We will prove the following:

If \(I = \displaystyle \int_{0}^{\frac{\pi}{2}} ln(sin(x))\,dx\) , then:

(1) \(I = -\frac{\pi ln(2)}{2}\)

(2) \(I =\displaystyle \int_{0}^{\frac{\pi}{2}} ln(cos(x))\,dx\)

(3) \(\displaystyle \int_{0}^{\frac{\pi}{2}} ln(tan(x))\,dx=0\)

Proof:

We may rewrite \(sin(x)\) as:

\(sin(x) = 2cos(\frac{x}{2})sin(\frac{x}{2})\)

Then \(I = \displaystyle \int_{0}^{\frac{\pi}{2}} ln(2cos(\frac{x}{2})sin(\frac{x}{2}))\,dx\)

\(\Rightarrow\) \(I = \displaystyle \int_{0}^{\frac{\pi}{2}} ln(2)\,dx + \displaystyle \int_{0}^{\frac{\pi}{2}} ln(cos(\frac{x}{2}))\,dx + \displaystyle \int_{0}^{\frac{\pi}{2}} ln(sin(\frac{x}{2}))\,dx\)

\(\Rightarrow\) \(I= \frac{\pi ln(2)}{2} + \displaystyle \int_{0}^{\frac{\pi}{2}} ln(cos(\frac{x}{2}))\,dx + \displaystyle \int_{0}^{\frac{\pi}{2}} ln(sin(\frac{x}{2}))\,dx\)

Letting \(u=\frac{x}{2}\)

\(\Rightarrow\) \(du =\frac{dx}{2}\)

\(\Rightarrow\) \(I = \frac{\pi ln(2)}{2} + 2\displaystyle \int_{0}^{\frac{\pi}{4}} ln(cos(u))\,du + 2\displaystyle \int_{0}^{\frac{\pi}{4}} ln(sin(u))\,du\)

Now consider \(2\displaystyle \int_{0}^{\frac{\pi}{4}} ln(cos(u))\,du \). If we let \(u = \frac{\pi}{2} - v\), \(du = -dv\), then:

\(2\displaystyle \int_{0}^{\frac{\pi}{4}} ln(cos(u))\,du = -2\displaystyle \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} ln(cos(\frac{\pi}{2} - v))\,dv = 2\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} ln(sin(v))\,dv\)

Then:

\(I= \frac{\pi ln(2)}{2} + 2\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} ln(sin(v))\,dv + 2\displaystyle \int_{0}^{\frac{\pi}{4}} ln(sin(u))\,du\)

Now we must note that the choice of variables \(u\) and \(v\) is arbitrary. The integrand is exactly the same, hence we may replace both with \(x\), arriving at:

\(I = \frac{\pi ln(2)}{2} +2\displaystyle \int_{0}^{\frac{\pi}{2}} ln(sin(x))\,dx\)

\(\Rightarrow\) \(I= \frac{\pi ln(2)}{2} + 2I\)

From which it follows that:

\(I = -\frac{\pi ln(2)}{2}\)

This proves (1).

Now, note that:

\(\displaystyle \int_{0}^{\frac{\pi}{2}} ln(sin(x))\,dx = \displaystyle \int_{0}^{\frac{\pi}{2}} ln(sin(\frac{\pi}{2}-x))\,dx =\displaystyle \int_{0}^{\frac{\pi}{2}} ln(cos(x))\,dx\)

This proves (2).

Now, since (2) has been proven we may write:

\(\displaystyle \int_{0}^{\frac{\pi}{2}} ln(sin(x))\,dx - \displaystyle \int_{0}^{\frac{\pi}{2}} ln(cos(x))\,dx = 0\)

\(\Rightarrow\) \(\displaystyle \int_{0}^{\frac{\pi}{2}} ln\left(\frac{sin(x)}{cos(x)}\right)\,dx = 0\)

\(\Rightarrow\) \(\displaystyle \int_{0}^{\frac{\pi}{2}} ln(tan(x))\,dx = 0\)

Which proves (3).

QED

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