Integrating from $\int_0^\infty \frac{\sin(x)}x dx$ with year $10$ mathematics?

Question

I am currently a year $10$ student self studying a level further maths. One day, my friend sent me this funny photo, from the photo we can easily know that glass beer = $10$, burger = $5$, and cup beer = $2$, and the next part is the challenging part. From the integral question provided, we know that we need to do $$\int_{0}^{\infty}\frac{10\sin x}{2x}dx$$

where it means $5\int_{0}^{\infty}\frac{\sin x}{x}dx$ after simplifying, but the problem is how to integrate it. After searching online, I realized that $\int_{0}^{\infty}\frac{\sin x}{x}dx$ is simply $\frac{\pi}{2}$, so the answer is $\frac{5\pi}{2}$. But why?! I am a person with a lot of curiosity haha and after searching online, I found out that lot of people use Laplace transforms or Feynman's Technique (by blackpenredpen which is a yt channel I like so much), but I don’t really understand those with my current knowledge. Then, I suddenly thought of a thing; can I do a Taylor series expansion, integrate it, and then find out where it converges?

For example, $\sin x = \sum\limits_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$, when you divide it by $x$, it is $\sum\limits_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n+1)!}$, and after you integrate it, it is $\sum\limits_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)(2n+1)!}$. But what's next, here is where I am currently stucking at. I found out that this formula is actually a bit like the taylor series of $\arctan $, just need to remove the $(2n+1)!$ and they are the same. When you $\lim_{x\to \infty}$, their answers are both $\frac{\pi}{2}$. Does anyone here know how to evaluate and where does $$\lim_{x\to \infty}\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)(2n+1)!}$$ or $$\lim_{x\to \infty}x-\frac{x^{3}}{18}+\frac{x^{5}}{600}-\frac{x^{7}}{35280} \cdot\cdot\cdot $$ converge? If anybody here could help me out, I would really be so thankful because I have spent days on it and still can't find the answer. Also, at the end sorry for my bad English as I am not a native speaker so the language maybe a bit odd!

Answer 1

This problem is solvable with fairly elementary (high school level) methods by way of Laplace transform. Let $$F(s)=\int_0^\infty \mathrm e^{-st}\frac{\sin t}{t}\mathrm dt$$ You may notice, using the Leibniz rule, that $$F'(s)=-\int_0^\infty \mathrm e^{-st}\sin(t)\mathrm dt$$ This integral is solvable using integration by parts. You can in fact calculate explicitly $$F'(s)=\frac{-1}{1+s^2}$$ This is a first order differential equation and can be integrated easily. Then using the condition $\lim _{s\to \infty} F(s)=0$ you can determine $$F(s)=\operatorname{arccot}(s)$$ And in particular $$\int_0^\infty \frac{\sin t}{t}\mathrm dt= \lim_{s\to 0^+}F(s)=\lim_{s\to 0^+}\operatorname{arccot}(s)=\frac{\pi}{2}$$ I'll let you fill in the details of the calculations yourself.

Answer 2

I do not understand what "Year 10" means without context, but I think it is safe to assume that you are familiar with differentiating and integrating functions, taking limits, and evaluating infinite sums. If you're willing to accept the infinite Euler product below

$$\frac {\sin x}x=\prod\limits_{n=1}^{+\infty}\left(1-\frac {x^2}{\pi^2 n^2}\right)\tag1$$

Then the rest of this answer should be understandable for a university student. If you are not familiar with $(1)$, I have included an elementary proof at the bottom of this answer.

Taking the natural log of both sides of $(1)$ and differentiating gives an infinite sum for $\cot x$

$$\cot x=\frac 1x+\sum\limits_{n=1}^{+\infty}\left(\frac 1{x-\pi n}+\frac 1{x+\pi n}\right)\tag2$$

Using the relationship $\tan x=-\cot\left(x-\frac {\pi}2\right)$, then we also have

$$\tan x=-\sum\limits_{n=1}^{+\infty}\left(\frac 1{x-\pi (2n-1)/2}+\frac 1{x+\pi(2n-1)/2}\right)\tag3$$

Next, observe that we have the following trigonometric identity relating $\csc x$ with tangent and cotangent.

$$\csc x=\frac 12\tan\left(\frac x2\right)+\frac 12\cot\left(\frac x2\right)\tag4$$

The proof of that can be done with the double angle identity for $\csc x$. Using $(2)$ and $(3)$, we can get an infinite series expansion for $\csc x$ in $(4)$ as

$$\csc x=\frac 1x+\sum\limits_{n=1}^{+\infty}(-1)^n\left(\frac 1{x-\pi n}+\frac 1{x+\pi n}\right)\tag5$$

Now, onto the actual integral. Denoting the integral as $I$, split the integral up into an infinite sum of integrals with a width of $\pi/2$.

\begin{align*} I\equiv\int\limits_0^{+\infty}\frac {\sin x}x\,\mathrm dx & =\sum\limits_{n=0}^{+\infty}\int\limits_{\pi n/2}^{\pi(n+1)/2}\frac {\sin x}x\,\mathrm dx\\ & =\int\limits_0^{\pi/2}\frac {\sin x}x\,\mathrm dx+\sum\limits_{n=1}^{+\infty}\int\limits_{\pi n/2}^{\pi(n+1)/2}\frac {\sin x}x\,\mathrm dx \end{align*}

A re-indexing of the sum is needed because we will try to employ $(5)$ into our integral. Taking a closer look at our infinite sum in the equation above, split the sum up over its even and odd contributions and enforce the substitution $x=\pi n-t$ on the odd contributions and $x=\pi n+t$ on the even contributions.

\begin{align*} \sum\limits_{n=1}^{+\infty}\int\limits_{\pi n}^{\pi(n+1)/2}\frac {\sin x}x\,\mathrm dx & =\sum\limits_{n=1}^{+\infty}\int\limits_{\pi(2n-1)/2}^{\pi n}\frac {\sin x}x\,\mathrm dx+\sum\limits_{n=1}^{+\infty}\int\limits_{\pi n}^{\pi(2n+1)/2}\frac {\sin x}x\,\mathrm dx\\ & =\sum\limits_{n=1}^{+\infty}(-1)^n\int\limits_0^{\pi/2}\frac {\sin t}{t-\pi n}\,\mathrm dt+\sum\limits_{n=1}^{+\infty}\int\limits_0^{\pi/2}\frac {\sin t}{t+\pi n}\,\mathrm dt \end{align*}

Thus, our original integral has become

$$I=\int\limits_0^{\pi/2}\frac {\sin x}x\,\mathrm dx+\sum\limits_{n=1}^{+\infty}(-1)^n\int\limits_0^{\pi/2}\frac {\sin t}{t-\pi n}+\frac {\sin t}{t+\pi n}\,\mathrm dt$$

Switching the order of the sum and integral, then it can be easily seen that the infinite sum is what we derived in $(5)$.

\begin{align*} I & =\int\limits_0^{\pi/2}\sin t\left[\frac 1t+\sum\limits_{n=1}^{+\infty}(-1)^n\left(\frac 1{t-\pi n}+\frac 1{t+\pi n}\right)\right]\,\mathrm dt\\ & =\int\limits_0^{\pi/2}\mathrm dt\\ & =\frac {\pi}2 \end{align*}

Thus, proving

$$\int\limits_0^{+\infty}\frac {\sin x}x\,\mathrm dx\color{blue}{=\frac {\pi}2}$$

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Proof of Euler's Infinite Product

For completeness, here is an elementary proof of $(1)$ using the double angle identity and a couple of limits. Observe that

$$\sin x=2\sin\left(\frac x2\right)\cos\left(\frac x2\right)=2\sin\left(\frac x2\right)\sin\left(\frac {\pi}2+\frac x2\right)\tag6$$

Next, we can use the double angle identity to further expand $\sin\left(\frac x2\right)$ and $\sin\left(\frac {\pi}2+\frac x2\right)$ as

\begin{align*} & \sin\left(\frac x2\right)=2\sin\left(\frac x4\right)\sin\left(\frac {\pi}2+\frac x4\right)\tag7\\ & \sin\left(\frac {\pi}2+\frac x2\right)=2\sin\left(\frac {\pi}4+\frac x4\right)\sin\left(\frac {3\pi}4+\frac x4\right)\tag8 \end{align*}

Substituting $(7)$ and $(8)$ into $(6)$ gives us

$$\sin x=2^3\sin\left(\frac x4\right)\sin\left(\frac {\pi+x}4\right)\sin\left(\frac {2\pi+x}4\right)\sin\left(\frac {3\pi+x}4\right)\tag9$$

Continuing this expansion indefinitely to $n$, we get the general formula written in the product notation.

$$\sin x=2^{n-1}\prod\limits_{k=1}^n\sin\left(\frac {\pi(k-1)+x}n\right)\tag{10}$$

A simplification can be made by exploiting the identity $\sin(\pi-x)=\sin x$. For example, the last factor in $(10)$ becomes

$$\sin\left(\frac {\pi(n-1)+x}n\right)=\sin\left(\frac {\pi-x}n\right)$$

And the penultimate factor, similarly, becomes

$$\sin\left(\frac {\pi(n-2)+x}n\right)=\sin\left(\frac {2\pi-x}n\right)$$

The plan is to use the product-to-sum formula on the "complementary" terms. That is, apply the sum-to-product formula on

$$\sin\left(\frac {\pi+x}n\right)\sin\left(\frac {\pi-x}n\right)=\sin^2\left(\frac {\pi}n\right)-\sin^2\left(\frac xn\right)$$

And so on with the rest of the terms. We stop applying the product-to-sum formula at the half-way point, when $n\mapsto n/2$. The middle term can be rewritten as

$$\sin\left(\frac {\pi n/2+x}n\right)=\cos\left(\frac xn\right)$$

Putting everything together, then our product becomes

$$\sin x=2^{n-1}\sin\left(\frac xn\right)\cos\left(\frac xn\right)\prod\limits_{k=1}^{n/2-1}\left[\sin^2\left(\frac {k\pi}n\right)-\sin^2\left(\frac xn\right)\right]\tag{11}$$

We are now one-half of the way complete. Equation $(11)$ is one of two equations needed to fully prove the expansion. The other can be derived off of equation $(11)$ by dividing both sides by $\sin\left(\frac xn\right)$ and taking the limit as $x\to0$.

$$n=2^{n-1}\prod\limits_{k=1}^{n/2-1}\sin^2\left(\frac {k\pi}n\right)\tag{12}$$

For the final step, we divide $(11)$ by $(12)$ and take the limit as $n$ gets infinitely large.

$$\sin x=n\sin\left(\frac xn\right)\cos\left(\frac xn\right)\prod\limits_{k=1}^{n/2-1}\left[1-\frac {\sin^2\left(\frac xn\right)}{\sin^2\left(\frac {k\pi}n\right)}\right]$$

Using the two limits below

\begin{align*} & \lim\limits_{n\to+\infty} n\sin\left(\frac xn\right)=x\\ & \lim\limits_{n\to+\infty}\frac {\sin^2\left(\frac xn\right)}{\sin^2\left(\frac {k\pi}n\right)}=\frac {x^2}{k^2\pi^2} \end{align*}

Then we prove the infinite product expansion

$$\sin x\color{blue}{=x\prod\limits_{n=1}^{+\infty}\left(1-\frac {x^2}{\pi^2 n^2}\right)}$$