4
$\begingroup$

One can easily show that $\int\limits_0^{\pi}\frac{\cos(2k+1)x}{\cos x} dx = 2 \int\limits_0^{\frac{\pi}{2}}\frac{\cos(2k+1)x}{\cos x} dx = (-1)^k \pi$.

But is there a closed form for $\int\limits_0^{\frac{\pi}{4}}\frac{\cos(2k+1)x}{\cos x} dx$ ?

So far, I tried by induction and by IBP but couldn't come to a simple closed form.

Thanks for your hints !

$\endgroup$
1
  • 1
    $\begingroup$ Here is the example for $k=14$. It seems that the answer is $$(-1)^k \left( \frac{\pi}{4} - \sum_{n \le k, n\mbox{ odd}} \frac{(-1)^n}{n} \right)$$ $\endgroup$
    – Crostul
    Jan 15, 2019 at 19:10

2 Answers 2

4
$\begingroup$

A sketch:

Write $$ \frac{\cos((2k+1) x)}{\cos(x)} = \mathrm{e}^{2 k \mathrm{i} x} \frac{1+\mathrm{e}^{-(2k+1) 2 \mathrm{i} x}}{1+\mathrm{e}^{-2\mathrm{i}x}} = \sum \limits_{l=0}^{2k} (-1)^l \mathrm{e}^{2 (k-l) \mathrm{i} x} = (-1)^k \left[1+2\sum \limits_{m=1}^k (-1)^m \cos(2mx)\right]$$ to obtain \begin{align} I_k &\equiv \int \limits_0^{\pi/4} \frac{\cos((2k+1) x)}{\cos(x)} \, \mathrm{d} x = (-1)^k \left[\frac{\pi}{4} + \sum \limits_{m=1}^k \frac{(-1)^m}{m} \sin\left(\frac{m\pi}{2}\right)\right] \\ &= (-1)^k \left[\frac{\pi}{4} - \sum \limits_{n=0}^{\lfloor \frac{k-1}{2}\rfloor} \frac{(-1)^n}{2n+1}\right] \end{align} for $k \in \mathbb{N}_0$ in agreement with Crostul's comment. In particular, $ \lim_{k \to \infty} I_k = 0$ .

$\endgroup$
0
$\begingroup$

It is worthwhile to recall that for $n\in\Bbb N$, $$\cos nx=T_n(\cos x)$$ Where $T_n(x)$ is the Chebyshev polynomial of the first kind, defined as $$T_n(x)=\frac{n}2\sum_{r=0}^{\lfloor n/2\rfloor}\frac{(-1)^r}{n-r}{n-r\choose r}(2x)^{n-2r}$$ Hence $$I_k=\int_0^{\pi/4}\frac{\cos[(2k+1)x]}{\cos x}dx$$ $$I_k=\int_0^{\pi/4}\frac{2k+1}{2\cos x}\sum_{r=0}^{\lfloor \frac{2k+1}2\rfloor}\frac{(-1)^r}{2k+1-r}{2k+1-r\choose r}(2\cos x)^{2k-2r+1}dx$$ $$I_k=\frac{2k+1}{2}\sum_{r=0}^{\lfloor \frac{2k+1}2\rfloor}\frac{(-1)^r}{2k+1-r}{2k+1-r\choose r}2^{2k-2r+1}\int_0^{\pi/4}\cos(x)^{2k-2r}dx$$ $$I_k=(2k+1)\sum_{r=0}^{k}\frac{(-1)^r4^{k-r}}{2k-r+1}{2k-r+1\choose r}\int_0^{\pi/4}\cos(x)^{2k-2r}dx$$ Then we focus on the integral $$J(r,k)=\int_0^{\pi/4}\cos(x)^{2(k-r)}dx$$ Unfortunately, the closest to a closed form for this guy is something in terms of the incomplete Beta function. Here's how:

Consider the integral $$H(k;a,b)=\int_0^k\sin(x)^{a}\cos(x)^{b}dt$$ For some $0\leq k\leq 1$. Making the substitution $t=\sin(x)^2$, we see that $$H(k;a,b)=\frac12\int_0^{\sin(k)^2}t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}dt$$ $$H(k;a,b)=\frac12\int_0^{\sin(k)^2}t^{\frac{a+1}2-1}(1-t)^{\frac{b+1}2-1}dt$$ Then Recall the definition of the incomplete Beta function: $$\mathrm{B}(x;a,b)=\int_0^x t^{a-1}(1-t)^{b-1}dt$$ Thus $$H(k;a,b)=\frac12\mathrm{B}\bigg(\sin(k)^2;\frac{a+1}2,\frac{b+1}2\bigg)$$ And $$J(r,k)=H(\pi/4;0,2k-2r)=\frac12\mathrm{B}\bigg(\frac12;\frac{1}2,k-r+\frac{1}2\bigg)$$ Thus $$I_k=\frac{2k+1}2\sum_{r=0}^{k}\frac{(-1)^r4^{k-r}}{2k-r+1}{2k-r+1\choose r}\mathrm{B}\bigg(\frac12;\frac{1}2,k-r+\frac{1}2\bigg)$$ Which could be considered a closed form.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.