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The equation $$\int_{-\pi}^xt\sin(t)\,dt=1$$ is given. Calculating the integral and simplifying a bit, it reduces to the equation $$-x\cos(x)+\sin(x)+\pi-1=0$$ I've tried rewriting the trig functions using complex exponentials, but that got me nowhere. Solving it numerically isn't too challenging, the Newton-Raphson method can be used with $$a_0=\pi n+\frac{\pi}{2},\:n\in\mathbb{N}$$ which is always close enough to a root, such that it converges to a root (with the exception of $n=0$).

As far as solving it analytically goes, I'm pretty clueless at this point. Any help is appreciated.

Maj Zobrani
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  • I'm afraid that this is impossible (unless you are willing to accept solutions in infinite series or approximations). There are many equations which do not have closed form solutions (even simple ones such as $x\cos(x)=1$). See e.g. Transcendental equation. – projectilemotion Jan 15 '22 at 13:00

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You want to find the zero's of function$$f(x)=-x\cos(x)+\sin(x)+\pi-1$$ As you wrote, the solutions are close to $(2n+1)\frac \pi 2$. Make a series expansion around this point and then series reversion to get $$x_{(n)}\sim (2n+1)\frac \pi 2+2t_n-\frac{4 }{ (2 n+1)\pi}t_n^2+\frac{4 \left((2n+1) \pi )^2+12\right) }{3 (2 n+1)\pi )^2}t_n^3+O(t_n^4)$$ where $$t_n=\frac{(-1)^n \left(1-(-1)^n+\pi \right)}{(2 n+1)\pi }$$ which implies $$t_{2n}=\frac \pi{(2 n+1)\pi }\qquad \text{and}\qquad t_{2n+1}=-\frac {\pi+2}{(2 n+1)\pi } $$ Some results $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 1 & 4.9511052 & 4.9509461 \\ 2 & 7.4326103 & 7.4305312 \\ 3 & 11.099098 & 11.099096 \\ 4 & 13.911342 & 13.911246 \\ 5 & 17.344751 & 17.344751 \\ 6 & 20.265315 & 20.265300 \\ 7 & 23.610365 & 23.610365 \\ 8 & 26.585359 & 26.585355 \\ 9 & 29.883366 & 29.883366 \\ 10 & 32.891203 & 32.891202 \end{array} \right)$$ We could do much better with a few terms more in the series expansion and reversion.

Claude Leibovici
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