Krdytkyu

I wish to evaluate $$I(2018)=int_{0}^{pi}frac{cos(2018x)}{5-4cos x} dx$$ Considering $$X=I(k)+iJ(k)=int_{-pi}^{pi}frac{cos{kx}}{5-4cos x} dx +iint_{-pi}^{pi}frac{sin{kx}}{5-4cos x} dx=int_{-pi}^{pi}frac{e^{ikx}}{5-4cos x} dx$$ let us substitute $$e^{ix}=zrightarrow dx=frac{dz}{iz} , ,|z|=1$$ Due to Euler's formula we can rewrite $$cos x=frac{z^2+1}{2z}$$ $$X=oint_{|z|=1} frac{z^k}{5-4frac{z^2+1}{2z}}frac{dz}{iz}=frac{1}{i}oint_{|z|=1} frac{z^k}{-2z^2+5z-2}dz$$ $$-2z^2+5z-2=-frac{1}{2}((2z)^2-5(2z)+4)=-frac{1}{2}(2z-4)(2z-1)=-2(z-2)(z-frac{1}{2})$$ Now let us notice that in our contour $|z|=1,$ only the pole $z_2=frac{1}{2}$ is found. Thus the integral we seek to evaluate is $$frac{1}{i} cdot 2pi i , text{Res} (f(z),z_2)$$ where $f(z)=frac{z^k}{-2(z-2)(z-frac{1}{2})}$ $$X=2pi lim_{zto z_2} (z-z_2)frac{z^k}{-2(z-2)(z-z_2)}=frac{2}{3}pi frac{1}{2^k}$$ therefore $$I(k)=Re (X) =frac{2pi}{3}frac{1}{2^k}$$ And $$int_{0}^{pi}frac{cos(2018 x)}{5-4cos x} dx=frac{pi}{3}cdotfrac{1}{2^{2018}}.$$ Now someone told me that the answer is $0$ and I am wrong (also wolfram gives $0$ as an answer). Could you please clarify? Or maybe give another solution to this integral if it's $0$ or another answer?

You are correct! This is an elementary approach without complex analysis.

Since $$cos((n+1)x)+cos((n-1)x)=2cos(nx)cos(x)$$ then for $ngeq 1$, we have the linear recurrence
$$begin{align}
I(n-1)+I(n+1)&=int_{0}^{pi}frac{2cos(nx)cos(x)}{5-4cos(x)} dx\
&=
-frac{1}{2}int_{0}^{pi}frac{cos(nx)(-5+5-4cos(x))}{5-4cos(x)} dx\
&=
frac{5}{2}I(n)-frac{1}{2}int_{0}^{pi}cos(nx) dx=frac{5}{2}I(n).
end{align}$$
Then
$$I(n)=A2^n+frac{B}{2^n}$$
for some constants $A$ and $B$. Since $I(n)$ is bounded it follows that $A=0$ and
$$I(n)=frac{I(0)}{2^n}=frac{1}{2^n}int_{0}^{pi}frac{dx}{5-4cos(x)} =frac{1}{2^n}left[frac{2arctan(3tan(x/2))}{3}right]_{0}^{pi}=frac{pi/3}{2^n}.$$
P.S. In order to show that $I(n)$ is not zero for any $n$ it suffices to say that $I(0)>0$ since it is the integral of a positive continuous function.

$$
frac{1}{5-4cosleft(xright)}=frac{1}{5-2e^{ix}-2e^{-ix}}=frac{e^{ix}}{5e^{ix}-2e^{2ix}-2}
$$
Let $X=e^{ix}$ then
$$
-2X^2+5X-2=-left(2X-1right)left(X-2right)
$$
Now we do a partial decomposition
$$-frac{1}{left(2X-1right)left(X-2right)}=frac{2}{3}frac{1}{2X-1}-frac{1}{3}frac{1}{X-2}
$$
So far we have
$$
frac{1}{5-4cosleft(xright)}=frac{1}{3}frac{1}{2e^{ix}-1}-frac{2}{3}frac{1}{e^{ix}-2}
$$
We'll express this as a series, so we transform it into an adapted form
$$
frac{1}{5-4cosleft(xright)}=frac{1}{3}left(frac{1}{2e^{ix}-1}-frac{2}{e^{ix}-2}right)=frac{1}{3}left(frac{frac{1}{2}e^{-ix}}{displaystyle {1-frac{1}{2}e^{-ix}}}+frac{1}{1-frac{1}{2}e^{ix}}right)$$

Hence
$$
frac{1}{5-4cosleft(xright)}=frac{1}{3}left(sum_{n=1}^{+infty}left(frac{1}{2}right)^ne^{-inx}+sum_{n=0}^{+infty}left(frac{1}{2}right)^ne^{inx}right)
$$
which finally gave us
$$
frac{1}{5-4cosleft(xright)}=frac{1}{3}left(1+sum_{n=1}^{+infty}left(frac{1}{2}right)^{n-1}cosleft(nxright)right)$$
Hence using normal convergence
$$
int_{0}^{pi}frac{cosleft(2018xright)}{5-4cosleft(xright)}text{d}x=int_{0}^{pi}frac{cosleft(2018 xright)}{3}text{d}x+sum_{n=1}^{+infty}left(frac{1}{2}right)^{n-1}int_{0}^{pi}cosleft(nxright)cosleft(2018xright)text{d}x
$$
Using that
$$
int_{0}^{pi}cosleft(Kxright)text{d}x=0
$$
for all $K in mathbb{Z}$, we have

$$
int_{0}^{pi}frac{cosleft(2018xright)}{5-4cosleft(xright)}text{d}x=frac{pi}{3}left(frac{1}{2}right)^{2018}
$$