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# Corrected exercise: Wallis integral

Here is the statement of an exercise that allows you to study different properties of Wallis integrals. It is an exercise on the border between the chapter of integrals and that suites. This is a completely feasible exercise in the first year of higher education. Here is the statement:

And let's start the correction right away

## Question 1

For this question, we will make a change of variable and ask

u = \dfrac{\pi}{2}-t

We then obtain

\begin{array}{l} W_n = \displaystyle \int_0^{\frac{\pi}{2}} \sin^n(t) dt \\ =\displaystyle\int_{\frac{\pi}{2 }}^{0} \sin^n(\frac{\pi}{2}-u) (-du)\\ =\displaystyle \int_0^{\frac{\pi}{2}} \cos^n (t) dt \end{array}


We used the properties of sines and cosines. This easily answers this first question (which is not a harder one)

Now let's move on to the second question!

## Question 2

We show that the sequence (Wn) is decreasing. We have :

\forall t \in [0, \frac{\pi}{2}], 0\leq \sin(t) \leq 1

By multiplying each side by sinn(t), we have

\forall t \in [0, \frac{\pi}{2}], 0\leq \sin^{n+1}(t) \leq \sin^n(t)

And integrating on each side, we then get

\begin{array}{l} \displaystyle \int_0^{\frac{\pi}{2}} 0dt \leq \int_0^{\frac{\pi}{2}}\sin^{n+1}( t) dt\leq \int_0^{\frac{\pi}{2}}\sin^n(t)dt\\ \Leftrightarrow 0 \leq W_{n+1}\leq W_n \end{array}

The sequence (Wn) is therefore indeed decreasing. We have also just obtained that it was lowered by 0.
So as a decreasing and minor sequence, the sequence (Wn) converges.
Now find its limit.
be

a \in \left[0, \dfrac{\pi}{2} \right] \text{ and } \varepsilon >0

Let's do the following manipulation:

\begin{array}{l} 0 \leq W_n \\ =\displaystyle \int_0^a \sin^{n}t dt+\int_a^{\frac{\pi}{2}} \sin^{n}t dt\\ \leq \displaystyle \int_0^a \sin^{n}a dt+\int_a^{\frac{\pi}{2}} 1 dt\\ \leq a\sin^{n}a + \left ( \frac{\pi}{2} - a \right) \end{array}

Now we choose a such that

0 < \left( \frac{\pi}{2} - a \right) < \dfrac{\varepsilon}{2}

Once this has been fixed, we know that from a certain rank, we have:

0 \leq a\sin^{n}a \leq \dfrac{\varepsilon}{2}

This gives us the following framework:

0 \leq W_n \leq a\sin^{n}a + \left( \frac{\pi}{2} - a \right) \leq \dfrac{\varepsilon}{2} + \dfrac{\varepsilon} {2} \varepsilon

## Question 3

We are going to do an integration by parts. For this, we write the following decomposition:

\forall n \geq 2, W_n = \int_0^{\frac{\pi}{2} } \sin(t) \sin^{n-1}(t) dt

We integrate the left member and we derive the right member:

W_n = \left[-\cos(t) \sin^{n-1}(t)\right]_0^{\frac{\pi}{2}} + (n-1)\int_0^{\frac {\pi}{2} } \cos^2(t) \sin^{n-2}(t) dt

We then use the trigonometry formulas and simplify the first term:

\begin{array}{l} W_n = \displaystyle (n-1)\int_0^{\frac{\pi}{2} } (1-\sin^2(t) )\sin^{n-2} (t) dt\\ \Leftrightarrow W_n = \displaystyle (n-1)\left(\int_0^{\frac{\pi}{2} } \sin^{n-2}(t) dt-\int_0^ {\frac{\pi}{2} } \sin^{n}(t) dt\right)\\ \Leftrightarrow W_n = (n-1)(W_{n-2}-W_n) \\ \Leftrightarrow nW_n = (n-1)W_{n-2} \\ \Leftrightarrow W_n = \dfrac{n-1}{n}W_{n-2} \\ \end{array}

Which is indeed a recurrence relation, of order 2.

## Question 4

Let’s calculate the first 2 values ​​of the sequence:

W_0 = \int_0^{\frac{\pi}{2}} \sin^0(t) dt = \int_0^{\frac{\pi}{2}} 1 dt = \dfrac{\pi}{2 }

Let's calculate W1

W_1 = \int_0^{\frac{\pi}{2}} \sin^1(t) dt =[-cos(t)]_0^{\frac{\pi}{2}}= 1

W_{2n} = \dfrac{2n-1}{2n}W_{2n-2} = \ldots = \dfrac{\prod_{k=1}^n (2k-1)}{\prod_{k=1 }^n (2k)}W_0

We multiply the even terms in the numerator and in the denominator so that the numerator contains all the terms between 1 and 2n.

W_{2n} = \dfrac{\prod_{k=1}^{2n} k}{\prod_{k=1}^n (2k)^2}W_0 = \dfrac{(2n)!}{2^ {2n}n!^2}\dfrac{\pi}{2}

We then do the same process with the odd terms:

W_{2n+1} = \dfrac{2n}{2n+1}W_{2n-1} = \ldots = \dfrac{\prod_{k=1}^n (2k)}{\prod_{k=1 }^n (2k+1)}W_1

Then we multiply the numerator and the denominator by all the even terms so that the denominator contains all the terms between 1 and 2n+1:

W_{2n+1} = \dfrac{\prod_{k=1}^n (2k)^2}{\prod_{k=1}^{2n+1} k}W_1= \dfrac{2^{2n}n!^2}{(2n+1)!}

## Question 5

Let's demonstrate a relationship that will help us. We have :

\begin{array}{l} W_n = \dfrac{n-1}{n}W_{n-2}\\ \Leftrightarrow nW_n = (n-1)W_{n-2}\\ \Leftrightarrow nW_nW_{n -1} = (n-1)W_{n-1}W_{n-2} \end{array}

The sequence (nWnWn-1) is therefore a constant sequence. So we have :

nW_nW_{n-1} = 1 W_1W_0 = \dfrac{\pi}{2}

Urban artist

\begin{array}{l} W_{n} \leq W_{n-1}\leq W_{n-2}\\ \Leftrightarrow W_{n} \leq W_{n-1}\leq \dfrac{n }{n-1}W_{n}\\ \Leftrightarrow 1 \leq \dfrac{W_{n-1}}{W_n}\leq \dfrac{n}{n-1} \end{array}

Which gives us the following equivalent:

W_n \sim W_{n-1}

So, taking our equality:

\begin{array}{l} \dfrac{\pi}{2} = nW_nW_{n-1} \sim n W_n^2\\ \Rightarrow W_n \sim \sqrt{\dfrac{\pi}{2n}} \end{array}

This concludes our question and therefore our exercise. We have seen several properties of Wallis integrals.

Did you like this exercise? Find out how this exercise can help calculate the Stirling's formula !