Can You Evaluate This Definite Integral with Trig Functions?

[anemone]

Evaluate the definite integral $\displaystyle \int_{-\pi}^{\pi} \dfrac{\sin nx}{(1+2^x)\sin x}\,dx$, where $n$ is a natural number.

 

[alyafey22]

Spoiler

$$I=\displaystyle \int_{-\pi}^{\pi} \dfrac{\sin nx}{(1+2^x)\sin x}\,dx $$

Use the substitution $t = -x$

$$I = \displaystyle \int_{-\pi}^{\pi} \dfrac{2^t\sin nt}{(1+2^t)\sin t}\,dt = \int_{-\pi}^{\pi} \dfrac{2^x\sin nx}{(1+2^x)\sin x}\,dx $$

Since $t$ is just a dummy variable and by summing the two formulas

$$2I =\int_{-\pi}^{\pi} \dfrac{\sin nx}{\sin x}\,dx = \int_{0}^{2\pi} \dfrac{\sin nx}{\sin x}\,dx $$

Let us consider two different cases when $n$ is even

$$I_k = \int^{2\pi}_0\frac{\sin(2kx)}{\sin(x)}\,dx$$

$$I_k-I_{k-1} = \int^{2\pi}_0\frac{\sin(2kx)-\sin(2kx-2x)}{\sin(x)}\,dx$$

$$I_k-I_{k-1} =2 \int^{2\pi}_0\frac{\sin(x)\cos(2kx-x)}{\sin(x)}\,dx = \int^{2\pi}_0\cos(2kx-x)\,dx = 0$$

Hence

$$I_k=I_{k-1} = I_1 = 2\int^{2\pi}_0 \cos(x) \,dx= 0 $$

If $n$ is odd , let

$$I_j = \int^{2\pi}_0\frac{\sin((2j+1)x)}{\sin(x)}\,dx$$

$$I_j-I_{j-1} = \int^{2\pi}_0\frac{\sin((2j+1)x)-\sin((2j-1)x)}{\sin(x)}\,dx$$

$$I_j-I_{j-1} = \int^{2\pi}_0\frac{\sin(x)\cos(2jx)}{\sin(x)}\,dx = 0$$

Similarily

$$I_j = I_{j-1} = I_0 = \int^{2\pi}_0 \frac{\sin(x)}{\sin(x)} \,dx= 2\pi$$

Finally we get that

$$I= \int_{-\pi}^{\pi} \dfrac{\sin nx}{(1+2^x)\sin x}\,dx= \begin{cases}
0 & \text{n is even} \\
\pi & \text{n is odd}
\end{cases}$$

 

[Prove It]

Spoiler

$$I=\displaystyle \int_{-\pi}^{\pi} \dfrac{\sin nx}{(1+2^x)\sin x}\,dx =\displaystyle \int_{-\pi}^{\pi} \dfrac{2^x\sin nx}{(1+2^x)\sin x}\,dx $$

$$I =\frac{1}{2}\int_{-\pi}^{\pi} \dfrac{\sin nx}{\sin x}\,dx = \frac{1}{2}\int_{0}^{2\pi} \dfrac{\sin nx}{\sin x}\,dx $$

The integral is $0$ if $n$ is even and $2\pi$ if it is odd. I will prove it later.

I'm having trouble seeing how any of these integrals are equal?

Where did the extra factor of $\displaystyle \begin{align*} 2^x \end{align*}$ in the integrand come from? How did that all become a factor of $\displaystyle \begin{align*} \frac{1}{2} \end{align*}$? Also how did you change the bounds?

 

[anemone]

Spoiler

$$I=\displaystyle \int_{-\pi}^{\pi} \dfrac{\sin nx}{(1+2^x)\sin x}\,dx =\displaystyle \int_{-\pi}^{\pi} \dfrac{2^x\sin nx}{(1+2^x)\sin x}\,dx $$

$$I =\frac{1}{2}\int_{-\pi}^{\pi} \dfrac{\sin nx}{\sin x}\,dx = \frac{1}{2}\int_{0}^{2\pi} \dfrac{\sin nx}{\sin x}\,dx $$

The integral is $0$ if $n$ is even and $2\pi$ if it is odd. I will prove it later.

I'm having trouble seeing how any of these integrals are equal?

Where did the extra factor of $\displaystyle \begin{align*} 2^x \end{align*}$ in the integrand come from? How did that all become a factor of $\displaystyle \begin{align*} \frac{1}{2} \end{align*}$? Also how did you change the bounds?

Hi Prove It,

I think given Zaid has mentioned that he would prove how such and such are true, we will wait for his second post for more clarification...(Smile)

 

[alyafey22]

I just edited my post. I added more clarifications and the proof of the last integral.