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Graduate POTW

dwsmith

Well-known member
Feb 1, 2012
1,673
So on the recent graduate problem of the week, I saw that $\int_0^{\infty}\frac{\sin x}{x}dx = \frac{\pi}{2}$, but so does, $\int_0^{\infty}\frac{\sin^2 x}{x^2}dx = \frac{\pi}{2}$.
How can they both be the same?
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
So on the recent graduate problem of the week, I saw that $\int_0^{\infty}\frac{\sin x}{x}dx = \frac{\pi}{2}$, but so does, $\int_0^{\infty}\frac{\sin^2 x}{x^2}dx = \frac{\pi}{2}$.
How can they both be the same?
Let us use integration by parts to compute $\displaystyle\int_0^{\infty}\frac{\sin^2 x}{x^2}\,dx$. At the end, we will need to use the fact that $\displaystyle\int_0^{\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}$

Let $u=\sin^2x$ and $\,dv=\dfrac{\,dx}{x^2}$. Then $\,du=2\sin x\cos x\,dx=\sin(2x)\,dx$ and $v=-\dfrac{1}{x}$. Therefore,
\[\int_0^{\infty}\frac{\sin^2 x}{x^2}\,dx = \left[-\frac{\sin^2 x}{x}\right]_0^{\infty}+\int_0^{\infty}\frac{\sin(2x)}{x}\,dx=\int_0^{\infty}\frac{\sin(2x)}{x}\,dx.\]
(We note that $|\sin x|\leq 1\implies |\sin^2 x|\leq 1$ and thus $\displaystyle\lim_{x\to\infty} \frac{\sin^2 x}{x}\sim \lim_{x\to\infty} \frac{1}{x}=0$; We also note that $\displaystyle\lim_{x\to 0}\frac{\sin^2 x}{x}=\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\sin x=0$. Hence, that's why the $\displaystyle\left[-\frac{\sin^2 x}{x}\right]_0^{\infty}$ term goes to zero.)

Now let $t=2x\implies\,dt=2\,dx$. Therefore,
\[\int_0^{\infty}\frac{\sin(2x)}{x}\,dx\xrightarrow{t=2x}{} \int_0^{\infty}\frac{\sin t}{t/2}\frac{\,dt}{2}=\int_0^{\infty}\frac{\sin t}{t}=\frac{\pi}{2}.\]
And thus, we also have that $\displaystyle\int_0^{\infty}\frac{\sin^2 x}{x^2}\,dx =\frac{\pi}{2}$.

I hope this makes sense!
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
$$F(a)=\int^{\infty}_0\frac{\sin^2(ax)}{x^2}$$

Differentiate w.r.t a :

$$F'(a)=\int^{\infty}_0 \frac{\sin(2ax)}{x}$$

Let 2ax=t

$$F'(a)=\int^{\infty}_0 \frac{\sin(t)}{t}=\frac{\pi}{2}$$

$$F(a)=\frac{\pi}{2}a+C$$

Putting a =0 we get C = 0 hence

$$\int^{\infty}_0\frac{\sin^2(ax)}{x^2}=\frac{\pi \cdot a}{2}$$

So for a =1 we get our result :

$$\int^{\infty}_0\frac{\sin^2(x)}{x^2}=\frac{\pi}{2}$$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
If your question is why such thing happen , then I don't know , to me it is pretty strange !

If you see the graph of both functions , then you have no indications ...
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I just thought it was strange. When I took Theory of Complex Variables, I had the $\int_0^{\infty}\frac{\sin^2x}{x^2}dx = \frac{\pi}{2}$ exercise so I was surprised to see that $\frac{\sin x}{x}$ lead to the same conclusion.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
In complex analysis $$\int^{\infty}_0 \dfrac{1-\cos(x)}{x^2}$$ and $$\int^{\infty}_0 \frac{\sin(x)}{x}$$ are conventional exercises to solve by contour integration ....