Evaluate integral $ \int_{\mathbb{R} } \frac{\sin^4(\frac{t}{2}) }{t^2}$

Question

Let $f(x)= (1-2|x|)\chi_{[-\frac{1}{2}, \frac{1}{2}]} (x) $. Its Fourier transform is given by $ \hat{f} (x) = \frac{8\sin^2(\frac{t}{4})}{t^2} $. Based on this, I need to evaluate the integral $ \int_{\mathbb{R} } \frac{\sin^4(\frac{t}{2}) }{t^2}$, but I don't know where to start. I tried Parseval's identity - I wrote $\frac{\sin^4(\frac{t}{2}) }{t^2} = \frac{\sin^2(\frac{t}{2}) }{t^2} \cdot \sin^2(\frac{t}{2})$, and then I would need to find the function whose Fourier transform is $\sin^2(t)$. I don't know if this is right way (seems too complicated), so is there any simple way to evaluate integral? I would be very grateful for help.

Answer 1

Let's consider $$I(k,q)=\int_0^\infty\frac{\sin^k(x)}{x^q}dx$$ where $k$ is positive integer and $q$ is rational.

Using the formula $\frac{1}{x^q}=\frac{1}{\Gamma(x)}\int_0^\infty{t}^qe^{-xq}dx$

we get $I(k,q)=\frac{1}{\Gamma(x)}\int_0^\infty{t}^qdt\int_0^\infty\sin^k(x){e}^{-xq}dx$, where $\Gamma(x)$ is Gamma-function.

Integrating by part and grouping the terms with the same power of $\sin(x)$ we get:

$$I(k,q)=\frac{k!}{\Gamma(q)}\int_0^{\infty}\frac{t^{q-2}}{(t^2+2^2)(t^2+4^2)...(t^2+k^2)}dt, k=2n, q>1, n=1, 2,3...$$

$$I(k,q)=\frac{k!}{\Gamma(q)}\int_0^{\infty}\frac{t^{q-1}}{(t^2+1^2)(t^2+3^2)...(t^2+k^2)}dt, k=2n-1, q>0, n=1, 2,3...$$

To evaluate the integrals let's consider the following contour: along the cut from $0$ to ${\infty}$ (along the axis), big circle of radius $R$ (counter clockwise), along the other bank of the cut back to $0$ and along a small circle $r$ around $0$ (around a brunch point of $t^q$ - $q$ is rational) to the starting point.

Making a full round, on the lower bank of the cut the integrand gets additional factor $e^{2\pi{iq}}$. Integral along this closed contour is equal to the sum of residuals in simple poles, and integrals along big and small circle set to zero as $R\to\infty$ and $r\to0$: $$I(k,q)(1-e^{2\pi{iq}})=2\pi{i}\frac{k!}{\Gamma(q)}\sum{Res}\frac{t^{q-2}}{(t^2+2^2)(t^2+4^2)...(t^2+k^2)}, k=2n$$ $$I(k,q)(1-e^{2\pi{iq}})=2\pi{i}\frac{k!}{\Gamma(q)}\sum{Res}\frac{t^{q-1}}{(t^2+1^2)(t^2+3^2)...(t^2+k^2)}, k=2n-1$$ Finally, for even ($k=2n$), $q>1$ ($q$ - rational, not necessarily integer); $a=2, b=4, ...=k$ $$I(k,q)=-\frac{\pi{k!}}{2\Gamma(q)\cos\frac{\pi{q}}{2}}\left(\frac{a^{q-3}}{(b^2-a^2)(c^2-a^2)..(k^2-a^2)}+\frac{b^{q-3}}{(a^2-b^2)(c^2-b^2)..(k^2-b^2)}+..\right)$$

... and for odd ($k=2n-1$), $q>0$, $a=1, b=3, ...=k$ $$I(k,q)=\frac{\pi{k!}}{2\Gamma(q)\sin\frac{\pi{q}}{2}}\left(\frac{a^{q-2}}{(b^2-a^2)(c^2-a^2)..(k^2-a^2)}+\frac{b^{q-2}}{(a^2-b^2)(c^2-b^2)..(k^2-b^2)}+..\right)$$

Now we can easily evaluate $\int_{\mathbb{R} } \frac{sin^4(\frac{t}{2})}{t^2}=I(k=4,q=2)=\int_0^\infty\frac{sin^4(t)}{t^2}dt$

$k$ is even and $k=4$, therefore we have only $a=2, b=4$. $$I(4,2)=-\frac{\pi{4!}}{2\Gamma(2)\cos\frac{\pi{2}}{2}}\left(\frac{2^{-1}}{(4^2-2^2)}+\frac{4^{-1}}{(2^2-4^2)}\right)=\frac{12\pi}{4^2-2^2}(\frac{1}{2}-\frac{1}{4})=\frac{\pi}{4}$$

Some additional comments:

1. While calculating residuals right phases to poles should be appointed: $i=e^{\frac{\pi{i}}{2}}$ and $-i=e^{\frac{3\pi{i}}{2}}$ (one quarter and three quarters of total turn counter clockwise correspondingly). $i^q=e^{\frac{\pi{iq}}{2}}$ and $(-i)^q=e^{\frac{3\pi{iq}}{2}}$
2. At some $q$ there are uncertainties (when $\cos\frac{\pi{q}}{2}$ or $\sin\frac{\pi{q}}{2}=0$ in the denominators) that need to be evaluated. For instance, $I(k=3,q=2)=\int_0^\infty\frac{\sin^3(x)}{x^2}dx$. We have $a=1, b=3$, we set $q=2+\epsilon$ and $$I(3,2)=\lim_{\epsilon\to0}\frac{6\pi}{2}\frac{1}{\sin\frac{\pi(2+\epsilon)}{2}}\left(\frac{1^\epsilon}{3^2-1^2}+\frac{3^\epsilon}{1^2-3^2}\right)=$$$$=-\lim_{\epsilon\to0}\frac{6}{\epsilon}\left(\frac{1^0}{3^2-1^2}+\frac{\epsilon\log(1)}{3^2-1^2}+\frac{3^0}{1^2-3^2}+\frac{\epsilon\log(3)}{1^2-3^2}\right)=\frac{3\log3}{4}$$
3. Any integral $I(k,q)=\int_0^\infty\frac{\sin^k(x)}{x^q}dx$ for not too big integer $k$ can easily be evaluated. Just few examples:

$I(k=4,q=4)=\int_0^\infty\frac{\sin^4(x)}{x^4}dx=-\frac{4!\pi2^2}{2^4\Gamma(4)}\left(\frac{1}{2^2-1^2}+\frac{2}{1^2-2^2}\right)=-\frac{\pi}{3}(1-2)=\frac{\pi}{3}$ $I(k=3,q=\frac{1}{2})=\int_0^\infty\frac{\sin^3(x)}{\sqrt{x}}dx=\frac{6\pi\sqrt{2}}{2\sqrt{\pi}}\left(\frac{1^-\frac{3}{2}}{3^2-1^2}+\frac{3^-\frac{3}{2}}{1^2-3^2}\right)=-\frac{\sqrt\pi}{4\sqrt6}(3\sqrt3-1)$ $I(k=5,q=\frac{3}{2})=\int_0^\infty\frac{\sin^5(x)}{x^\frac{3}{2}}dx=\frac{5!\pi}{2\Gamma(\frac{3}{2})\sin\frac{3\pi}{4}}\left(\frac{1}{\sqrt1(3^2-1^2)(5^2-1^2)}+\frac{1}{\sqrt3(1^2-3^2)(5^2-3^2)}+\frac{1}{\sqrt5(1^2-5^2)(3^2-5^2)}\right)=\sqrt{\frac{5\pi}{2}}\frac{2\sqrt5-\sqrt15+1}{24}$

etc...

Answer 2

First a simple substitution $$\int_{\Bbb{R}}\frac{\sin^4(t/2)}{t^2}\mathrm{d}t=\int_{0}^\infty \frac{\sin^4(s)}{s^2}\mathrm{d}s$$ There is a known formula $$\int_{0}^\infty\frac{\sin(x)^m}{x^n}\mathrm{d}x$$ $$=\frac{\pi^{1-p}(-1)^{\lfloor(m-n)/2\rfloor}}{2^{m-p}(n-1)!}\sum_{k=0}^{\lfloor m/2\rfloor-p}(-1)^k \mathrm{C}(m,k)(m-2k)^{n-1}\log(m-2k)^p$$ $p=\operatorname{mod}(m-n,2)$, which holds for $m,n\in\Bbb{N}$ and $m\geq n>p$. When using this expression we make the definition $0^0:=1$. In our case, $m=4,n=2,p=0$ so $$\int_{0}^\infty \frac{\sin^4(s)}{s^2}\mathrm{d}s=\frac{-\pi}{2^4}\sum_{k=0}^{2}(-1)^k \mathrm{C}(4,k)(4-2k)$$ $$=\frac{-\pi}{2^4}\cdot(-4)=\frac{\pi}{4}$$

Answer 3

I would guess it's a typo and there was supposed to be a $t^4$ in the denominator and you were supposed to use Plancherel's theorem. But you can do the original integral as follows:

Note that the integrand is even and that $\sin^2 (t/2) = {1 - \cos t \over 2}$. So your integral is $$2\int_0^{\infty} {(1 - \cos t)^2 \over 4t^2}\,dt$$ You can integrate this by parts, integrating ${1 \over 4t^2}$ and differentiating $(1 - \cos t)^2$ (being careful that the improper integrals converge). Your integral becomes $$2 \int_0^{\infty} {2(1 - \cos t)\sin t \over 4t}\,dt$$ Using $\sin 2t = 2\sin t \cos t$, this is $$\int_0^{\infty} {\sin t \over t}\,dt - {1 \over 2} \int_0^{\infty}{\sin 2t \over t}\,dt$$ Changing variables to $2t$ in the latter integral, this becomes $$\int_0^{\infty} {\sin t \over t}\,dt - {1 \over 2} \int_0^{\infty}{\sin t \over t}\,dt$$ $$= {\pi \over 2} - {\pi \over 4}$$ $$= {\pi \over 4}$$