Take a look at the averaging sum

$$\frac{\pi}{n}\sum_{k=1}^n\;\exp{(-\sin\theta_k)}\cdot \sin(\theta_k +\cos\theta_k)\, \qquad\text{where }\;\theta_k=(2k-1)\frac{\pi}{2n}$$

depending on $n\in\mathbb{N}$.

How could one analyse convergence for $n\rightarrow\infty$, and possibly compute its limit? ...

but 'ignoring' the fact that it is equal to the point evaluation of the Cosine transform $$\int_{-\infty}^\infty\; \frac{\cos(\omega x)}{x^{2n}+1}\, dx$$ for $\omega = 1$, cf the Math. SE answer. If $n\to\infty$ then $1/(x^{2n}+1)$ converges pointwise to the characteristic function $\chi_{[-1,1]}$ (ignoring $x=\pm 1$). Hence the sought limit should equal $$\int^1_{-1}\cos x\, dx = 2\sin(1)\approx 1.6829$$

Let me repeat my question as "How to tame the above expression when encountered in the wild?"
Thus deliberately excluding the ‘Fourier background’, I'd like to know appropriate methods to determine the limit.
The above formulation "limit should equal" could then be replaced with the statement "is convergent with limit equal to".

  • $\begingroup$ Look at sine as imaginary part of complex exponent and then consider your sum as a Riemann sum in points $\theta_k$ $\endgroup$ Jun 9, 2016 at 15:05

1 Answer 1


Fedor's hint works. $$ \frac{\pi}{n}\sum_{k=1}^n\;\exp{(-\sin\theta_k)}\cdot \sin(\theta_k +\cos\theta_k)\, \qquad\text{where }\;\theta_k=(2k-1)\frac{\pi}{2n}$$ is a Riemann sum for the integral $$\int_0^\pi \exp{(-\sin t)}\cdot \sin(t +\cos t)\;dt = 2\sin(1)$$ This integral is easy because of the antiderivative $$\int \exp{(-\sin t)}\cdot \sin(t +\cos t)\;dt = -e^{-\sin t}\sin(\cos t) + C$$

  • $\begingroup$ From OP's perspective this is a concise & satisfactory answer, Gerald Edgar, many thx! But I'm still struggling with your "$\pi$ in the denominator" because the Riemann sum sample spacing is $\frac{\pi}{n} = \theta_{k+1} - \theta_k$ (leaving aside the endpoints of the integration interval). Numerical Evaluation (Simpson's rule within Python) back the expression as of above in the question, having $\pi$ in the nominator. Could you check this, and eventually edit the answer? $\endgroup$
    – Hanno
    Jun 24, 2016 at 8:36
  • $\begingroup$ You are correct, the $\pi$ should be in the numerator. $\endgroup$ Jun 24, 2016 at 12:55
  • 2
    $\begingroup$ Of course you mean @FedorPetrov, not 'Fodor'. $\endgroup$
    – LSpice
    Jun 24, 2016 at 18:11

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