Integrand with no Indefinite Integral (2)

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

Note by Ethan Robinett
3 years, 8 months ago

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