So I tried to solve this integral:


By turning it into a contour integral with a small semi-circle to avoid the singularity at z=0, and taking the imaginary part:


The answer I got is $\pi$, but Wolfram Alpha can’t evaluate the integral. How can I check my answer ?

  • 5
    $\begingroup$ "How can I check my answer ?" -- Short of doing numerical approximations, you claiming the answer is $\pi$ without explaining in more detail why makes it borderline impossible for us to assess the validity of your work in the first place. $\endgroup$ Oct 13, 2022 at 0:25
  • $\begingroup$ How does a semicircle avoid $0?$ $\endgroup$ Oct 13, 2022 at 0:26
  • $\begingroup$ Explain how you showed the off-axis part of the integral goes to $0$. Probably better is $$\oint\frac{e^{\sin(z)i}-1}{z}dz$$since it has a removable singularity at $z=0$. $\endgroup$
    – GEdgar
    Oct 13, 2022 at 0:35
  • $\begingroup$ I avoided zero by substituting the path $re^{ti}$ directly in the integral and taking the limit as r approaches zero. $\endgroup$ Oct 13, 2022 at 0:44
  • $\begingroup$ What contour are you using? $\endgroup$ Oct 13, 2022 at 2:12

2 Answers 2


Just to show up what the correct answer looks like. By Lobachevsky's integral formula, $$I:=\int_{-\infty}^\infty\frac{\sin\sin x}{x}\,dx=\int_0^\pi\frac{\sin\sin x}{\sin x}\,dx$$ can be written as $\pi\int_0^1 J_0(z)\,dz$ using the Bessel function $J_0(z)$ and its integral representation; this has a closed form expression in terms of Struve functions (and Bessel functions). Alternatively, $$I=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\int_0^\pi\sin^{2n}x\,dx=\pi\sum_{n=0}^\infty\frac{(-1/4)^n}{n!^2(2n+1)},$$ useful for numerics: $I\approx2.889418299621118615441122131089\cdots$


After @metamorphy's elegant solution and answers, we can also write $$I=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\int_0^{\frac\pi2}\sin^{2n}(x)\,dx= \sqrt \pi \sum_{n=0}^\infty (-1)^n\frac{ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1) \Gamma (2 n+2)}$$ $$I=\pi \,\,\, _1F_2\left(\frac{1}{2};1,\frac{3}{2};-\frac{1}{4}\right)$$

The convergence is very fast since we have an alernating series and $$a_n=\frac{ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1) \Gamma (2 n+2)}\implies \frac{a_{n+1}}{a_n}=\frac{2 n+1}{4 (n+1)^2 (2 n+3)}=\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)$$

If we truncate the sum as $$I=\sqrt \pi \sum_{n=0}^p (-1)^n\frac{ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1) \Gamma (2 n+2)}+\sqrt \pi \sum_{n=p+1}^\infty (-1)^n\frac{ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1) \Gamma (2 n+2)}$$ for, a given accuracy we search for $p$ such that $$\frac{\sqrt{\pi }\,\, \Gamma \left(p+\frac{3}{2}\right)}{\Gamma (p+2) \Gamma (2 p+4)} \leq \epsilon$$ an approximation (in fact an overestimate) is given by $$p \sim \frac{1}{2} \exp\left(1+W\left(-\frac{\log (16 \epsilon )}{e}\right)\right)$$ where $W(.)$ is the principal branch of Lambert function.

For $\epsilon=10^{-50}$, this gives, as a real, $p=20.648$ while the "exact" solution is $p=19.022$


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