integration of chebyshev polynomials of first kind with an exponential funcion In my class, my tutor raised a question of the following integral:
$\int T_n(x)*\exp(a*x)dx,$ where $T_n(x)$ is an n power of Chebyshev polynomials of first kind and a is a constant. The hint he gave us is using the recurrence relation of Chebyshev polynomials. I have tried a few times but I didn't get lucky. I think the question is wrong, isn't it. 
 A: The question isn't wrong, the hint is... a bit misleading, I'd say. You might use
$$T_n(x)=\frac{1}{2}(U_n(x)-U_{n-2}(x)),$$ for $n\ge2$ together with
$$T'_n(x)=n\,U_{n-1}(x)$$ and partial integration, to get a recurrence relation for your integrals (the $U_n$ are Chebyshev polynomials of second kind).
Here are more details: Let's define $$I_n(x)=\int T_n(x)\,e^{ax}\,dx.$$ Since $$T_n(x)=\frac{1}{2}\left(\frac{T'_{n+1}(x)}{n+1}-\frac{T'_{n-1}(x)}{n-1}\right),$$ multiplying by $e^{ax}$ and integrating (by parts) gives
$$I_n(x)=\frac{1}{2}\left(\frac{T_{n+1}(x)}{n+1}-\frac{T_{n-1}(x)}{n-1}\right)\,e^{ax}-\frac{a}{2}\left(\frac{I_{n+1}(x)}{n+1}-\frac{I_{n-1}(x)}{n-1}\right).$$ I'll omit integration constants, because they cancel out when calculating definite integrals. If people insist in them, feel free to add any constant you like best. Now you can solve for $I_{n+1}(x)$, and calculate it
starting from $n=2$ recursively from $I_{n-1}(x)$ and $I_n(x).$ To prime the pump, you'll have to calculate $I_0(x)$, $I_1(x)$ and $I_2(x)$ manually, but with $T_0(x)=1$, $T_1(x)=x$ and $T_2(x)=2x^2-1$ that's a very simple task I won't explain here.
