Integration by Parts | Racquel Sanderson

Category Archives: Integration by Parts

November 20, 2024 · 6:57 am

More Integration

I went down a rabbit hole while reading An Imaginary Tale by Paul J Nahin and I decided I wanted to do this…

$\begin{equation*}\int_0^1{x^x dx}\end{equation}$

$\begin{equation*}x^x=e^{ln(x^x)}\end{equation}$

$\begin{equation*}x^x=e^{xln(x)}\end{equation}$

The power series expansion of $e^x$ is

$\begin{equation*}e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+...\end{equation}$

$\begin{equation*}\therefore e^{xln(x)}=1+xln(x)+\frac{1}{2!}(xln(x))^2+\frac{1}{3!}(xln(x))^3+...\end{equation}$

$\begin{equation*}\therefore e^{xln(x)}=\Sigma_{n=0}^{\infty}(\frac{1}{n!}(xln(x))^n)\end{equation}$

Hence $\int_0^1{x^x dx}=\int_0^1(\Sigma_{n=0}^{\infty}(\frac{1}{n!}(xln(x))^n dx)$

$\begin{equation*}=\Sigma_{n=0}^{\infty}(\frac{1}{n!}\int_0^1xln(x))^n dx)\end{equation}$

Let’s consider the integral

(1) $\begin{equation*}\int_0^1 x^n(ln(x))^n dx\end{equation*}$

Let $u=ln(x)$ then $\frac{du}{dx}=\frac{1}{x}$ and $dx=x du$ where $x=e^u$

When $x=0, u=-\infty$ and when $x=1, u=0$

(2) $\begin{equation*}\int_0^1 x^n(ln(x))^n dx=\int_{-\infty}^0 e^{nu}u^ne^u du\end{equation*}$

(3) $\begin{equation*}\int_0^1 x^n(ln(x))^n dx=\int_{-\infty}^0 e^{(n+1)u}u^n du\end{equation*}$

Integrate by parts using the tabular method.

Sign	Differentiate	Integrate
+	$u^n$	$e^{(n+1)u}$
–	$nu^{n-1}$	$\frac{e^{(n+1)u}}{n+1}$
+	$n(n-1)u^{n-2}$	$\frac{e^{(n+1)u}}{(n+1)^2}$
–	$n(n-1)(n-2)u^{n-3}$	$\frac{e^{(n+1)u}}{(n+1)^3}$
+	$\frac{n!}{(n-4)!}u^{n-4}$	$\frac{e^{(n+1)u}}{(n+1)^4}$
	$\vdots$	$\vdots$
	$\frac{n!}{(n-n)!}u^{n-n}$	$\frac{e^{(n+1)u}}{(n+1)^{n}}$
$(-1)^n$	$0$	$\frac{e^{(n+1)u}}{(n+1)^{n+1}}$

When we substitute $u=-\infty$ or $u=0$ the differentiation column is zero except for $\frac{n!}{(n-n)!}u^{n-n}$ , which is $n!$ ,

Thus $\int_0^1 x^n(ln(x))^n dx=\frac{e^{(n+1)u}}{(n+1)^{n+1}}}]_{-\infty}^0$

$\begin{equation*}=n!\times\frac{e^0}{(n+1)^{n+1}}-0\end{equation}$

$\begin{equation*}=\frac{n!}{(n+1)^{n+1}}\end{equation}$

Now we just need to think about the sign.

$\begin{equation*}=(-1)^n\frac{n!}{(n+1)^{n+1}}\end{equation}$

The integral is now

$\int_0^1{x^x dx}=(\Sigma_{n=0}^{\infty}(\frac{1}{n!}( (-1)^n\frac{n!}{(n+1)^{n+1}})$

So $\int_0^1{x^x dx}=\Sigma_{n=0}^{\infty}( (-1)^n\frac{1}{(n+1)^{n+1}}$

Let’s work out some partial sums

$\int_0^1{x^x dx}=0.778343$

Integration by Parts using the Tabular Method

(1) $\begin{equation*}\int_0^1x^2e^xdx\end{equation*}$

I am going to do this integral in two ways; the traditional method and the tabular method.

Traditional Method

Remember $\int{u dv}=u\times v-\int{v du}$

Let $u=x^2$ and $dv=e^x$

Then $du=2x$ and $v=\int{e^x dx}=e^x$

$\begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-\int_0^1{e^x 2x dx}\end{equation}$

Now we need to do integration by parts on $\int_0^1{e^x 2x dx}$

Let $u=2x$ and $dv=e^x$

Then $du=2$ and $v=e^x$

$\begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-(2x\times e^x]_0^1-\int_0^1{e^x 2 dx})\end{equation}$

$\begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-(2x\times e^x]_0^1-(2e^x )]_0^1)\end{equation}$

$\begin{equation*}\int_0^1{x^2e^x dx}=e-(2e-(2e-2))\end{equation}$

$\begin{equation*}\int_0^1{x^2e^x dx}=e+2\end{equation}$

Tabular Integration

Similar to before, select a $u$ and a $dv$ , $u=x^2$ and $dv=e^x$

Sign	D(ifferentiate)	I(ntegrate)
+	$x^2$	$e^x$
–	$2x$	$e^x$
+	$2$	$e^x$
–	$0$	$e^x$

Stop when the differentiating column reaches zero.

Then we multiply diagonally

$(+x^2)(e^x)+(-2x)(e^x)+(+2e^x)$

$=x^2e^x-2xe^x+2e^x]_0^1$

$=e-2e+2e-(0-0-2)$

$=e+2$

It is only worth using this method if integration by parts is required more than once. Also, the $u$ has to eventually differentiate to $0$ .

Let’s try another one

(2) $\begin{equation*}\int{x^3cos(2x) dx}\end{equation*}$

Let $u=x^3$ and $dv=cos(2x)$

Sign	D	I
+	$x^3$	$cos(2x)$
–	$3x^2$	$\frac{1}{2}sin(2x)$
+	$6x$	$-\frac{1}{4}cos(2x)$
–	$6$	$-\frac{1}{8}sin(2x)$
+	$0$	$\frac{1}{16}cos(2x)$

$\int{x^3cos(2x) dx}=(x^3)(\frac{1}{2}sin(2x))+(-3x^2)(-\frac{1}{4}cos(2x))+(6x)(-\frac{1}{8}sin(2x))+(-6)(\frac{1}{16}cos(2x))+c$

$=\frac{x^3}{2}sin(2x)+\frac{3x^2}{4}cos(2x)-\frac{3x}{4}sin(2x)-\frac{3}{8}cos(2x)+c$

$n$	$((-1)^n\frac{1}{(n+1)^{n+1}})$
$5$	$=0.78343$
$10$	$=0.78343$
$20$	$=0.78343$
$100$	$=0.78343$

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