How do i find and what is the recurrence formula for (x^n)(1+x)^1/2.dx between the x terminals 0 and -1 ? Thanks Nath
First you problably mean that,Originally Posted by nath_quam
$\displaystyle 0\leq x\leq 1$ otherwise the function is not defined, $\displaystyle x=-1$ which means that ,
$\displaystyle \int_{-1}^0 \frac{x^n}{\sqrt{1+x}}dx$ makes no sense at all!.
Assuming you mean,
$\displaystyle \int_0^1\frac{x^n}{\sqrt{1+x}}dx$
Use integration by parts,
let, $\displaystyle u=x^n$ then, $\displaystyle u'=nx^{n-1}$ and, $\displaystyle v'=\frac{1}{\sqrt{1+x}}$ then, $\displaystyle v=2\sqrt{1+x}$
Thus,
$\displaystyle \left 2x^n\sqrt{1+x} \right|^1_0-\int_0^1 2nx^{n-1}\sqrt{1+x}dx$
Thus,
$\displaystyle 2\sqrt{2}-2n\int_0^1x^{n-1}\sqrt{1+x}dx$
The trick is to express,
$\displaystyle \sqrt{1+x}$ as, $\displaystyle \frac{1+x}{\sqrt{1+x}}$
Thus,
$\displaystyle 2\sqrt{2}-2n\int_0^1\frac{x^{n-1}(1+x)}{\sqrt{1+x}}dx$
Working with the integral we have,
$\displaystyle \int_0^1\frac{x^{n-1}(1+x)}{\sqrt{1+x}}dx=\int_0^1\frac{x^{n-1}+x^n}{\sqrt{1+x}}$=$\displaystyle \int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx+\int^1_0\frac{x^n}{\sqrt{1+x}}dx$
Let,
$\displaystyle \chi=\int_0^1\frac{x^n}{\sqrt{1+x}}$
Then, as we have shown,
$\displaystyle \chi=2\sqrt{2}-2n\int_0^1\frac{x^{n-1}(1+x)}{\sqrt{1+x}}dx$
Which is,
$\displaystyle \chi=2\sqrt{2}-2n\left(\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx+\chi\right)$
Thus,
$\displaystyle \chi=2\sqrt{2}-2n\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx-2n\chi$
Thus,
$\displaystyle \chi(2n+1)=2\sqrt{2}-2n\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx$
Thus,
$\displaystyle \chi=\frac{2\sqrt{2}}{2n+1}-\frac{2n}{2n+1}\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx$
Thus, the recurrence formula is,
$\displaystyle \boxed{ \int_0^1\frac{x^n}{\sqrt{1+x}}dx =\frac{2\sqrt{2}}{2n+1}-\frac{2n}{2n+1}\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx}$
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Credits:
I learend the box trick from Soroban
Let:Originally Posted by CaptainBlack
$\displaystyle I_n=\int_{-1}^0 x^n\sqrt{1+x}\ dx$.
Now we are going to use integration by parts, which has the schema:
$\displaystyle \int u\ dv=uv-\int v\ du$.
Here we let: $\displaystyle dv=\sqrt{1+x}$, and $\displaystyle u=x^n$, so:
$\displaystyle v=\frac{2}{3}(1+x)^{3/2}=\frac{2}{3}(1+x)\sqrt{1+x}$,
and:
$\displaystyle du=n\ x^{n-1}$.
So
$\displaystyle
\int x^n\ \sqrt{1+x} \ dx=x^n\ \frac{2}{3}(1+x)^{3/2}$$\displaystyle - \int \frac{2}{3}(1+x)\sqrt{1+x}\ n\ x^{n-1} dx
$.
Putting the limits back in:
$\displaystyle
I_n=\int_{-1}^0 x^n\ \sqrt{1+x} \ dx=\left[x^n\ \frac{2}{3}(1+x)^{3/2}\right]_{-1}^0$$\displaystyle - \int_{-1}^0 \frac{2}{3}(1+x)\sqrt{1+x}\ n\ x^{n-1} dx
$.
Simplifying:
$\displaystyle
I_n=\int_{-1}^0 x^n\ \sqrt{1+x} \ dx=$$\displaystyle
- \int_{-1}^0 \frac{2}{3}(1+x)\sqrt{1+x}\ n\ x^{n-1} dx
$
$\displaystyle.
= - \frac{2}{3}n \left[\int_{-1}^0 \sqrt{1+x}\ x^{n-1} dx+\int_{-1}^0 \sqrt{1+x}\ x^{n} dx\right]
$
$\displaystyle
=- \frac{2}{3}n [I_{n-1}+I_n]
$
So the recurrence relation for $\displaystyle I_n$ is:
$\displaystyle
I_n=-\frac{\frac{2}{3}n}{1-\frac{2}{3}n}I_{n-1}
$
RonL
(you will need to check this carefully as there are too many steps to
guarantee there is no mistake)