Sine Integral mixed with 'e'
Sine Integral mixed with 'e'
Show that:
$$\int_{0}^{\infty}\left(\frac{\sin(x)}{x}\right)^{2}e^{-2ax}dx=a\log\left(\frac{a}{\sqrt{1+a^{2}}}\right)+\cot^{-1}(a)$$
$$\int_{0}^{\infty}\left(\frac{\sin(x)}{x}\right)^{2}e^{-2ax}dx=a\log\left(\frac{a}{\sqrt{1+a^{2}}}\right)+\cot^{-1}(a)$$
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Re: Sine Integral mixed with 'e'
Hello C,galactus wrote:Show that:
$$\int_{0}^{\infty}\left(\frac{\sin(x)}{x}\right)^{2}e^{-2ax}dx=a\log\left(\frac{a}{\sqrt{1+a^{2}}}\right)+\cot^{-1}(a)$$
I think I have this. For $a>0$ define the function:
$$f(a)= \int_{0}^{\infty}e^{-2ax}\left ( \frac{\sin x}{x} \right )^2 \, {\rm d}x$$
and differentiate with respect to $a$ twice. Hence:
$$f''(a)=4\int_{0}^{\infty}e^{-2ax}\sin^2 x \, {\rm d}x = \frac{4}{4a^3+4a}= \frac{1}{a^3+a}$$
where the laplace trasform of $\sin^2 x$ is quite known and one can find it in tables.
Integrate back twice (don't forget the contants while doing that), hence:
$$f(a)=\int_{0}^{\infty}e^{-2ax} \left ( \frac{\sin x}{x} \right )^2 \, {\rm d}x= a \log a - \frac{1}{2}a \log \left ( a^2+1 \right )+\arccot a$$
Edit: ($17/01/2016$, $18:23$ pm) Fixed an error with the $\arctan$. Indeed it is $\arccot$.
Imagination is much more important than knowledge.
Re: Sine Integral mixed with 'e'
Wow T! That's neat.
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Re: Sine Integral mixed with 'e'
Actually using the above result one can easily deduce the very known result:galactus wrote:Show that:
$$\int_{0}^{\infty}\left(\frac{\sin(x)}{x}\right)^{2}e^{-2ax}dx=a\log\left(\frac{a}{\sqrt{1+a^{2}}}\right)+\cot^{-1}(a)$$
$$\int_0^{\infty} \frac{\sin^2 x}{x^2}\, {\rm d}x =\frac{\pi}{2}$$
Also, the identity
$$\int_0^{\infty} e^{-ax} \frac{\sin x}{x}\, {\rm d}x =\arccot a \tag{1} \label{*}$$
is derived by the same way, just by differentiating with respect to $a$ once. Using \eqref{*} we also deduce that:
$$\int_0^{\infty} \frac{\sin x}{x}\, {\rm d}x = \frac{\pi}{2}$$
but we have to be very careful since the last integral does not converge absolutely.
Imagination is much more important than knowledge.
Re: Sine Integral mixed with 'e'
I suppose I could add the contour way of dealing with it (although it's not so elegant as T's solution).
Integrating by parts we have:
\begin{align*}I&=\int_0^{\infty} \frac{\sin^2 x}{x^2}e^{-2ax}\,dx \\&= \int_0^{\infty} \frac{1}{x}\left(e^{-2ax}\sin 2x - ae^{-2ax}(1-\cos 2x)\right)\,dx\\&= \int_0^{\infty}\frac{\sin 2x}{x}e^{-2ax}\,dx - \frac{a}{2}\int_0^{\infty} \frac{2e^{-2ax} - e^{-2(a+i)x} - e^{-2(a-i)x}}{x}\,dx\\&= \int_0^{\infty}\frac{\sin (a^{-1}x)}{x}e^{-x}\,dx - \frac{a}{2}\int_0^{\infty}\frac{2e^{-x} - e^{-(1+ia^{-1})x} - e^{-(1-ia^{-1})x}}{x}\,dx\end{align*}
The integrals of the form $\displaystyle \int_0^{\infty} \frac{e^{-x} - e^{-(1+ik)x}}{x}\,dx, \text{ for } k \in \mathbb{R}$, can be evaluated by integrating the function $f(z) = \dfrac{e^{-z}}{z}$ on a closed contour, $\gamma := A = r,B = R, C = R(1+ik), D = r(1+ik)$ for real $r,R > 0$. Since, $f(z)$ has no poles in $\gamma$, we have:
$$0 = \int_{\gamma} \frac{e^{-z}}{z}\,dz = \int_r^{R} \frac{e^{-t}}{t}\,dt + \int_{BC} \frac{e^{-z}}{z}\,dz +\int_{R}^{r} \frac{e^{-(1+ik)t}}{t}\,dz +\int_{DA} \frac{e^{-z}}{z}\,dz$$
i.e., letting $r \to 0$ and $R \to \infty$,
\begin{align*}\int_0^{\infty} \frac{e^{-x} - e^{-(1+ik)x}}{x}\,dx &= \left.\operatorname{Log} (z)\right|_{z = r}^{z = r(1+ik)}+\lim_{r \to 0}\int_{AD}\frac{e^{-z}-1}{z}\,dz\\&= \log (\sqrt{1+k^2})+i\arctan{k}\end{align*}
Thus, $$\int_0^{\infty}\frac{2e^{-x} - e^{-(1+ia^{-1})x} - e^{-(1-ia^{-1})x}}{x}\,dx = 2\log \left(\sqrt{1+a^{-2}}\right)$$
and, $$\int_0^{\infty}\frac{\sin (a^{-1}x)}{x}e^{-x}\,dx = \frac{1}{2i}\int_0^{\infty} \frac{e^{-(1-a^{-1}i)} - e^{-(1+a^{-1}i)}}{x}\,dx = \arctan \frac{1}{a}$$
Hence, $$I = a\log \left(\frac{a}{\sqrt{1+a^2}}\right)+\arctan \frac{1}{a}$$
Integrating by parts we have:
\begin{align*}I&=\int_0^{\infty} \frac{\sin^2 x}{x^2}e^{-2ax}\,dx \\&= \int_0^{\infty} \frac{1}{x}\left(e^{-2ax}\sin 2x - ae^{-2ax}(1-\cos 2x)\right)\,dx\\&= \int_0^{\infty}\frac{\sin 2x}{x}e^{-2ax}\,dx - \frac{a}{2}\int_0^{\infty} \frac{2e^{-2ax} - e^{-2(a+i)x} - e^{-2(a-i)x}}{x}\,dx\\&= \int_0^{\infty}\frac{\sin (a^{-1}x)}{x}e^{-x}\,dx - \frac{a}{2}\int_0^{\infty}\frac{2e^{-x} - e^{-(1+ia^{-1})x} - e^{-(1-ia^{-1})x}}{x}\,dx\end{align*}
The integrals of the form $\displaystyle \int_0^{\infty} \frac{e^{-x} - e^{-(1+ik)x}}{x}\,dx, \text{ for } k \in \mathbb{R}$, can be evaluated by integrating the function $f(z) = \dfrac{e^{-z}}{z}$ on a closed contour, $\gamma := A = r,B = R, C = R(1+ik), D = r(1+ik)$ for real $r,R > 0$. Since, $f(z)$ has no poles in $\gamma$, we have:
$$0 = \int_{\gamma} \frac{e^{-z}}{z}\,dz = \int_r^{R} \frac{e^{-t}}{t}\,dt + \int_{BC} \frac{e^{-z}}{z}\,dz +\int_{R}^{r} \frac{e^{-(1+ik)t}}{t}\,dz +\int_{DA} \frac{e^{-z}}{z}\,dz$$
i.e., letting $r \to 0$ and $R \to \infty$,
\begin{align*}\int_0^{\infty} \frac{e^{-x} - e^{-(1+ik)x}}{x}\,dx &= \left.\operatorname{Log} (z)\right|_{z = r}^{z = r(1+ik)}+\lim_{r \to 0}\int_{AD}\frac{e^{-z}-1}{z}\,dz\\&= \log (\sqrt{1+k^2})+i\arctan{k}\end{align*}
Thus, $$\int_0^{\infty}\frac{2e^{-x} - e^{-(1+ia^{-1})x} - e^{-(1-ia^{-1})x}}{x}\,dx = 2\log \left(\sqrt{1+a^{-2}}\right)$$
and, $$\int_0^{\infty}\frac{\sin (a^{-1}x)}{x}e^{-x}\,dx = \frac{1}{2i}\int_0^{\infty} \frac{e^{-(1-a^{-1}i)} - e^{-(1+a^{-1}i)}}{x}\,dx = \arctan \frac{1}{a}$$
Hence, $$I = a\log \left(\frac{a}{\sqrt{1+a^2}}\right)+\arctan \frac{1}{a}$$
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Re: Sine Integral mixed with 'e'
Nicely done RD. Of course the result found by RD is equivelant to that of the two previous solutions. Simply we recall the fact that:r9m wrote: $$I = a\log \left(\frac{a}{\sqrt{1+a^2}}\right)+\arctan \frac{1}{a}$$
$$\arctan \frac{1}{a}= \arccot a$$
and that's all.
Imagination is much more important than knowledge.
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 26 guests