Tutorial note 3

First of all, I suggest memorizing the following easy formula! For any $a,b,c$ with $a<b$ and $c\neq 0$, we have \begin{equation}\label{multiple}

\boxed{\dis \int_a^b f(x)\,dx = \int_{ca}^{cb}f\brac{\frac{x}{c}}\,\frac{1}{c}dx}

\end{equation} whenever the notation makes sense. Together with the formula that I urge to memorize:\[

\boxed{\dis\int_a^bf(x)\,dx=\int_{a+c}^{b+c}f(x-c)\,dx}

\] you will be able to treat definite integral flexibly.

Problem 4 (Further Discussion). Let's see an interesting application of the result in part (b). Recall that we have shown that for positive integer $n$ and $T$-periodic function $f:\R\to \R$, \[

\int_0^{nT}e^{-x}f(x)\,dx=\frac{1-e^{-nT}}{1-e^{-T}} \int_0^T e^{-x} f(x)\,dx.

\] Later in tutorial we will know the following limit \[

 \lim_{t\to \infty} \int_0^{t} e^{-x}f(x)\,dx

\] exists when $f$ is nice enough (broad class of functions make the above limit exists!). Conventionally we denote this limit by $\int_0^\infty e^{-x}f(x)\,dx$, moreover, \begin{align}

 \nonumber \int_0^\infty e^{-x}f(x)\,dx&= \limn \int_0^{nT} e^{-x}f(x)\,dx\\
\nonumber& = \limn \frac{1-e^{-nT}}{1-e^{-T}} \int_0^T e^{-x} f(x)\,dx \\
\label{disappear}&= \frac{1}{1-e^{-T}} \int_0^T e^{-x} f(x)\,dx.

\end{align} To explain (\ref{disappear}), note that by period we usually mean a positive constant, hence $T>0$, so $e^T>1$, and hence $0<e^{-T}<1$, thus the result follows.

Let's construct an artificial example, let $f(x)=|\sin x|$, then $f$ is $\pi$-periodic, hence \[

\int_0^\infty e^{-x}|\sin x|\,dx= \frac{1}{1-e^{-\pi}} \int_0^\pi e^{-x} |\sin x|\,dx=\frac{1+e^{\pi}}{2(e^\pi-1)} \approx 0.545166.

\] Note that unlike $\int e^{-x}\sin x\,dx$, $\int e^{-x}|\sin x|\,dx$ cannot be computed in general.

Problem 5. We first consider $n\ge 0$. Recall that $I_n=\dis \int_{-\pi/2}^{\pi/2}\frac{\sin nx}{(2^x+1)\sin x}\,dx$, thus by (\ref{multiple}) with $c=-1$, one has (recall that $\sin (-x)=-\sin x$) \begin{align*}

 I_n&=\int_{\pi/2}^{-\pi/2} \frac{\sin nx}{(2^{-x}+1)\sin x}(-1)\,dx\\
&=\int_{-\pi/2}^{\pi/2}\frac{{\color{blue}2^{\color{blue}x}}\sin n x}{{\color{blue}2^{\color{blue} x}}(2^{-x}+1)\sin x}\,dx\\
&=\int_{-\pi/2}^{\pi/2}\frac{(1+2^x-1)\sin nx}{(1+2^x)\sin x}\,dx\\
&=\int_{-\pi/2}^{\pi/2}\brac{\frac{\sin nx}{\sin x} - \frac{\sin nx}{(1+2^x)\sin x}}\,dx\\
&= \int_{-\pi/2}^{\pi/2}\frac{\sin nx}{\sin x}\,dx - I_n\\
2I_n&=\int_{-\pi/2}^{\pi/2}\frac{\sin nx}{\sin x}\,dx\\
I_n&=\frac{1}{2} \int_{-\pi/2}^{\pi/2}\frac{\sin nx}{\sin x}\,dx\\
&= \int_{0}^{\pi/2}\frac{\sin nx}{\sin x}\,dx,

\end{align*} the last equality follows from the property that $\int_{-a}^af(x)\,dx=2\int_0^af(x)\,dx$ if $f$ is even. This equality is a great improvement, we have got rid of the term $2^x+1$ in the denominator. From the last equality we also have two immediate consequence:\[

I_0 = 0,\quad I_1=\int_0^{\pi/2}\,dx=\pi/2.

\] This is somewhat surprising, we have shown that $\dis I_1 =\int_{-\pi/2}^{\pi/2} \frac{dx}{2^x+1} = \frac{\pi}{2}$, this suggests that the value of the integral only depends on the domain. In fact with the same proof we can show a little more: Let $a\in \R$ and $b\ge 0$, then one always have \[

\int_{-a}^a \frac{dx}{b^x+1}=a,

\] I leave the proof of this interesting fact to you.

Since the case that $n=0$ and $n=1$ are done, let's suppose that $n\ge 2$. Recall the sum to product formula:\[

\sin x-\sin y= 2\cos \frac{x+y}{2}\sin \frac{x-y}{2},

\] hence we have \[

\sin nx-\sin (n-2)x = 2\cos (n-1)x\sin x,

\] thus \begin{align}

\nonumber I_n-I_{n-2}&=\int_0^{\pi/2}\frac{\sin nx-\sin(n-2)x}{\sin x}\,dx\\
\nonumber &=\int_0^{\pi/2}2\cos(n-1)x\,dx\\
\label{need this}&=-\frac{2}{n-1}\cos \frac{n\pi}{2}.

\end{align} Hence when $n$ is odd, from (\ref{need this}) we have $I_{n}=I_{n-2}$, since $n-2$ is still odd, we have \[

I_n=I_{n-2}=I_{n-4}=\cdots=I_1=\frac{\pi}{2}.

\]
When $n$ is even, say $n=2p$ for some $p\ge 1$, by (\ref{need this}) again we have for every integer $k$, \[

 I_{2k}-I_{2k-2}=-\frac{2}{2k-1}\cos\frac{2k\pi}{2}=-\frac{2}{2k-1}(-1)^k.

\] We take summation $\sum_{k=1}^p$ on both sides to get \[

I_n=I_{2p}-I_0=-\sum_{k=1}^p(-1)^k\frac{2}{2k-1} = \sum_{k=1}^{n/2}(-1)^{k+1}\frac{1}{k-1/2}.

\]
Here we have used the telescoping technique to compute the summation. For a sequence of numbers $a_0,a_1,a_2,\dots,a_n$, we have \begin{align*}

\sum_{k=1}^n (a_{k}-a_{k-1}) = &a_n-a_{n-1}+\\
&a_{n-1}-a_{n-2}+\\
&a_{n-2}-a_{n-3}+\\
&a_{n-3}-a_{n-4}+\\
&\qquad \vdots\\
&a_1-a_0\\
=& a_n-a_0.

\end{align*} In our case we have chosen $a_k=I_{2k}$.


When $n<0$, since \[

I_n= \int_{-\pi/2}^{\pi/2}\frac{\sin nx}{(2^x+1)\sin x}\,dx =- \int_{-\pi/2}^{\pi/2}\frac{\sin (-nx)}{(2^x+1)\sin x}\,dx = -I_{-n}

\] and $-n>0$, we are done.