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An electrical engineering friend in an introductory signals and systems class asked me for advice on calculating

$$\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T(\sin(t+1)-1)^4\,dt$$

by hand. I don't see a better way to do this than writing $\sin(t+1)$ using complex exponentials and then doing the binomial expansion. Is there a slicker approach?

symplectomorphic
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3 Answers3

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Note that the integrand is periodic, hence:

$$I=\lim_{T\to\infty}\frac1{2T}\int_{-T}^T(\sin(t+1)-1)^4\ dt=\frac1{2\pi}\int_0^{2\pi}(\sin(t+1)-1)^4\ dt\\I=\frac1\pi\int_0^\pi(\cos(t)+1)^4\ dt$$

Binomial expanding and Chebyshev polynomials of the first kind:

$$\begin{align}8(c(t)+1)^4&=8c^4(t)+32c^3(t)+48c^2(t)+32c(t)+8\\&=c(4t)+8c(3t)+28c(2t)+56c(t)+35\end{align}$$

Thus,

$$\begin{align}8\pi I&=\int_0^\pi c(4t)+8c(3t)+28c(2t)+56c(t)+35\ dt\\&=\frac14s(4t)+\frac83s(3t)+14s(2t)+56s(t)+35t\bigg|_{t=0}^\pi\\&=35\pi\end{align}$$

Thus,

$$I=\frac{35}8$$

Simply Beautiful Art
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  • We might be able to answer this by letting $t \to 2t$ and then using the power reduction formula for $\cos^2(x)$ in reverse. I'll check out this route later. Should remove the need for binomial expansion if done right. Regardless, good answer: a +1 from me – Brevan Ellefsen Feb 19 '17 at 02:33
  • Thanks, this is slick. I'm having a bit of trouble, though, proving rigorously that we can replace the original problem with the average value over a period. Here is the obvious first step. Consider $F(T):=\frac{1}{2T}\int_{-T}^Tf(x),dx$ where $f$ is periodic with period $P$. Setting $T=nP/2$ for $n\in\mathbb{N}$ gives $F(nP/2)=\frac{1}{nP}\cdot n\int_{-P/2}^{P/2}f(x),dx=\frac{1}{P}\int_{-P/2}^{P/2}f$. So as $n\to\infty$, $F(nP/2)$ converges to (in fact, is always equal to) the average over a period. But in general $g(x_n)\to L$ as $x_n\to\infty$ does not imply $g(x)\to L$ as $x\to\infty$. – symplectomorphic Feb 19 '17 at 23:28
  • (Continued) How do we repair or finish this argument to get the conclusion? – symplectomorphic Feb 19 '17 at 23:29
  • @symplectomorphic Hint:$$\int_{-T}^Tf(x)\ dx=\int_{2\pi-T}^Tf(x)\ dx+\underbrace{\int_T^{2\pi}f(x)\ dx}_{Bounded?}$$ – Simply Beautiful Art Feb 19 '17 at 23:39
  • @symplectomorphic Er, the right most integral should have $T+2\pi$ as the upper bound. – Simply Beautiful Art Feb 20 '17 at 00:00
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It is clear we can transform our integral into $$I=\frac1\pi\int_0^\pi(\cos(t)+1)^4\ dt$$ See @SimplyBeautifulArt's answer for a proof. Once we are here we let $x \to 2x$ $$I=\frac2\pi\int_0^{\pi/2}(\cos(2x)+1)^4\ dx$$ $$I=\frac{2^5}{\pi}\int_0^{\pi/2}\cos^8(x)\ dx$$ By symmetry this is simply $$I=\frac{2^3}{\pi}\int_0^{2\pi}\cos^8(x)\ dx$$ We now have a fairly standard integral, for which many clever solutions can be found. A fairly general one is to apply the common reduction formulas over repeatedly to find that $$\int_0^{2\pi}\cos^n(x) dx = \frac{2\pi}{2^n}{n \choose n/2}$$ Applying this closed form here, we find that $$I=\frac{1}{2^4}{8 \choose 4}= \frac{70}{16}=\frac{35}{8}$$ Of course, we could always use the fact that $\cos^2(x) = 1-\sin^2(x)$ to reduce the power of our integral, or substitute $u=\cos(x)$ or $u=\cos^8(x)$, or any other number of tricks to solve without resorting to the general closed form. I only use the general form because it is relatively well known. See here for a few proofs.

Community
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Brevan Ellefsen
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take the area of $ (sin(t)+1)^4$, over $2 \pi$, once you expand the $ (sin(t)+1)^4$, sine terms to the odd power give zero area. so you have to find the area of $sin^4x + 6 Sin^2x+1$. It comes to $70\pi/8$. dividing it by $2\pi$, gives $35/8$.

jnyan
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