Definite | Racquel Sanderson

Category Archives: Definite

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

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

Effect of Function Transformations on Integration

My year 12 Mathematical Methods students have questions like this

Given that $f(x)$ is continuous everywhere and that $\int_{4}^{10} f(x) dx=-10$ , find:

(a) $\int_{4}^{10}2x +f(x) dx$

(b) $\int_{5}^{11} f(x-1) dx$

(c) $\int_{1}^{3} f(3x+1) dx$

(d) $\int_{-10}^{-4} -f(-x) dx$

(e) $\int{10}^{22} f(\frac{x-2}{2}) dx$

(f) $\int_{-3}^{-9} f(1-x) dx$

OT Lee Mathematics Methods Textbook Ex 8.3 question 6

For the most part these questions aren’t too difficult, but the horizontal dilations cause issues.

\int_{4}^{10} 2x dx +\int_{2}^{10} f(x) dx

Once you get the hang of it, you can skip the change of variable and multiply the value of the definite integral by the scale factor of the horizontal dilation (only if the integration bounds are also changed).

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

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