Cracking the Hardest Integral: A Step-by-Step Guide Using Dilogarithms

In this article, we will explore the intricacies of finding the integral of (int ln(1-sin x) , dx). This integral is considered difficult due to its trigonometric and logarithmic components, and it benefits significantly from the use of (text{dilogarithm function}) in the solution. Let's dive into the detailed steps and the final result.

The Problem at Hand

The integral we are dealing with is:

(int ln(1-sin x) , dx)

Initial Replacements and Simplifications

First, we start by making a substitution to simplify the integral. Let’s use the identity (sin x 1 - 2cos^2 y) which gives us:

(int ln(1-sin x) , dx -2 int ln(1-cos 2y) , dy -2 int ln(2cos^2 y) , dy -2 int ln(2) cdot ln(cos y) , dy 2y cdot ln(2))

Applying the Dilogarithm Function

To further simplify the integral, we use the dilogarithm function (text{Li}_2(z)) which is defined as:

(text{Li}_2(z) -int_0^z frac{ln(1-t)}{t} , dt)

By applying this function, we can express:

(int ln(1-cos 2y) , dy int ln left[2e^{2yi}left(1 - e^{-2yi}right)right] , dy int ln(2) , dy int ln left(1 - e^{-2yi}right) , dy)

Then, we can integrate the second part using the (text{dilogarithm function}):

(int ln(1 - e^{-2yi}) , dy -frac{i}{2} text{Li}_2left(-e^{-2yi}right))

Putting it all together, we get:

(int ln(1-sin x) , dx -2 left(yi - frac{i}{2} text{Li}_2left(-e^{-2yi}right)right) 2y ln(2))

The Final Form

By substituting back the original variable, we obtain:

(int ln(1-sin x) , dx -4 text{Im} left(frac{1}{2} text{Li}_2left(i e^{xi}right)right) - x ln(2))

This is the final form of the integral using dilogarithms.

Challenging the Integral of Squared Logarithm

Now, let’s tackle the integral of the squared logarithm:

(int frac{ln^2(1-sin x)}{1-sin x} , dx)

We begin by integrating by parts, where:

(int frac{1}{1-sin x} , dx int left(sec^2 x - tan x sec xright) , dx tan x - sec x)

Then applying integration by parts:

(I int ln(1-sin x) cdot tan x - sec x , dx)

We continue with further substitutions and simplifications, eventually leading to:

(int frac{ln^2(1-sin x)}{1-sin x} , dx tan x - sec x cdot ln^2(1-sin x) - 4 int frac{ln(1-sin x)}{1-sin x} , dx - 2 int ln(1-sin x) , dx)

After simplifying, we reach a final form that combines the dilogarithm function and trigonometric expressions:

(int frac{ln^2(1-sin x)}{1-sin x} , dx tan x - sec x left[ln^2(1-sin x) cdot 4 left(frac{sin x - 1}{cos x} - xright)right] - 2 left[sum_{k1}^{infty} frac{i^{k-1}}{k 2^k} left[sum_{r0}^k binom{k}{r} (-1)^r frac{e^{(k-2r)xi}}{k-2r}right]right])

This result represents the final, more complex, form of the integral.

Conclusion

The integral (int ln(1-sin x) , dx) is elegantly cracked using dilogarithms, showcasing the power of this special function in solving complex integrals. The integral of the squared logarithm, however, requires further simplifications and combines both trigonometric and special function results. By understanding each step, one can apply similar techniques to solve advanced integrals.