I would like to evaluate the following integral:

$$I(x_0)=\int_{-1}^1 \left(\sum_{k=0}^n a_k \, T_k(x)\right) \, \mathrm{e}^{b(x-x_0)}\,\mathrm{d}x$$

with $a_k$ some constant coefficients comming from an Chebyshev interpolation of a function $f(x)$, $T_k(x)$ the Chebyshev polynomials, $b\in\mathbb{C}$, $n\in\mathbb{N}$ and $x_0>1$.

I assume, I can rewrite the integral as

$$I(x_0)=\mathrm{e}^{-b\,x_0}\sum_{k=0}^n a_k \left(\int_{-1}^1 T_k(x)\, \mathrm{e}^{b\,x}\,\mathrm{d}x\right)$$

In this question this integral is solved by a recurrence relation. I used this recurrence relation to evaluate the sum, but sadly the recurrence formular is not stable for large $n$. I also tryed to adopt the Clenshaw algorithm (Link) to overcome this problem, but still the evaluation diverges.

Is there a possibility to write this integral in explicit form?

Edit: I am wondering if replacing

$$\sum_{k=0}^n a_k \, T_k(x)$$ by the barycentric interpolation formula $$\sum_{j=0}^N a_j \, T_j(x) = \frac{\displaystyle \sum_{j=0}^N \frac{w_j}{x-x_j}f_j}{\displaystyle \sum_{j=0}^N \frac{w_j}{x-x_j}}, $$ where $$ w_j = \left\{ \begin{array}{cc} (-1)^j/2, & j=0\text{ or }j=N,\\ (-1)^j, & \text{otherwise} \end{array} \right. $$ is any better!


In "The Fourier Transforms of the Chebyshev and Legendre Polynomials" (Fokas&Smitheman https://arxiv.org/abs/1211.4943 ) Eq. (2.4) gives an expression for $\hat T_m(\lambda) = \int_{-1}^1 \exp(-i\lambda x) T_m(x) dx$.

There's another expression for $\hat T_m(\lambda)$ in "The finite Fourier transform of classical polynomials" (Dixit,Jiu,Moll,Vignat https://arxiv.org/abs/1402.5544) in Thm 4.1 which looks slightly more inviting:

$$ \hat T_n(\lambda) = \sum_{k=0}^n (-1)^{k+1} \frac{n 2^k (n+k)! k!}{(2k+1)!(n-k)!} \frac{(-1)^{n-k} e^{-i\lambda} - e^{i\lambda} }{ (-2i\lambda)^{k+1}} $$

(If you can work in the Legendre basis, the transform of a Legendre polynomial is a half-integer Bessel, which would at least look nicer)

  • $\begingroup$ Thank you for this information. I am wondering if it is possible to reduce the double sum to something simpler from a computational point of view. $\endgroup$ – Michael_K Mar 29 at 21:31

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