$\int_0^1 \frac{1}{x}\sin\frac{1}{x^3}\,dx=\frac{\pi}{6}-\frac{1}{3}\int_0^1 \frac{\sin t}{t}\,dt$

Question

I am trying to prove

$$\int_0^1 \frac{1}{x}\sin\frac{1}{x^3}\,dx=\frac{\pi}{6}-\frac{1}{3}\int_0^1 \frac{\sin t}{t}\,dt$$

This result is from slide 4 of Tim Sullivan's presentation here.

I have tried integrating by parts, both with $u=\sin\frac{1}{x^3}$ and $v'=\frac{1}{x}$, and with $u=-\frac{1}{3}x$ and $v'=-\frac{3}{x^2}\sin\frac{1}{x^3}$, but in both cases I do not get anything close to the right-hand side (since I end up with terms involving $\log$ or $\cos$).

Does anyone have any suggestions for how to prove this?

Answer 1

Substitute $u=\frac{1}{x^3}$.

From there, $du=-\frac{3}{x^4}dx = -3u \frac{1}{x} dx$

Hence, $\frac{1}{3u}du = \frac{1}{x}dx$

$$\int_0^1 \frac{1}{x} \sin{\frac{1}{x^3}}dx = -\int_{\infty}^1\frac{\sin{u}}{3u}du = \frac{1}{3}\int_1^{\infty} \frac{\sin u}{u}du$$

Now $$\frac{\pi}{6}=\frac{1}{3}\int_0^{\infty} \frac{\sin u}{u}du=\frac{1}{3}\int_1^{\infty} \frac{\sin u}{u}du+\frac{1}{3}\int_0^1 \frac{\sin u}{u}du$$ where $\int_0^{\infty} \frac{\sin u}{u}du=\pi/2$ is the Dirchlet Integral.

So $$\frac{1}{3}\int_1^{\infty} \frac{\sin u}{u}du=\frac{\pi}{6}-\frac{1}{3}\int_0^1 \frac{\sin u}{u}du$$

Answer 2

Making the substitution $t=\frac{1}{x^3}$ in the integral on the left-hand side we get: $$\int_{0}^{1} \frac{1}{x}\sin \left(\frac{1}{x^3}\right)dx=\frac{1}{3}\int_{1}^{\infty}\frac{\sin t}{t}dt=\frac{1}{3}\int_{1}^{0}\frac{\sin t}{t}dt+\frac{1}{3}\int_{0}^{\infty}\frac{\sin t}{t}dt \rightarrow (1)$$ The first integral on the right-hand side of $(1)$ appears in the equation to be proven, so I'll leave it as it is. While for the second integral, Lobachevsky's famous integral formula, I'll use some interesting properties of the Laplace transform, of which I've talked in two of my previous answers. It can be defined as follows:- $$ L\{f(x)\}=\int_{0}^\infty e^{-px}f(x)dx = F(p)$$ Here we are transforming the function $f$ with domain $x$ to a function $F$ with domain $p$ with a unilateral integral transform of kernel $e^{-px}$. Now consider the general Laplace transform formula given above. Differentiating both sides with respect to $p$ we get: $$F'(p)=\int_{0}^\infty e^{-px}(-x)f(x)dx=-L\{xf(x)\} \rightarrow (2)$$ Now put $G(p)$ as Laplace transform of $f(x)/x$ and obtain its differentiation using equation $(2)$:- $$G(p)=L\left\{\frac{f(x)}{x}\right\}\Rightarrow G'(p)=-L\{f(x)\}=-F(p)\rightarrow (3)$$ Using the Fundamental Theorem of Calculus (relationship between derivative and integral) for $(3)$: $$G(p)=-\int_{a}^p F(p)dp \Rightarrow \int_{0}^\infty e^{-px}\frac{f(x)}{x}dx=-\int_{a}^p F(p)dp \rightarrow (4)$$ Note that $a$ here is some constant. If $G(p) \rightarrow 0$ as $p \rightarrow \infty$ then we put $a = \infty$ and obtain the following:- $$\int_{0}^\infty e^{-px}\frac{f(x)}{x}dx=\int_{p}^\infty F(p)dp \rightarrow (5)$$ If we let $p \rightarrow 0$ on both sides of equation $(5)$ we get the following: $$\int_{0}^\infty \frac{f(x)}{x}dx=\int_{0}^\infty F(p)dp \rightarrow (6)$$ This is useful for us for finding the improper integral of various functions of the form $f(x)/x$ where the transform $F(p)$ is known. Now it has become easy to compute the second integral on the right-hand side of equation $(1)$: $$\int_{0}^{\infty}\frac{\sin t}{t}dt=\int_{0}^{\infty}\frac{\sin x}{x}dx = \int_{0}^{\infty}L\{\sin x\}dp \rightarrow (7)$$ We know the transform of $\sin x$ as follows (which can be proved using integration by parts): $$L\{\sin x\}=\frac{1}{p^2+1} (p>0) \rightarrow (8)$$ Using equation $(7)$ in $(8)$ we get that: $$\int_{0}^{\infty}\frac{\sin t}{t}dt=\int_{0}^{\infty}\frac{1}{p^2+1}dp=\left[\tan^{-1} p\right]_{0}^{\infty}=\frac{\pi}{2} \Rightarrow \frac{1}{3}\int_{0}^{\infty}\frac{\sin t}{t}dt=\frac{\pi}{6} \rightarrow (9)$$ Therefore substituting (9) in (1) we conclude: $$\int_{0}^{1} \frac{1}{x}\sin \left(\frac{1}{x^3}\right)dx= \frac{\pi}{6} - \frac{1}{3}\int_{0}^{1}\frac{\sin t}{t}dt$$