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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.

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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.

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  • $\begingroup$ Hey professor Vector, would you mind giving me more steps such that I can follow your answer? Many thanks in advance. $\endgroup$ – Ray Jun 15 '17 at 20:44
  • $\begingroup$ In principle, the above derivation is valid for $n=1,$ too, but there are caveats: it's valid if you assume $U_{-1}(x)=0.$ That's in agreement with the definition $$U_n(\cos t)=\frac{\sin(n+1)t}{\sin t},$$ and even $T'_0(x)=0\cdot U_{-1}(x)$ is true, you just can't divide both sides by $0,$ you have to omit the terms originating from $U_{-1}$. So the above recurrence for $n=1$ becomes $$I_1(x)=\frac{1}{4}T_2(x)\,e^{ax}-\frac{a}{4}I_2(x).$$ $\endgroup$ – Professor Vector Jun 16 '17 at 7:16
  • $\begingroup$ Thanks Professor Vector. $\endgroup$ – Ray Jun 17 '17 at 0:55

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