Finding $f(2019)$ in definite integration

Question

If $\displaystyle f(n)=\int^{1}_{0}(1+x+x^2+\cdots +x^{n-1})(1+3x+5x^2+\cdots +(2n-1)x^{n-1})dx$. Then $f(2019)$ is

What I tried:

$$1+x+x^2+\cdots +x^{n-1}=\frac{1-x^n}{1-x}$$

and $$1+3x+5x^2+\cdots +(2n-1)x^{n-1}=\frac{1}{1-x}+\frac{2(1-x^n)}{1-x}-\frac{(2n-1)x^n}{1-x}$$

How do I solve it? Help me, please.

score 12 · Accepted Answer · answered Mar 17 '19 at 08:13

12

By making change of variable $x=t^2$, we obtain $$\begin{align*} f(n)&=2\int_0^1 t(1+t^2+\cdots +t^{2n-2})(1+3t^2+5t^4+\cdots +(2n-1)t^{2n-2})dt. \end{align*}$$ Further making substitution $u=t+t^3+\cdots +t^{2n-1}$, we get $$ f(n)=2\int_0^n u\ du=\left[u^2\right]^n_0=n^2. $$ So $f(2019)=2019^2$ follows.

answered Mar 17 '19 at 08:13

Myunghyun Song

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+1 for this excellent answer, and also for the many other excellent answers you have posted. – Toby Mak Mar 17 '19 at 08:26
@TobyMak Your kind words make me feel rewarded for participating in the MSE community. I really appreciate it :) – Myunghyun Song Mar 17 '19 at 09:20
This is such an elegant answer! You are a genius! +1 – clathratus Mar 18 '19 at 23:30
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@clathratus Thank you very much for your praise :-) But there are many people on the MSE who are marvelous at integrals, and most of the time I'm just learning from them, including you! – Myunghyun Song Mar 20 '19 at 02:04

Finding $f(2019)$ in definite integration

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