Finding the value of $\int^{1}_{0}(1+x)^{m}(1-x)^{n} \,\mathrm d x$

Question

Find the value of $$\displaystyle \int^{1}_{0} (1+x)^{m}(1-x)^{n} \,\mathrm d x$$ where $m, n \geq 1$ and $m, n \in \mathbb{N}$.

Let

$$\displaystyle I_{m,n} = \int^{1}_{0}(1+x)^m(1-x)^n \,\mathrm d x = \int^{1}_{0}(2-x)^m\cdot x^n \,\mathrm d x$$

Put $x=2\sin^2 \theta$ and $dx = 4\sin \theta \cos \theta \,\mathrm d \theta$ and changing limits, We get

$$\displaystyle I_{m,n} = 2^{m+n+2}\int^{\frac{\pi}{2}}_{0}\cos^{2m+1}\theta\cdot \sin^{2n+1}\theta \,\mathrm d \theta$$

How can I form a recursive relation? Could someone help me? Thanks.

Answer 1

By parts,

$$I_{m,n}:=\int_0^1(1+x)^m(1-x)^ndx\\ =\left.\frac{(1+x)^{m+1}(1-x)^n}{m+1}\right|_0^1+\frac n{m+1}\int_0^1(1+x)^{m+1}(1-x)^{n-1}dx.$$

This gives you the recurrence relation

$$I_{m,n}=-\frac1{m+1}+\frac n{m+1}I_{m+1,n-1}.$$

From this, $$I_{m,n}=-\frac1{m+1}+\frac n{m+1}\left(-\frac1{m+2}+\frac{n-1}{m+2}I_{m+2,n-2}\right)\\ =-\frac1{m+1}-\frac n{(m+1)(m+2)}+\frac{n(n-1)}{(m+1)(m+2)}I_{m+2,n-2}\\ =-\frac1{m+1}-\frac n{(m+1)(m+2)}-\frac{n(n-1)}{(m+1)(m+2)(m+3)}+\frac{n(n-1)(n-2)}{(m+1)(m+2)(m+3)}I_{m+3,n-3}\\ =\cdots\\ =-\sum_{k=0}^n\frac{m!n!}{(m+k+1)!(n-k)!}+\frac{n!}{(m+n+1)!}I_{m+n,0}.$$

The final integral is elementary, $\dfrac1{m+n+1}$. I don't know how to simplify the summation.

Answer 2

We first express the integrand as a product of $\sin \theta$ and $\cos \theta $ by

letting $x=\cos 2\theta $, yields $$I = \int_{\frac{\pi}{4}}^{0}\left(2 \cos ^{2} \theta\right)^{m}\left(2 \sin ^{2} \theta\right)^{n}\left(-2 \sin ^{2} \theta\right) d \theta =2^{m+n+2} \int_{0}^{\frac{\pi}{4}} \cos ^{2 m+1} \theta \sin^{2n+1} \theta d\theta $$

Putting $s=\sin \theta$ gives $$ I=2^{m+n+2} \int_{0}^{\frac{1}{\sqrt{2}}}\left(1-s^{2}\right)^{m} s^{2 n+1} d s . $$

Using Binomial Expansion gives $$ \begin{aligned} I&=2^{m+n+2} \int_{0}^{\frac{1}{\sqrt{2}}} \sum_{k=0}^{m}\left(\begin{array}{c} m \\ k \end{array}\right)(-1)^{k} s^{2 k} s^{2 n+1} d s\\ &=2^{m+n+2} \int_{0}^{\frac{1}{\sqrt{2}}} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c} m \\ k \end{array}\right) s^{2 k+2 n+1} d s\\ &=2^{m+n+2} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c} m \\ k \end{array}\right) \int_{0}^{\frac{1}{\sqrt{2}}} s^{2 k+2 n+1} d s\\ &=2^{m+n+2} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{l} m \\ k \end{array}\right)\left[\frac{s^{2 k+2 n+2}}{2 k+2 n+2}\right]_{0}^{\frac{1}{\sqrt{2}}}\\ &=2^{m+n+1} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c} m \\ k \end{array}\right) \frac{1}{(k+n+1) 2^{k+n+1}}\\ &=2^{m} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c} m \\ k \end{array}\right) \frac{1}{(k+n+1) 2^{k}} \end{aligned} $$