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Evaluate: $$\underline{\quad\int _0 ^\pi x^3 \ln (\sin x) \mathrm d x\quad\:}\\{\int _0 ^\pi x^2 \ln(\sqrt 2 \sin x) \mathrm d x}$$

I tried expressing the integral in the numerator as a multiple of the integral in the denominator. To do this, I used the following property of definite integrals:

\begin{align} \int _a^b f(x) \mathrm d x &= \int _a ^b f(a+b-x) \mathrm d x \\ &\Downarrow \\ 2 \int _a^b f(x) \mathrm d x &= \int _a^b f(x) \mathrm d x + \int _a ^b f(a+b-x) \mathrm d x \tag{1} \end{align}

Applying this property to the integral in the numerator, we get:

\begin{align} \int _0 ^\pi x^3 \ln (\sin x) \mathrm d x &= \frac 1 2 \int _0 ^\pi ((\pi-x)^3 + x^3) \ln (\sin x) \mathrm d x\\ &=\frac 1 2 \int _0 ^\pi (\pi ^3 -3\pi^2 x +3\pi x^2)\ln (\sin x) \mathrm d x \tag{2} \end{align}

Now if we apply property $(1)$ to the integral

$$\int _0 ^\pi x^2 \ln (\sin x) \mathrm d x$$

then we see that

$$\int _0 ^\pi x^2 \ln (\sin x) \mathrm d x = \int _0 ^\pi (\pi-x)^2 \ln (\sin x) \mathrm d x$$

Thus upon simplifying, we get

$$\frac{\pi}{2}\int _0 ^\pi \ln (\sin x) \mathrm d x= \int _0 ^\pi x \ln (\sin x) \mathrm d x \tag{3}$$

Now substituting this result in equation $(3)$, we get

$$\int _0 ^\pi x^3 \ln (\sin x) \mathrm d x =\frac 1 2 \int _0 ^\pi \left(3\pi x^2 - \frac{\pi^3}{2}\right) \ln (\sin x) \mathrm d x$$

Now I have reduced the integral in the numerator in two two terms, one ($x^2$ term) which is a multiple of the term in the denominator and other which isn't. And sadly, I am not able to go any further at simplifying it. So how do I express the integral in the numerator as a multiple of the integral in the denominator? Is this approach of converting to a multiple, the only approach? Can we directly compute the integrals and then take their ratio? I also realize that computing the exact value of both of the integrals and then taking their ratio will be cumbersome and would also involve applying a few limits as the givven integrals are improper integrals.

I am looking for answers which continue the solution from the point where I left and lead me to the final answer. And just in case if you need to know the answer, then it is:

$$\frac{3\pi}{2}$$

1 Answers1

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After a bit scribbling and help from the folks at the h bar, I found the solution. First of all, I am going to use the fact that

$$\int _0 ^π \ln(\sin x) \, \mathrm d x = -\pi \ln 2$$

Using this, we can simplify the second term (the term without any powers of $x$). So,

$$\int_0 ^\pi \frac{\pi^3}{2} \ln(\sin x) \, \mathrm d x = -\frac{\pi^4}{2} \ln 2 = -\pi^4 \ln \sqrt 2$$

Now we can express this in terms of a definite integral,

$$-3\pi\int_0 ^\pi x^2 \ln \sqrt 2 \, \mathrm d x= -\pi^4 \ln \sqrt 2 $$

Substituting this integral in the integral in the numerator, we get

$$\frac{3\pi}{2}\int _0 ^\pi x^2 \ln(\sqrt 2 \sin x) \, \mathrm d x$$

And thus the ratio comes out to be $3\pi/2$, as expected.

Michael Hardy
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