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I tried to solve this integral and find its original value, unfortunately I did not succeed. I made some variable changes, but I think it has become more complicated. Numerical Solution: 1.15745 Other: 0.99 This solution My solution is as follows:

$$I=\int_0^{\infty}\frac{\sin(\tan(x))}{x}dx$$ $$I(a)=\int_0^{\infty}\frac{\sin(a\tan(x))}{x}dx$$ $$I'(a)=\int_0^{\infty}\frac{\tan(x)\cos(a\tan(x))}{x}dx$$ $$u=tan(x) \iff \arctan{u}=x \Rightarrow \frac{du}{1+u^2}=dx$$ $$I'(a)=\int_0^{\infty}\frac{u\cos(a*u)}{\arctan{u}}\frac{dx}{1+u^2}$$

And the other hand: $$I(a,b)=\int_0^{\infty}\frac{\sin(\tan(bx))}{x}e^{-ax}dx$$

Jorge Rodriguez
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2 Answers2

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The integrand function is even, hence $$ I = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{\sin(\tan x)}{x}\,dx $$ which due to the periodicity of the tangent function equals $$ I = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\sin(\tan x)\sum_{k\in\mathbb{Z}}\frac{dx}{x+k\pi}.$$ The sum over the integers has to be intended in the symmetric sense, i.e. as $$ \frac{1}{x}+\sum_{n\in\mathbb{N}}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right)=\cot(x) $$ where the last equality follows from considering the logarithmic derivative of the Weierstrass product for the sine function (or directly from Herglotz' trick). As a consequence, $$ I = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{\sin(\tan x)}{\tan(x)}\,dx = \int_{0}^{\pi/2}\frac{\sin(\tan x)}{\tan x}\,dx = \int_{0}^{+\infty}\frac{\sin(u)}{u(1+u^2)}\,du = \frac{\pi}{2}-\int_{0}^{+\infty}\frac{u\sin u}{1+u^2}\,du $$ where the last equality follows from a partial fraction decomposition and Dirichlet's integral $\int_{0}^{+\infty}\frac{\sin(u)}{u}\,du=\frac{\pi}{2}$. By the residue theorem, or by the self-adjointness of the Laplace transform, the last integral equals $\frac{\pi}{2e}$. The conclusion is that

$$ \boxed{\int_{0}^{+\infty}\frac{\sin(\tan x)}{x}\,dx = \color{red}{\frac{\pi}{2}\left(1-\frac{1}{e}\right)}\approx 0.992933} $$

and I wonder if there are slick ways to prove $I<1$ without resorting to explicit computations (or, at least, using only a very limited amount of them). Update: I found one. The Archimedean approximation $\pi < \frac{22}{7}$ can be deduced from the fact that the integral $\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,dx$ is positive and it equals $\frac{22}{7}-\pi$. The inequality $e<\frac{11}{4}$ can be deduced from the fact that $$ e=2+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\ldots < 2+\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{3^2}+\ldots\right)=\frac{11}{4}.$$ These inequalities lead to $\frac{\pi}{2}<\frac{11}{7}$ and $1-\frac{1}{e} < \frac{7}{11}$, so $\frac{\pi}{2}\left(1-\frac{1}{e}\right)<1$.

Jack D'Aurizio
  • 353,855
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Lobachevsky Integral Formula: If $f(x)$ meet $f(x+π)=-f(x)$ and $f(−x)=f(x)$.

Then $$\int_0^\infty f(x)\cdot\frac{\sin x}{x}dx=\int_0^\frac{\pi}{2}f(x)\cdot \cos xdx$$

so $$\int_0^\infty\frac{\sin(\tan x)}{x}dx=\int_0^\infty\frac{\sin(\tan x)}{\sin x}\cdot\frac{\sin x}{x}dx$$ where $$f(x)=\frac{\sin(\tan x)}{\sin x}$$

so $$\int_0^\infty\frac{\sin(\tan x)}{x}dx=\int_0^\frac{\pi}{2}\frac{\sin(\tan x)}{\sin x}\cdot \cos xdx=\int_0^\frac{\pi}{2}\frac{\sin(\tan x)}{\tan x}dx$$ $$=\int_0^\infty\frac{\sin \theta}{\theta(1+\theta^2)}d\theta =\int_0^\infty\frac{\sin\theta}{\theta}d\theta-\int_0^\infty\frac{\theta\sin\theta}{1+\theta^2}d\theta=\frac{\pi}{2}-\frac{\pi}{e\pi}=\frac{\pi}{2}\Big(1-\frac{1}{e}\Big)$$

anhui-xiaocaigui
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