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To solve the integral:

$I={\int }_0^{\pi }\log \left(\sin \right(x)dx$

We need to proceed carefully. This integral is famous for having a nontrivial solution, and the derivation involves several steps, including symmetry considerations and the use of advanced techniques. Below is a detailed step-by-step solution.

Step 1: Symmetry of the Function

The sine function is symmetric about $\mathrm{\pi }/2$ in the interval $0$ to $\pi$. This symmetry allows us to rewrite the integral over half the interval:

$I=2{\int }_0^{\frac{\pi }{2}}\log \left(\sin \right(x)dx$

This step reduces the limits of integration from 0 to π/2 instead of 0 to π, making the next steps more manageable.

Step 2: Use of a Known Identity

There is a standard trick used to deal with integrals involving logarithms and trigonometric functions. We introduce a known identity for integrals involving logarithmic functions:

${\int }_0^{\frac{\pi }{2}}\log \left(\sin \right(x)dx={\int }_0^{\frac{\pi }{2}}\log (\cos \left(x\right)dx$

This symmetry between sine and cosine integrals will be useful later.

Step 3: Combining the Integrals of log(sin(x)) and log(cos(x))

Now, consider the sum of these two integrals:

$S={\int }_0^{\frac{\pi }{2}}\log \left(\sin \right(x)dx+{\int }_0^{\frac{\pi }{2}}\log (\cos \left(x\right)dx$

Using the identity $\mathrm{sin(x)\; cos(x)\; =\; (1/2)sin(2x)}$, we get:

$S={\int }_0^{\frac{\pi }{2}}\log \left(\frac{1}{2}\mathrm{sin(2x)}\right)dx$

This can be expanded as:

$S={\int }_0^{\frac{\pi }{2}}\log \left(\frac{1}{2}\right)dx+{\int }_0^{\frac{\pi }{2}}\log \left(\mathrm{sin(2x)}\right)dx$

Step 4: Simplifying the Result

The first term is straightforward:

${\int }_0^{\frac{\pi }{2}}\log \left(\frac{1}{2}\right)dx=-\frac{\pi }{2}\mathrm{log(2)}$

The second term involves a change of variables. Let $u = my aniket roy August 28, 2023 Contact Us Email : aniketkroy@gmail.com Send Message Search Powered by Blogger. Blog Archive August 2023 (2) Report Abuse Home $