Evaluate $\int_0^\infty \frac{t^2 - \sin^2 t}{t^4} dt$.

The integrand is clearly positive as for all $t > 0, t > |\sin t|.$ I'm not sure if finding the Fourier series of some functions might be useful. Or maybe a Taylor expansion or telescoping sum might be useful? $\sin^2 t$ has period $\pi$, so we can split the integral into the sum $\sum_{k=0}^{\infty} \int_{k\pi}^{(k+1)\pi} \frac{t^2 - \sin^2 t}{t^4} dt.$ This sum, even if valid, doesn't seem very useful.

In hindsight, it seems the question was fairly straightforward, provided one can accept the fact that $\int_0^\infty (\sin x)/xdx = \pi/2.$ Many proofs are provided here for instance.

  • $\begingroup$ hm Let me just brute force this lol $\endgroup$ Commented Sep 24, 2022 at 2:18
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    $\begingroup$ Are you allowed to use Laplace Transform? Because if so, then using$$\int\limits_0^{+\infty} f(t)g(t)\,\mathrm dt=\int\limits_0^{+\infty}\mathcal{L}[f(t)]\mathcal{L}^{-1}[g(t)]\,\mathrm dt$$reduces the integral to$$\frac 43\int\limits_0^{+\infty}\frac {\mathrm dt}{t^2+4}$$ $\endgroup$
    – Frank W
    Commented Sep 24, 2022 at 2:31
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    $\begingroup$ @Max0815 I honestly have no idea, I watched a YouTube video that proved the result and have seen a couple of answers on AoPS and MSE using that identity. If the OP knows fourier transforms, then I presume they are also aware of Laplace transforms, granted this approach probably isn't the elementary way they were looking for $\endgroup$
    – Frank W
    Commented Sep 24, 2022 at 4:49
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    $\begingroup$ @Max0815 youtube.com/watch?v=bUlbBq7F9ng&t=187s $\endgroup$
    – Frank W
    Commented Sep 24, 2022 at 6:26
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    $\begingroup$ @Max0815 Indeed, that is a well-known property of the Laplace transform based on the convolution property of it. $\endgroup$
    – KStarGamer
    Commented Sep 24, 2022 at 12:06

3 Answers 3


Integrate by parts repeatedly to reduce the integral

\begin{align} \int_0^\infty \frac{t^2 - \sin^2 t}{t^4} dt =& \ -\frac13\int_0^\infty ({t^2 - \sin^2 t})\ d(\frac1{t^3})\\ \overset{ibp}=& \ -\frac16\int_0^\infty ({2t- \sin 2t})\ d(\frac1{t^2})\\ \overset{ibp} =& \ -\frac13 \int_0^\infty ({1- \cos 2t})\ d(\frac1{t})\\ \overset{ibp} =& \ \frac23\int_0^\infty \frac{ \sin 2t}t\ dt =\frac23 \cdot \frac \pi2=\frac\pi3 \end{align} where $\int_0^\infty \frac{ \sin 2t}tdt\overset{2t\to t}= \int_0^\infty \frac{ \sin t}tdt=\frac\pi2 $.


I'll brute force this. There are probably smarter ways though. Let's find the antiderivative. $$\int\frac{t^2}{t^4}\text{ d}t-\int \frac{\sin^2(t)}{t^4} \text{ d}t$$ $$\implies -\frac1t-\frac12\int (1-\cos(2t))\cdot\frac1{t^4} \text{ d}t$$ $$\implies -\frac1t+\frac1{6t^3}-\frac12\underbrace{\int\frac{\cos(2t)}{t^4} \text{ d}t}_A$$ Okay, let's solve $A$. We can integrate by parts, with $u$(derivative side) being the trig function and $v$(integral side) being the denominator. You can quickly calculate this using tabular, and we get $$A=-\frac1{3t^3}\cos(2t)+\frac{1}{3t^2}\sin(2t)+\frac2{3t}\cos(2t)+\frac43\operatorname{Si}(2t)$$ where Si is the sine integral.

Hence the full antiderivative is $$I(t)=-\frac1t+\frac1{6t^3} -\frac1{6t^3}\cos(2t)+\frac{1}{6t^2}\sin(2t)+\frac1{3t}\cos(2t)+\frac23\operatorname{Si}(2t)$$

We now need to take two limits, first to $0$, then to $\infty$.

$$\lim_{t\to0}I(t)=0$$ It's very very bashy so I'll use words to describe it in case you want to work this out (probably not).

We find this by splitting the Si out first, that goes to $0$. Write every other term in a common denominator, use le hospital rule. Now, you would take out several terms using addition rule, spam le hospital and you have the answer.

$$\lim_{t\to\infty}I(t)=\frac{\pi}3$$ For this one, completely split every term via addition rule. The only term that gives a value is the Si term, which spits out our limit value.

We now have shown your integral is $$\int^{\infty}_0\frac{t^2-\sin^2(t)}{t^4} \text{ d}t=\frac{\pi}3$$

this is a dumb method ig, there's probably much smarter ones like the lapalce one noted in the comments or some other stuff


One more method, using the residue theorem: since the integrand is even, the integral is$$\Re\int_{\Bbb R}f(z)dz,\,f(z):=\frac{2z^2-1+e^{2iz}-2iz+\frac43iz^3}{4z^4}.$$I've defined $f$ so as to have no poles; in particular, $f\in O(1)$ for small $z$. A semicircular contour on the upper half-plane integrates to $0$, so the linear part is $-1$ times the arc part, i.e.$$\Re\int_{\Bbb R}f(z)dz=-\Re\lim_{R\to\infty}\int_0^\pi f(Re^{i\theta})iRe^{i\theta}d\theta.$$The only nonvanishing contribution is from the cubic term in $f$'s numerator, so$$\Re\int_{\Bbb R}f(z)dz=-\Re\int_0^\pi\lim_{R\to\infty}\frac{\frac43i(Re^{i\theta})^3}{4(Re^{i\theta})^4}iRe^{i\theta}d\theta=\frac{\pi}{3}.$$This problem is a very instructive example of the residue theorem, as some readers may only be familiar with cases where the arc vanishes so the line is the contour, but here the contour vanishes so the line cancels a nonvanishing arc contribution.


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