assumption, we can apply the hypothesis.
• Oct 9th 2013, 06:46 PM
vidomagru
Re: Use Mathematical Induction and Integration by Parts to prove integral
Quote:

Originally Posted by SlipEternal
Ok, I get it now. The dt was throwing me. With dx, it actually seems fairly trivial, so I may be missing something.

Base case: n = 0

$\int_0^1(1-x)^y~dx = \left. -\dfrac{(1-x)^{y+1}}{y+1} \right]_0^1 = \dfrac{1}{y+1} = \dfrac{0!}{y+0+1}$

So, it satisfies the base case. Suppose it is true for all nonnegative integers less than $n>0$. I want to show it is true for $n$.

The integration by parts is very straightforward: $u = x^n, dv = (1-x)^y~dx \Rightarrow du = nx^{n-1}~dx, v = -\dfrac{(1-x)^{y+1}}{y+1}$.

Now, we have:

$\int_0^1 x^n(1-x)^y~dx = \left[ -\dfrac{x^n(1-x)^{y+1}}{y+1} \right]_0^1 + \dfrac{n}{y+1}\int_0^1 x^{n-1}(1-x)^{y+1}~dx$

The first term is zero at both 1 and 0. For the second term, since $y+1 \in \mathbb{R}$ and $n-1$ is a nonnegative integer less than $n>0$, so by the induction assumption, we can apply the hypothesis.

Yes I did verify with my professor that it should be a dx not a dt.
• Oct 15th 2013, 01:34 PM
vidomagru
Re: Use Mathematical Induction and Integration by Parts to prove integral
Quote:

Originally Posted by SlipEternal
Ok, I get it now. The dt was throwing me. With dx, it actually seems fairly trivial, so I may be missing something.

Base case: n = 0

$\int_0^1(1-x)^y~dx = \left. -\dfrac{(1-x)^{y+1}}{y+1} \right]_0^1 = \dfrac{1}{y+1} = \dfrac{0!}{y+0+1}$

So, it satisfies the base case. Suppose it is true for all nonnegative integers less than $n>0$. I want to show it is true for $n$.

The integration by parts is very straightforward: $u = x^n, dv = (1-x)^y~dx \Rightarrow du = nx^{n-1}~dx, v = -\dfrac{(1-x)^{y+1}}{y+1}$.

Now, we have:

$\int_0^1 x^n(1-x)^y~dx = \left[ -\dfrac{x^n(1-x)^{y+1}}{y+1} \right]_0^1 + \dfrac{n}{y+1}\int_0^1 x^{n-1}(1-x)^{y+1}~dx$

The first term is zero at both 1 and 0. For the second term, since $y+1 \in \mathbb{R}$ and $n-1$ is a nonnegative integer less than $n>0$, so by the induction assumption, we can apply the hypothesis.

So based on your help, this is my proof. Does this seem accurate?

Problem: $\int\limits_0^1{x^n(1-x)^ydx} = \frac{n!}{(y+1)(y+2)...(y+n+1)}$

where $n$ is a nonnegative integer and $y$ is any real number, which is not a negative integer.

[Hint: Use the method of mathematical induction and integration by parts.]

Proof: We will prove this by Mathematical Induction.
We need to show that it holds for n=0. Now,

$\int_0^1 (x^0)(1-x)^y~dx = \int_0^1 (1-x)^y~dx = \left. -\dfrac{(1-x)^{y+1}}{y+1} \right|_0^1$ and this simplifies to:

$\left. -\dfrac{(1-x)^{y+1}}{y+1} \right|_0^1 = -\dfrac{(1-1)^{y+1}}{y+1} + \dfrac{(1-0)^{y+1}}{y+1} = 0 + \dfrac{1}{y+1} = \dfrac{0!}{y+0+1}$.

Hence it holds for n=0. Now, suppose that it is true for $n=k$, we need to show that it holds for $n = k+1$.

First, evaluate the integral using integration by parts. Let $u = x^n, dv = (1-x)^y~dx \Rightarrow du = nx^{n-1}~dx, v = -\dfrac{(1-x)^{y+1}}{y+1}$. So now, we have:

$\int_0^1 x^n(1-x)^y~dx = \left[ -\dfrac{x^n(1-x)^{y+1}}{y+1} \right]_0^1 + \dfrac{n}{y+1}\int_0^1 x^{n-1}(1-x)^{y+1}~dx$.

Now, $\left. -\dfrac{x^n(1-x)^{y+1}}{y+1} \right|_0^1 = -\dfrac{1^n(1-1)^{y+1}}{y+1} + \dfrac{0^n(1-0)^{y+1}}{y+1}$ = 0 + 0 = 0.

Now by our assumption we have $\int_0^1 x^k(1-x)^y~dx = \dfrac{k}{y+1}\int_0^1 x^{k-1}(1-x)^{y+1}~dx = \frac{k!}{(y+1)(y+2)...(y+k+1)}$, so now consider $n = k+1$.

$\dfrac{k+1}{y+1}\int_0^1 x^{k+1-1}(1-x)^{y+1}~dx = \dfrac{k+1}{y+1}\int_0^1 x^{k}(1-x)^{y+1}~dx$.

Where do I go about applying the hypothesis of n = k?
• Oct 15th 2013, 04:21 PM
SlipEternal
Re: Use Mathematical Induction and Integration by Parts to prove integral
Quote:

Originally Posted by vidomagru
First, evaluate the integral using integration by parts. Let $u = x^n, dv = (1-x)^y~dx \Rightarrow du = nx^{n-1}~dx, v = -\dfrac{(1-x)^{y+1}}{y+1}$. So now, we have:

$\int_0^1 x^n(1-x)^y~dx = \left[ -\dfrac{x^n(1-x)^{y+1}}{y+1} \right]_0^1 + \dfrac{n}{y+1}\int_0^1 x^{n-1}(1-x)^{y+1}~dx$.

Now, $\left. -\dfrac{x^n(1-x)^{y+1}}{y+1} \right|_0^1 = -\dfrac{1^n(1-1)^{y+1}}{y+1} + \dfrac{0^n(1-0)^{y+1}}{y+1}$ = 0 + 0 = 0.

Now by our assumption we have $\int_0^1 x^k(1-x)^y~dx = \dfrac{k}{y+1}\int_0^1 x^{k-1}(1-x)^{y+1}~dx = \frac{k!}{(y+1)(y+2)...(y+k+1)}$, so now consider $n = k+1$.

$\dfrac{k+1}{y+1}\int_0^1 x^{k+1-1}(1-x)^{y+1}~dx = \dfrac{k+1}{y+1}\int_0^1 x^{k}(1-x)^{y+1}~dx$.

Where do I go about applying the hypothesis of n = k?

You have
$\int_0^1 x^n(1-x)^y~dx = \dfrac{n}{y+1}\int_0^1 x^{n-1}(1-x)^{y+1}~dx$.

Since $n-1 = (k+1)-1 = k$ and $y\in \mathbb{R}\setminus \mathbb{Z}^- \Rightarrow y+1 \in \mathbb{R} \setminus \mathbb{Z}^-$, you can apply the induction hypothesis on $\int_0^1 x^{n-1}(1-x)^{y+1}~dx$.
• Oct 15th 2013, 06:24 PM
vidomagru
Re: Use Mathematical Induction and Integration by Parts to prove integral
Quote:

Originally Posted by SlipEternal
You have
$\int_0^1 x^n(1-x)^y~dx = \dfrac{n}{y+1}\int_0^1 x^{n-1}(1-x)^{y+1}~dx$.

Since $n-1 = (k+1)-1 = k$ and $y\in \mathbb{R}\setminus \mathbb{Z}^- \Rightarrow y+1 \in \mathbb{R} \setminus \mathbb{Z}^-$, you can apply the induction hypothesis on $\int_0^1 x^{n-1}(1-x)^{y+1}~dx$.

I am not sure I understand how you are applying the induction hypothesis. Are you saying that since

$n-1 = (k+1)-1 = k$ and $y\in \mathbb{R}\setminus \mathbb{Z}^- \Rightarrow y+1 \in \mathbb{R} \setminus \mathbb{Z}^-$, then

$\int_0^1 x^{n-1}(1-x)^{y+1}~dx = \dfrac{n-1}{y+1}\int_0^1 x^{n-2}(1-x)^{y+1}~dx = \frac{(n-1)!}{(y+1)(y+2)...(y+(n-1)+1)}$?

I am very confused on how this works.
• Oct 15th 2013, 09:21 PM
vidomagru
Re: Use Mathematical Induction and Integration by Parts to prove integral
Quote:

Originally Posted by SlipEternal
You have
$\int_0^1 x^n(1-x)^y~dx = \dfrac{n}{y+1}\int_0^1 x^{n-1}(1-x)^{y+1}~dx$.

Since $n-1 = (k+1)-1 = k$ and $y\in \mathbb{R}\setminus \mathbb{Z}^- \Rightarrow y+1 \in \mathbb{R} \setminus \mathbb{Z}^-$, you can apply the induction hypothesis on $\int_0^1 x^{n-1}(1-x)^{y+1}~dx$.

So what we are saying here is:

$\int_0^1 x^{k+1}(1-x)^{y}~dx = \dfrac{k+1}{y+1} \int_0^1 x^{k+1-1}(1-x)^{y+1}~dx$

$= \dfrac{k+1}{y+1} \int_0^1 x^{k}(1-x)^{y+1}~dx$

$= \dfrac{k+1}{y+1} \cdot \dfrac{k}{y+1} \int_0^1 x^{k-1}(1-x)^{y+1}~dx$

$= \dfrac{k+1}{y+1} \cdot \dfrac{k!}{(y+1)(y+2)...(y+k+1)}$

$= \dfrac{(k+1)!}{(y+1)(y+2)...(y+(k+1)+1)}$.

Am I missing something here?
• Oct 15th 2013, 09:41 PM
SlipEternal
Re: Use Mathematical Induction and Integration by Parts to prove integral
Quote:

Originally Posted by vidomagru
So what we are saying here is:

$\int_0^1 x^{k+1}(1-x)^{y}~dx = \dfrac{k+1}{y+1} \int_0^1 x^{k+1-1}(1-x)^{y+1}~dx$

$= \dfrac{k+1}{y+1} \int_0^1 x^{k}(1-x)^{y+1}~dx$

This is good, but the next line is not needed:

Quote:

Originally Posted by vidomagru
$= \dfrac{k+1}{y+1} \cdot \dfrac{k}{y+1} \int_0^1 x^{k-1}(1-x)^{y+1}~dx$

You can apply the induction hypothesis directly. You are not applying it correctly here:

Quote:

Originally Posted by vidomagru
$= \dfrac{k+1}{y+1} \cdot \dfrac{k!}{(y+1)(y+2)...(y+k+1)}$

$= \dfrac{(k+1)!}{(y+1)(y+2)...(y+(k+1)+1)}$.

Am I missing something here?

To apply it correctly,
\begin{align*}\dfrac{n}{y+1} \int_0^1 x^{n-1}(1-x)^{y+1}~dx & = \dfrac{n}{y+1}\dfrac{(n-1)!}{((y+1)+1)((y+1)+2)\cdots ((y+1)+(n-1)+1)} \\ & = \dfrac{n!}{(y+1)(y+2)\cdots (y+n+1)}\end{align*}