Question

How do you prove that the integral of ln(sinx) on the interval $ \left[ 0,\dfrac{\pi }{2} \right] $ is convergent?

Answer 1

Hint: In this question, we need to prove that integral of ln(sinx) on the interval $ \left[ 0,\dfrac{\pi }{2} \right] $ is convergent. For this, we will try to solve the integral $ \int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x \right)dx} $ and prove that it is equal to a finite value. We will use the following properties for solving the integral
(I) Substitution property.
(II) Property of logarithm: $ {{\log }_{b}}m+{{\log }_{b}}n={{\log }_{b}}\left( mn \right) $ .
(III) Formula for sine of double angle: sin2x = 2sinxcosx.
(IV) \[\int\limits_{a}^{b}{f\left( x \right)dx}+\int\limits_{a}^{b}{g\left( x \right)dx}=\int\limits_{a}^{b}{\left( f\left( x \right)+g\left( x \right) \right)dx}\].
(V) \[\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{c}{f\left( x \right)dx}+\int\limits_{c}^{b}{f\left( x \right)dx}\] where a(VI) \[\int\limits_{a}^{b}{f\left( x \right)dx}=-\int\limits_{b}^{a}{f\left( x \right)dx}\].

Complete step by step answer:
Here we need to prove that integral of ln(sinx) on the interval $ \left[ 0,\dfrac{\pi }{2} \right] $ is convergent.
Therefore we need to prove that $ \int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x \right)dx} $ is some finite value. Let us calculate it. We have the integral as $ \int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x \right)dx} $ . Suppose it to be equal to S. So, $ S=\int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x \right)dx}\ldots \ldots \ldots \left( 1 \right) $ .
Let us substitute $ t=\dfrac{\pi }{2}-x $ .
Taking derivative on both sides dt = -dx.
Putting x = 0, $ t=\dfrac{\pi }{2}-0\Rightarrow t=\dfrac{\pi }{2} $ .
Putting $ x=\dfrac{\pi }{2},t=\dfrac{\pi }{2}-\dfrac{\pi }{2}\Rightarrow t=0 $ .
Our integral becomes $ S=-\int\limits_{\dfrac{\pi }{2}}^{0}{\ln \left( \sin \left( \dfrac{\pi }{2}-x \right) \right)dx} $ .
We know that $ \sin \left( \dfrac{\pi }{2}-\theta \right) $ is equal to $ \cos \theta $ . So we get $ S=-\int\limits_{\dfrac{\pi }{2}}^{0}{\ln \left( \cos x \right)dx} $ .
Now changing the limits and removing negative sign using the formula, \[\int\limits_{a}^{b}{f\left( x \right)dx}=-\int\limits_{b}^{a}{f\left( x \right)dx}\] we get, $ S=\int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \cos x \right)dx}\ldots \ldots \ldots \left( 2 \right) $ .
Now let us add integrals (1) and (2) we get,
 $ S+S=\int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x \right)dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \cos x \right)dx} $ .
Applying \[\int\limits_{a}^{b}{f\left( x \right)dx}+\int\limits_{a}^{b}{g\left( x \right)dx}=\int\limits_{a}^{b}{\left( f\left( x \right)+g\left( x \right) \right)dx}\] we get,
 $ 2S=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \ln \left( \sin x \right)+\ln \left( \cos x \right) \right)dx} $ .
We know that $ \ln \left( m \right)+\ln \left( n \right)=\ln \left( mn \right) $ so we get,
 $ 2S=\int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x\cos x \right)dx} $ .
Let us use the formula for sine of double angle $ \sin 2x=2\sin \cos x\Rightarrow \dfrac{\sin 2x}{2}=\sin x\cos x $ we get,
 $ 2S=\int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \dfrac{\sin 2x}{2} \right)dx} $ .
Using $ \ln \left( m \right)+\ln \left( n \right)=\ln \left( mn \right) $ to separate $ \sin 2x\text{ and }\dfrac{1}{2} $ we get, $ \begin{align}
  & 2S=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \ln \left( \sin 2x \right)+\ln \left( \dfrac{1}{2} \right) \right)dx} \\
 & \Rightarrow 2S=\int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \dfrac{1}{2} \right)dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin 2x \right)dx}\cdots \cdots \cdots \left( 3 \right) \\
\end{align} $ .
Let us solve both integrals separately, $ \int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \dfrac{1}{2} \right)dx} $ .
Since $ \ln \left( \dfrac{1}{2} \right) $ is constant term, so we will have \[\int\limits_{0}^{\dfrac{\pi }{2}}{\operatorname{l}\cdot dx=\left[ x \right]_{0}^{\dfrac{\pi }{2}}}\] we get,
 $ \ln \left( \dfrac{1}{2} \right)\left[ x \right]_{0}^{\dfrac{\pi }{2}}\Rightarrow \ln \left( \dfrac{1}{2} \right)\left[ \dfrac{\pi }{2}-0 \right]\Rightarrow \dfrac{\pi }{2}\ln \left( \dfrac{1}{2} \right) $ .
Therefore, \[\int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \dfrac{1}{2} \right)dx=\dfrac{\pi }{2}\ln \left( \dfrac{1}{2} \right)\cdots \cdots \cdots \left( 4 \right)}\]
Now let us solve other terms,
 $ \int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2xdx} $ .
As the function is defined on $ x=\dfrac{\pi }{4} $ so we can apply the property \[\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{c}{f\left( x \right)dx}+\int\limits_{c}^{b}{f\left( x \right)dx}\], a\[\int\limits_{0}^{\dfrac{\pi }{2}}{\sin \left( 2x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\sin \left( 2x \right)dx}+\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\sin \left( 2x \right)dx}\].
In the first term substituting $ t=2x,dt=2dx\Rightarrow \dfrac{dt}{2}=dx $ .
Putting x = 0, t = 0.
 $ x=\dfrac{\pi }{4},t=2\times \dfrac{\pi }{4}\Rightarrow t=\dfrac{\pi }{2} $ .
In the second term, substituting $ t=2x-\dfrac{\pi }{2},dt=2dx\Rightarrow \dfrac{dt}{2}=dx $ .
Putting $ x=\dfrac{\pi }{4},t=2\times \dfrac{\pi }{4}-\dfrac{\pi }{2}=\dfrac{\pi }{2}-\dfrac{\pi }{2}=0 $ .
Putting $ x=\dfrac{\pi }{2},t=2\times \dfrac{\pi }{2}-\dfrac{\pi }{2}=\pi -\dfrac{\pi }{2}=\dfrac{\pi }{2} $ .
So our integral after all substitution becomes,
\[\begin{align}
  & \int\limits_{0}^{\dfrac{\pi }{2}}{\sin t}\dfrac{dt}{2}+\int\limits_{0}^{\dfrac{\pi }{2}}{\sin \left( t+\dfrac{\pi }{2} \right)\dfrac{dt}{2}} \\
 & \Rightarrow \dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin t}dt+\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin \left( t+\dfrac{\pi }{2} \right)dt} \\
\end{align}\].
We know $ \sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $ so we get,
\[\Rightarrow \dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin t}dt+\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\cos tdt}\].
Changing variable from t to x we get,
\[\Rightarrow \dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x}dx+\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\cos xdx}\].
From equation (1) and (2) we have,
\[\Rightarrow \dfrac{1}{2}S+\dfrac{1}{2}S\Rightarrow S\].
So our integral $ \int\limits_{0}^{\dfrac{\pi }{2}}{\sin \left( 2x \right)dx} $ becomes equal to S.
 $ \int\limits_{0}^{\dfrac{\pi }{2}}{\sin \left( 2x \right)dx}=S\cdots \cdots \cdots \left( 5 \right) $ .
Putting values from (4) and (5) into (3) we get,
 $ \begin{align}
  & 2S=\dfrac{\pi }{2}\ln \left( \dfrac{1}{2} \right)+S \\
 & \Rightarrow 2S-S=\dfrac{\pi }{2}\ln \left( \dfrac{1}{2} \right) \\
 & \Rightarrow S=\dfrac{\pi }{2}\ln \left( \dfrac{1}{2} \right) \\
\end{align} $ .
Therefore $ \int\limits_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x \right)dx}=\dfrac{\pi }{2}\ln \left( \dfrac{1}{2} \right) $ .
Since $ \dfrac{\pi }{2}\text{ and }\ln \left( \dfrac{1}{2} \right) $ are both finite values therefore value of this integral is finite.
Hence the integral of ln(sinx) in the interval $ \left( 0,\dfrac{\pi }{2} \right) $ is covergent.

Note:
Students should keep in mind all the integral and trigonometric properties for solving this sum. Make sure to mention each property wherever used. Note that, these is definite integral properties only and we are not using any constant term (C) after evaluating integral.