# Thread: Use Mathematical Induction and Integration by Parts to prove integral

1. ## Use Mathematical Induction and Integration by Parts to prove integral

Prove that

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

where $\displaystyle n$ is a nonnegative integer and $\displaystyle y$ 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?

2. ## Re: Use Mathematical Induction and Integration by Parts to prove integral

Is $\displaystyle x$ a variable that depends on $\displaystyle t$? What is the relationship?[/tex]

3. ## Re: Use Mathematical Induction and Integration by Parts to prove integral

Originally Posted by SlipEternal
Is $\displaystyle x$ a variable that depends on $\displaystyle t$? What is the relationship?[/tex]
It is not clear in the problem, but I believe so.

4. ## Re: Use Mathematical Induction and Integration by Parts to prove integral

Originally Posted by vidomagru
It is not clear in the problem, but I believe so.
Without a relationship, I must assume that $\displaystyle x$ is a constant, as well. So $\displaystyle \int x^n(1-x)^ydt = x^n(1-x)^yt + C$. I think something is missing...

5. ## 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.

6. ## Re: Use Mathematical Induction and Integration by Parts to prove integral

Originally Posted by vidomagru
Prove that

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

where $\displaystyle n$ is a nonnegative integer and $\displaystyle y$ 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.

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

-Dan

7. ## Re: Use Mathematical Induction and Integration by Parts to prove integral

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.

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

-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

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

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

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

Now, we have:

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

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

8. ## Re: Use Mathematical Induction and Integration by Parts to prove integral

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

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

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

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

Now, we have:

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

The first term is zero at both 1 and 0. For the second term, since $\displaystyle y+1 \in \mathbb{R}$ and $\displaystyle n-1$ is a nonnegative integer less than $\displaystyle n>0$, 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.

9. ## Re: Use Mathematical Induction and Integration by Parts to prove integral

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

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

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

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

Now, we have:

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

The first term is zero at both 1 and 0. For the second term, since $\displaystyle y+1 \in \mathbb{R}$ and $\displaystyle n-1$ is a nonnegative integer less than $\displaystyle n>0$, so by the induction assumption, we can apply the hypothesis.
So based on your help, this is my proof. Does this seem accurate?

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

where $\displaystyle n$ is a nonnegative integer and $\displaystyle y$ 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,

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

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

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

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

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

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

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

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

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

10. ## Re: Use Mathematical Induction and Integration by Parts to prove integral

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

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

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

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

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

Where do I go about applying the hypothesis of n = k?
You have
$\displaystyle \int_0^1 x^n(1-x)^y~dx = \dfrac{n}{y+1}\int_0^1 x^{n-1}(1-x)^{y+1}~dx$.

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

11. ## Re: Use Mathematical Induction and Integration by Parts to prove integral

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

Since $\displaystyle n-1 = (k+1)-1 = k$ and $\displaystyle y\in \mathbb{R}\setminus \mathbb{Z}^- \Rightarrow y+1 \in \mathbb{R} \setminus \mathbb{Z}^-$, you can apply the induction hypothesis on $\displaystyle \int_0^1 x^{n-1}(1-x)^{y+1}~dx$.
I am not sure I understand how you are applying the induction hypothesis. Are you saying that since

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

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

I am very confused on how this works.

12. ## Re: Use Mathematical Induction and Integration by Parts to prove integral

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

Since $\displaystyle n-1 = (k+1)-1 = k$ and $\displaystyle y\in \mathbb{R}\setminus \mathbb{Z}^- \Rightarrow y+1 \in \mathbb{R} \setminus \mathbb{Z}^-$, you can apply the induction hypothesis on $\displaystyle \int_0^1 x^{n-1}(1-x)^{y+1}~dx$.
So what we are saying here is:

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

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

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

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

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

Am I missing something here?

13. ## Re: Use Mathematical Induction and Integration by Parts to prove integral

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

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

$\displaystyle = \dfrac{k+1}{y+1} \int_0^1 x^{k}(1-x)^{y+1}~dx$
This is good, but the next line is not needed:

Originally Posted by vidomagru
$\displaystyle = \dfrac{k+1}{y+1} \cdot \dfrac{k}{y+1} \int_0^1 x^{k-1}(1-x)^{y+1}~dx$
You can apply the induction hypothesis directly. You are not applying it correctly here:

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

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

Am I missing something here?
To apply it correctly,
\displaystyle \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*}

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