6
$\begingroup$

I am trying to solve the following integral

\begin{equation} {\cal I}=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\frac{2\tau}{1+\omega^{2}\tau^{2}}\frac{\sin^{2}\left[\left(\omega+\Omega\right)t/2\right]}{\left[\left(\omega+\Omega\right)/2\right]^{2}} \end{equation} for finite $\Omega$. I could only solve it for $\Omega=0$, yielding

\begin{equation} {\cal I}=2 \left[\tau^{2}\left(e^{-t/\tau}-1\right)+\tau \thinspace t\right] \end{equation}

I also tried to solve using the fact below

\begin{equation} {\cal I}=\int_{0}^{t}\int_{0}^{t}dt_{1}dt_{2} e^{-|t_1-t_2|/\tau}e^{i\Omega\left(t_{1}-t_{2}\right)}=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\frac{2\tau}{1+\omega^{2}\tau^{2}}\frac{\sin^{2}\left[\left(\omega+\Omega\right)t/2\right]}{\left[\left(\omega+\Omega\right)/2\right]^{2}} \end{equation} although this did not help much. Any thoughts?

$\endgroup$

3 Answers 3

12
$\begingroup$

Let $$I_0=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\frac{2\tau}{1+\omega^{2}\tau^{2}}\frac{\sin^{2}\left[\left(\omega+\Omega\right)t/2\right]}{\left[\left(\omega+\Omega\right)/2\right]^{2}}$$ After easy transformations we can write the integral in the form $$I_0=\frac{t^3}{2\pi\tau}\int_{-\infty}^{\infty}\frac{dx}{a^2+(x-b)^2}\frac{\sin^2x}{x^2}$$ where $a=\frac{t}{2\tau} , \,b=\frac{\Omega t}{2}$.

We can also rewrite the integral in the form $$I_0=\frac{t^3}{2\pi\tau}\int_{-\infty}^{\infty}\frac{dx}{a^2+(x-b)^2}\frac{1-\cos2x}{2\,x^2}=\frac{t^3}{2\pi\tau}\Re \,I$$ where $$I=\frac{1}{2}\int_{-\infty}^{\infty}\frac{dx}{(x-b+ia)(x-b-ia)}\frac{1-e^{2ix}}{x^2}$$ To evaluate $I$ we go in the complex plane and make a closed contour - adding a big half-circle with the radius $R\to\infty$ (counter-clockwise) in the upper half-plane, and a small half-circle with the radius $r\to0$ around $z=0$ (clockwise), also in the upper half-plane. $$\oint=I+I_R+I_r=2\pi i\sum Res $$ We have a single simple pole inside our contour (at $z=b+ia$). It is easy to show that $I_R\to0$ at $R\to\infty$, therefore $$I=2\pi i Res_{z=b+ia}\frac{1}{2}\frac{1}{(x-b+ia)(x-b-ia)}\frac{1-e^{2ix}}{x^2}-I_r$$ But $-I_r$ give us $\,\,\pi i \,Res_{z=0}\frac{1}{2}\frac{1}{(x-b+ia)(x-b-ia)}\frac{1-e^{2ix}}{x^2}$.

Evaluating residues, we get $$I=\frac{2\pi i}{2}\frac{1-e^{2ib-2a}}{(b+ia)^2\,2ia}+\frac{\pi i}{2}\frac{-2i}{(b-ia)(b+ia)}$$

and, after easy transformation $$I=\frac{\pi}{a^2+b^2}\Big(1+\frac{1}{2a}\big(e^{-2i\tan^{-1}\frac{a}{b}}-e^{-2a+2ib-2i\tan^{-1}\frac{a}{b}}\big)\Big)$$ Taking the real part, putting $a$ and $b$ and switching to $I_0$, we finally get the answer: $$\bbox[5px,border:2px solid #C0A000]{I_0=\frac{2\tau t}{1+\Omega^2\tau^2}\Big[1+\frac{\tau}{t}\Big(e^{-\frac{t}{\tau}}\cos(\Omega t+2\tan^{-1}\Omega \tau)-\cos(2\tan^{-1}\Omega \tau)\Big)\Big]}$$ It is interesting to check the answer in specific cases:

  1. $\tau=0\,\,\Rightarrow \frac{I_0}{2\tau}=t$

On the other hand, $$\frac{I_0}{2\tau}=\int_{-\infty}^\infty\frac {d\omega}{2\pi}\frac{\sin^2(\frac{\omega+\Omega}{2}t)}{(\frac{\omega+\Omega}{2})^2}=\frac{1}{\pi}\int_{-\infty}^\infty\frac{\sin^2(xt)}{x^2}dx=t$$

  1. $\Omega=0$ $$I_0 (\Omega=0)=2\tau t\Big[1+\frac{\tau}{t}\Big(e^{-\frac{t}{\tau}}-1\Big)\Big]$$ what is in agreement with your partial solution.
  2. $t=0\,\, \Rightarrow I_0=0$

Indeed, $$I_0(t\to0)\to \frac{2\tau^2}{1+\Omega^2\tau^2}\big(e^{-\frac{t}{\tau}}-1\big)\cos(2\tan^{-1}\Omega\tau)\to 0$$ 4. $\tau\to\infty$ $$I_0\to\frac{2}{\Omega^2}\Big(\cos(\Omega t+\pi)-\cos\pi\Big)=\frac{2}{\Omega^2}\Big(1-\cos(\Omega t)\Big)$$ On the other hand, using $\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}=\pi\delta(x) $ (the Dirac' delta-function) $$I_0(\tau\to\infty)=\int_{-\infty}^\infty\frac{d\omega}{\pi}\frac{1/\tau}{\omega^2+(1/\tau)^2}\frac{\sin^{2}\left[\left(\omega+\Omega\right)t/2\right]}{\left[\left(\omega+\Omega\right)/2\right]^2}\to\frac{4\sin^2(\Omega t/2)}{\Omega^2}=\frac{2}{\Omega^2}\Big(1-\cos(\Omega t)\Big)$$

$\endgroup$
3
$\begingroup$

$$I=\iint\limits_{[0,t]^2}dt_1\,dt_2\,e^{-|t_1-t_2|/\tau}e^{i\Omega(t_1-t_2)}$$ notice that: $$|t_1-t_2|=\begin{cases}(t_1-t_2)&t_1>t_2\\0&t_1=t_2\\-(t_1-t_2)&t_1<t_2\end{cases}$$ so we could split it into two parts to make it easier: $$0\le t_1\le t,0\le t_2\le t_1$$ $$0\le t_2\le t,0\le t_1\le t_2$$ leaving: $$\int\limits_0^tdt_1\int\limits_0^{t_1}dt_2\,e^{-(t_1-t_2)/\tau}e^{i\Omega(t_1-t_2)}+\int\limits_0^tdt_2\int\limits_0^{t_2}dt_1\,e^{(t_1-t_2)/\tau}e^{i\Omega(t_1-t_2)}\\ =\frac{{\tau}\mathrm{e}^{-\frac{t}{{\tau}}}\left(\left(\left(\mathrm{i}{\omega}^3t+{\omega}^2\right){\tau}^3+\left({\omega}^2t-2\mathrm{i}{\omega}\right){\tau}^2+\left(\mathrm{i}{\omega}t-1\right){\tau}+t\right)\mathrm{e}^\frac{t}{{\tau}}+\left(-\mathrm{i}{\omega}^2\sin\left({\omega}t\right)-{\omega}^2\cos\left({\omega}t\right)\right){\tau}^3+\left(2\mathrm{i}{\omega}\cos\left({\omega}t\right)-2{\omega}\sin\left({\omega}t\right)\right){\tau}^2+\left(\mathrm{i}\sin\left({\omega}t\right)+\cos\left({\omega}t\right)\right){\tau}\right)}{\left({\omega}^2{\tau}^2+1\right)^2} -\frac{{\tau}\mathrm{e}^{-\frac{t}{{\tau}}}\left(\left(\left(\mathrm{i}{\omega}^3t-{\omega}^2\right){\tau}^3+\left(-{\omega}^2t-2\mathrm{i}{\omega}\right){\tau}^2+\left(\mathrm{i}{\omega}t+1\right){\tau}-t\right)\mathrm{e}^\frac{t}{{\tau}}+\left({\omega}^2\cos\left({\omega}t\right)-\mathrm{i}{\omega}^2\sin\left({\omega}t\right)\right){\tau}^3+\left(2{\omega}\sin\left({\omega}t\right)+2\mathrm{i}{\omega}\cos\left({\omega}t\right)\right){\tau}^2+\left(\mathrm{i}\sin\left({\omega}t\right)-\cos\left({\omega}t\right)\right){\tau}\right)}{\left({\omega}^2{\tau}^2+1\right)^2}$$ (I swapped the $\Omega$ for $\omega$ here just for speed). Whilst this result is quite long, it can be simplified and is definately do-able.

$\endgroup$
1
  • $\begingroup$ Can you please simplify the result? $\endgroup$ Sep 24, 2021 at 1:43
3
$\begingroup$

After some minor simplification, we have for the antiderivative

$$I=\frac{4\tau}\pi\int \frac{\sin ^2\left(\frac{1}{2} t (\omega +\Omega )\right)}{\left(\tau ^2 \omega ^2+1\right) (\omega +\Omega )^2}\,d\omega$$ Using $x=\frac{t (\omega +\Omega )}2$ gives $$\int \frac{\sin ^2\left(\frac{1}{2} t (\omega +\Omega )\right)}{\left(\tau ^2 \omega ^2+1\right) (\omega +\Omega )^2}\,d\omega=\frac {t^3}{8\tau^2}\int \frac{\sin^2(x) }{x^2(x^2+ax+b) }\,dx$$ with $a=-t \Omega$ and $b=\frac{t^2 \left(\tau ^2 \Omega ^2+1\right)}{4 \tau ^2}$.

Let $r$ and $s$ be the complex roots of the quadratic and use partial fraction decomposition $$\frac 1{x^2(x-r)(x-s)}=\frac{r+s}{r^2 s^2 x}+\frac{1}{r^2 (r-s) (x-r)}-\frac{1}{s^2 (r-s) (x-s)}+\frac{1}{r s x^2}$$ which make that we face two kinds of integratels $$J=\int \frac{\sin^2(x) }{x^2} \,dx=\text{Si}(2 x)+\frac{\cos (2 x)-1}{2 x}$$

$$K_c=\int \frac{\sin^2(x) }{x-c} \,dx=-\frac{1}{2} (\cos (2 c) \text{Ci}(2 x-2 c)+\sin (2 c) \text{Si}(2 c-2 x)-\log (x-c))$$

$\endgroup$

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.