$$I_n = \int_0^{\pi/2} x^n \cos(2x)dx,\,\,\, n = 0,1,2 \ldots$$

(a) Evaluate $I_0$

(b) By using integration by parts twice, or otherwise, show that a reduction formula for $I_n$ is

$$I_n = -\frac{\pi^{n-1}n}{2^{n+1}} - \frac{n(n-1)}{4} I_{n-2}$$

I have evaluated $I_0=0$. It's (b) I'm having issues with. I've come to the conclusion that I need to integrate by parts $\int_0^{\pi/2} (x\cos(2x))(x^{n-1}\cos(2x))dx$ twice. Is this correct?


  • 1
    $\begingroup$ You need to integrate by parts twice, the first time the cosine turns to a sine and the second you get the cosine back.... $\endgroup$ – marwalix Apr 11 '15 at 15:35

Let $$ f(t) = \int_{0}^{\pi/2}e^{tx}\cos(2x)\,dx = -\frac{t}{4+t^2}\left(1+e^{\frac{\pi t}{2}}\right)\tag{1}$$ where the last identity follows from considering $\cos(2x)$ as $\text{Re}(e^{2ix})$.

By differentiating under the integral sign, we have: $$ I_n = \frac{d^n f}{dt^n}(0) \tag{2}$$ from which it follows that: $$ f(t) = -\frac{t}{4+t^2}\left(1+e^{\frac{\pi t}{2}}\right) = \sum_{n\geq 0}\frac{I_n}{n!}\,t^n \tag{3}$$ and since the middle term vanishes for $t=0$, we have $I_0=0$.

Now we can write $(4+t^2)\,f(t)$ in two different ways, namely: $$ -t\left(1+e^{\frac{\pi t}{2}}\right) = -t-\sum_{n\geq 0}\frac{\pi^n}{2^n n!}t^{n+1} \tag{4} $$ and: $$ \sum_{n\geq 0}\frac{I_n}{n!}(4t^n+t^{n+2})\tag{5} $$ and by comparing the coefficients of $[t^n]$ in $(4)$ and $(5)$ our reduction formula follows.

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