3
$\begingroup$

I am studying something and encountered this:
" Let $R(\theta,T) = \int_{-T}^{T} \frac{(\sin \theta t)}{t}dt, S(T) = \int_0^T\frac{(\sin x)}{x}dx$, then for $\theta > 0$ and changing variables $t=x/\theta $ shows that

$R(\theta,T)=2\int_0^{T\theta}\frac{\sin x}{x}dx = 2S(T\theta)$ while for $\theta<0$, $R(\theta,T) = -R(|\theta|,T)$" which I don't understand.

If $\theta<0$, then $R(\theta,T)=2\int_{-T|\theta|}^{0}\frac{\sin x}{x}dx =2\int_0^{T|\theta|}\frac{\sin x}{x}dx = R(|\theta|,T)$ as $\frac{\sin x}{x}$ is an even function, right? I am missing something simple here, thanks and appreciate an explanation.

$\endgroup$
4
$\begingroup$

When $t=-T$ we get $x=t\theta =-T\theta =T|\theta|$ and not $-T|\theta|$.

$\endgroup$
  • 1
    $\begingroup$ Correct! Thanks. $\endgroup$ – manifolded Feb 27 at 8:39
3
$\begingroup$

$$\begin{align}\theta <0 & \Rightarrow R(\theta ,T)=\int_{-T}^T\frac{\sin (\theta t)}{t}dt\\ &\Rightarrow R(\theta ,T)=\int_{-T}^T\frac{\sin (-|\theta| t)}{t}dt\\ &\Rightarrow R(\theta ,T)=-\int_{-T}^T\frac{\sin (|\theta| t)}{t}dt \ \ \ [\because \sin(-x)=-\sin (x)\big]\\ &\Rightarrow R(\theta ,T)=-R(|\theta|,T)\\ \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.