What happens to Chebyshev polynomials integration when n=1

Question

The integration of Chebyshev polynomials of the first kind has the following value, $$\int T_{n}(x) \, dx = \frac{1}{2} \, \left( \frac{T_{n+1}(x)}{n+1} - \frac{T_{n-1}(x)}{n-1} \right)$$ what happens when n=1? We have zero division.

Best, Mohd

Answer 1

The integral really ought to read:

$$\int T_{n}(x) \, dx = \frac{1}{2} \, \left( \frac{T_{n+1}(x)}{n+1} - \frac{T_{n-1}(x)}{n-1} \right)+C$$

and we can rewrite this as

$$\int T_{n}(x) \, dx = \frac{1}{2} \, \left( \frac{T_{n+1}(x)}{n+1} - \frac{T_{n-1}(x)-1}{n-1} \right)+C'$$

Note now that L'Hopital's Rule gives us:

$$\lim_{n\to1}\left(\frac{T_{n-1}(x)-1}{n-1}\right)=\lim_{n\to1}\left(\frac{\cos((n-1)\arccos x)-1}{n-1}\right)=\lim_{n\to1}\left(\frac{-\arccos x\cdot\sin((n-1)\arccos x)}{1}\right)=0$$

Hence,

$$\int T_{1}(x) \, dx = \frac{1}{2} \, \left( \frac{T_2(x)}{2} - 0 \right)+C'=\frac{T_2(x)}{4}+C'$$

There technically ought to be a bit more rigor involved in taking $n\to1$, but hopefully this explains the motivation somewhat.