Evaluating the Definite Integral \(I_m = \int_0^\pi \frac{\sin 2{\rm{mx}}}{\sin {\rm{x}}}{\rm{dx}}\)

The problem asks us to find the value of the definite integral \({\rm{I}}_{\rm{m}} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{mx}}}}{{\sin {\rm{x}}}}{\rm{dx}},\) where \({\rm{m}}\) is a positive integer. This is a classic integral evaluation problem often encountered in calculus. We will analyze the integral by looking at small values of \({\rm{m}}\) and establishing a recursive relationship.

Step-by-Step Evaluation

Let's consider the integral for the first few positive integer values of \({\rm{m}}\).

• Case \({\rm{m}} = 1\): \[ {\rm{I}}_1 = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2(1){\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}} \] Using the double angle identity \(\sin 2{\rm{x}} = 2 \sin {\rm{x}} \cos {\rm{x}}\): \[ {\rm{I}}_1 = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{2 \sin {\rm{x}} \cos {\rm{x}}}{{\sin {\rm{x}}}}{\rm{dx}} \] For \({\rm{x}} \in (0, \pi)\), \(\sin {\rm{x}} \neq 0\), so we can cancel \(\sin {\rm{x}}\): \[ {\rm{I}}_1 = \mathop \smallint \nolimits_0^{\rm{\pi }} 2 \cos {\rm{x}} \, {\rm{dx}} \] Now, we evaluate the integral: \[ {\rm{I}}_1 = [2 \sin {\rm{x}}]_0^\pi = 2 \sin \pi - 2 \sin 0 = 2(0) - 2(0) = 0 \] So, \({\rm{I}}_1 = 0\).
• Case \({\rm{m}} = 2\): \[ {\rm{I}}_2 = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2(2){\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 4{\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}} \] This integral is not as straightforward to evaluate directly as the \({\rm{m}}=1\) case. Let's try to find a relationship between \({\rm{I}}_{\rm{m}}\) and \({\rm{I}}_{{\rm{m}}-1}\).

Establishing a Recursive Relation for the Integral

Consider the difference \({\rm{I}}_{\rm{m}} - {\rm{I}}_{{\rm{m}}-1}\) for \({\rm{m}} \ge 2\): \[ {\rm{I}}_{\rm{m}} - {\rm{I}}_{{\rm{m}}-1} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{mx}}}}{{\sin {\rm{x}}}}{\rm{dx}} - \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2({\rm{m}}-1){\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}} \] Since the limits of integration are the same, we can combine the integrals: \[ {\rm{I}}_{\rm{m}} - {\rm{I}}_{{\rm{m}}-1} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{mx}} - \sin 2({\rm{m}}-1){\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}} \] We use the trigonometric identity for the difference of sines: \(\sin {\rm{A}} - \sin {\rm{B}} = 2 \cos \left(\frac{{\rm{A}}+{\rm{B}}}{2}\right) \sin \left(\frac{{\rm{A}}-{\rm{B}}}{2}\right)\). Let \({\rm{A}} = 2{\rm{mx}}\) and \({\rm{B}} = 2({\rm{m}}-1){\rm{x}}\). Then \(\frac{{\rm{A}}+{\rm{B}}}{2} = \frac{2{\rm{mx}} + 2({\rm{m}}-1){\rm{x}}}{2} = \frac{(2{\rm{m}} + 2{\rm{m}} - 2){\rm{x}}}{2} = \frac{(4{\rm{m}}-2){\rm{x}}}{2} = (2{\rm{m}}-1){\rm{x}}\). And \(\frac{{\rm{A}}-{\rm{B}}}{2} = \frac{2{\rm{mx}} - 2({\rm{m}}-1){\rm{x}}}{2} = \frac{(2{\rm{m}} - 2{\rm{m}} + 2){\rm{x}}}{2} = \frac{2{\rm{x}}}{2} = {\rm{x}}\). So, \(\sin 2{\rm{mx}} - \sin 2({\rm{m}}-1){\rm{x}} = 2 \cos((2{\rm{m}}-1){\rm{x}}) \sin {\rm{x}}\). Substituting this back into the integral: \[ {\rm{I}}_{\rm{m}} - {\rm{I}}_{{\rm{m}}-1} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{2 \cos((2{\rm{m}}-1){\rm{x}}) \sin {\rm{x}}}{{\sin {\rm{x}}}}{\rm{dx}} \] For \({\rm{x}} \in (0, \pi)\), \(\sin {\rm{x}} \neq 0\), so we can cancel \(\sin {\rm{x}}\): \[ {\rm{I}}_{\rm{m}} - {\rm{I}}_{{\rm{m}}-1} = \mathop \smallint \nolimits_0^{\rm{\pi }} 2 \cos((2{\rm{m}}-1){\rm{x}}) \, {\rm{dx}} \] Now, we evaluate this integral: \[ {\rm{I}}_{\rm{m}} - {\rm{I}}_{{\rm{m}}-1} = \left[ \frac{2 \sin((2{\rm{m}}-1){\rm{x}})}{2{\rm{m}}-1} \right]_0^\pi \] \[ {\rm{I}}_{\rm{m}} - {\rm{I}}_{{\rm{m}}-1} = \frac{2 \sin((2{\rm{m}}-1)\pi)}{2{\rm{m}}-1} - \frac{2 \sin((2{\rm{m}}-1)0)}{2{\rm{m}}-1} \] Since \({\rm{m}}\) is a positive integer, \(2{\rm{m}}-1\) is an odd integer. For any integer \({\rm{k}}\), \(\sin({\rm{k}}\pi) = 0\). Thus, \(\sin((2{\rm{m}}-1)\pi) = 0\) and \(\sin(0) = 0\). \[ {\rm{I}}_{\rm{m}} - {\rm{I}}_{{\rm{m}}-1} = \frac{2(0)}{2{\rm{m}}-1} - \frac{2(0)}{2{\rm{m}}-1} = 0 - 0 = 0 \] So, we have established the recursive relation \({\rm{I}}_{\rm{m}} = {\rm{I}}_{{\rm{m}}-1}\) for all positive integers \({\rm{m}} \ge 2\).

Conclusion from the Recursive Relation

The relation \({\rm{I}}_{\rm{m}} = {\rm{I}}_{{\rm{m}}-1}\) means that the value of the integral is the same for all consecutive positive integers \({\rm{m}}\). Since \({\rm{I}}_1 = 0\), we can conclude:

• \({\rm{I}}_2 = {\rm{I}}_1 = 0\)
• \({\rm{I}}_3 = {\rm{I}}_2 = 0\)
• ... and so on.

Therefore, for any positive integer \({\rm{m}}\), the value of the integral \({\rm{I}}_{\rm{m}}\) is 0.

Summary of the Integral Evaluation

By evaluating the integral for \({\rm{m}}=1\) and finding a recursive relationship \({\rm{I}}_{\rm{m}} = {\rm{I}}_{{\rm{m}}-1}\) for \({\rm{m}} \ge 2\), we found that the value of the integral \({\rm{I}}_{\rm{m}}\) is constant for all positive integers \({\rm{m}}\). Since \({\rm{I}}_1 = 0\), it follows that \({\rm{I}}_{\rm{m}} = 0\) for all positive integers \({\rm{m}}\).

The final answer is 0.

Revision Table: Key Concepts for Integral Evaluation

Additional Information: Properties of Definite Integrals

Definite integrals have several useful properties that were implicitly or explicitly used in solving this problem. Understanding these properties helps in solving various integral problems.

• Linearity: \(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} (c f(x) + d g(x)) dx = c \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} f(x) dx + d \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} g(x) dx\). We used this when combining the two integrals for \({\rm{I}}_{\rm{m}} - {\rm{I}}_{{\rm{m}}-1}\).
• Additivity: \(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} f(x) dx = \mathop \smallint \nolimits_{\rm{a}}^{\rm{c}} f(x) dx + \mathop \smallint \nolimits_{\rm{c}}^{\rm{b}} f(x) dx\). (Not directly used here, but a fundamental property).
• Integral over zero length interval: \(\mathop \smallint \nolimits_{\rm{a}}^{\rm{a}} f(x) dx = 0\).
• Swapping limits: \(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} f(x) dx = - \mathop \smallint \nolimits_{\rm{b}}^{\rm{a}} f(x) dx\).

The integrand \(\frac{{\sin 2{\rm{mx}}}}{{\sin {\rm{x}}}}\) is well-behaved over the interval \([0, \pi]\) despite \(\sin {\rm{x}}\) being zero at the endpoints. As \({\rm{x}} \to 0\), \(\frac{{\sin 2{\rm{mx}}}}{{\sin {\rm{x}}}} \approx \frac{2{\rm{mx}}}{\rm{x}} = 2{\rm{m}}\). As \({\rm{x}} \to \pi\), let \({\rm{x}} = \pi - \epsilon\). \(\sin(2{\rm{m}}(\pi - \epsilon)) = \sin(2{\rm{m}}\pi - 2{\rm{m}}\epsilon) = \sin(2{\rm{m}}\pi)\cos(2{\rm{m}}\epsilon) - \cos(2{\rm{m}}\pi)\sin(2{\rm{m}}\epsilon) = 0 \cdot \cos(2{\rm{m}}\epsilon) - 1 \cdot \sin(2{\rm{m}}\epsilon) = -\sin(2{\rm{m}}\epsilon) \approx -2{\rm{m}}\epsilon\). \(\sin(\pi - \epsilon) = \sin \pi \cos \epsilon - \cos \pi \sin \epsilon = 0 \cdot \cos \epsilon - (-1) \cdot \sin \epsilon = \sin \epsilon \approx \epsilon\). So, \(\frac{\sin 2{\rm{mx}}}{\sin {\rm{x}}} \approx \frac{-2{\rm{m}}\epsilon}{\epsilon} = -2{\rm{m}}\) as \({\rm{x}} \to \pi^-\). Since the limits exist and are finite, the integral is proper.