Question

Employing Euler's formula \(\mathrm{e}^{\mathrm{ix}}=\cos \mathrm{x}+\mathrm{isin} \mathrm{x}\), prove that \(\int_{0}^{2 \pi} \mathrm{e}^{-\mathrm{inx}} \cdot \mathrm{e}^{\mathrm{imx} x} \mathrm{~d} \mathrm{x}=\left\\{\begin{array}{lc}0 & \text { for } \mathrm{m} \neq \mathrm{n} \\ 2 \pi & \text { for } \quad \mathrm{m}=\mathrm{n}\end{array}\right.\)

Short answer

Question: Prove that the integral \(\int_{0}^{2 \pi} \mathrm{e}^{-\mathrm{inx}} \cdot \mathrm{e}^{\mathrm{imx} x} \mathrm{~d} \mathrm{x}\) is equal to 0 when m ≠ n and 2π when m = n. Solution: We have shown that after rewriting the integrand using Euler's formula and integrating, the integral evaluates to 0 when \(m \neq n\) and 2π when \(m = n\). Therefore, we proved that the integral satisfies the given conditions.

Step-by-step solution

Rewrite the integrand using Euler's formula

We will first rewrite the given integrand \(\mathrm{e}^{-\mathrm{inx}} \cdot \mathrm{e}^{\mathrm{imx} x}\) using Euler's formula: \(\mathrm{e}^{\mathrm{ix}}=\cos\mathrm{x}+\mathrm{isin} \mathrm{x}\). Given \(\mathrm{e}^{-\mathrm{inx}} \cdot \mathrm{e}^{\mathrm{imx} x}\), we can rewrite it as: \(\mathrm{e}^{\mathrm{i}(m-n)x}\) Now, using Euler's formula, we get: \(\cos{(m-n)x} + \mathrm{i}\sin{(m-n)x}\)

Rewrite the integral and integrate

Now, rewrite the integral using the new integrand: \(\int_{0}^{2 \pi} (\cos{(m-n)x} + \mathrm{i}\sin{(m-n)x}) \mathrm{~d} \mathrm{x}\) We can split the integral into two parts, one for the real part and one for the imaginary part: \(\int_{0}^{2 \pi} \cos{(m-n)x} \mathrm{~d} \mathrm{x} + \mathrm{i} \int_{0}^{2 \pi} \sin{(m-n)x} \mathrm{~d} \mathrm{x}\) Now, we integrate both parts separately: \(\left[ \frac{\sin{(m-n)x}}{m-n} \right]_0^{2\pi} + \mathrm{i} \left[ -\frac{\cos{(m-n)x}}{m-n} \right]_0^{2\pi}\)

Evaluate the integral

We will now evaluate the integral and show that it satisfies the given conditions. First, consider the case when \(m \neq n\): \(\left[ \frac{\sin{(m-n)(2\pi)}}{m-n} - \frac{\sin{(m-n)(0)}}{m-n} \right] + \mathrm{i} \left[ -\frac{\cos{(m-n)(2\pi)}}{m-n} + \frac{\cos{(m-n)(0)}}{m-n} \right]\) Since \(m-n\) is an integer, both \(\sin{(m-n)(2\pi)}\) and \(\cos{(m-n)(2\pi)}\) are equal to the \(\sin{(0)}\) and \(\cos{(0)}\) terms, respectively, which evaluates the integral to 0: \(0 - 0 + \mathrm{i}\left[0 - 0\right]\) Now, let's consider the case when \(m = n\): Notice that the integral reduces to: \(\int_0^{2\pi}\cos{(0)}\mathrm{~d}\mathrm{x} + \mathrm{i}\int_0^{2\pi}\sin{(0)}\mathrm{~d}\mathrm{x}\) Let's evaluate this integral: \(\left[x\right]_0^{2\pi} + \mathrm{i}\left[-\cos{(0)}x\right]_0^{2\pi}\) Which simplifies to: \(2\pi-0+\mathrm{i}\left[-0+0\right]=2\pi\) Thus, the integral is equal to 0 when \(m \neq n\) and 2π when \(m = n\).