Evaluating the Integral of sin(x)/ln(x) Using Series Expansion and Feynman’s Trick

Integral evaluation often requires innovative methods when dealing with functions that are not straightforward to work with. In this article, we will evaluate the integral (int frac{sin x}{ln x} dx) by utilizing the series expansion of the sine function and Feynman’s trick. We will also explore the reasoning behind each step and discuss the nuances of the solution.

Introduction

The integral in question is:

[ I int frac{sin x}{ln x} dx ]

This integral is not an elementary one, meaning it cannot be expressed in terms of elementary functions. However, we can approach this problem by decomposing the sine function and using advanced techniques such as series expansion and Feynman’s parametrization method.

Series Expansion of sin(x)

The sine function can be expressed as a Taylor series:

[ sin x sum_{k0}^{infty} frac{(-1)^k x^{2k 1}}{(2k 1)!} ]

Substituting this into the integral, we obtain:

[ I int frac{sin x}{ln x} dx int frac{sum_{k0}^{infty} frac{(-1)^k x^{2k 1}}{(2k 1)!}}{ln x} dx ]

This can be rewritten as:

[ I sum_{k0}^{infty} frac{(-1)^k}{(2k 1)!} int frac{x^{2k 1}}{ln x} dx ]

Feynman’s Trick

To simplify the problem, we introduce a parameter (a) and consider a more general integral:

[ I(a) int frac{sin x}{ln x} x^a dx ]

By differentiating (I(a)) with respect to (a), we can simplify the integrand:

[ frac{d}{da} I(a) int sin x , ln x cdot x^a dx ]

Recognizing that the derivative of (sin x / ln x) with respect to (a) gives us a simpler expression to integrate, we proceed as follows:

Maclaurin Series and Indefinite Integrals

Using the Maclaurin series for (sin x), we can substitute and integrate term by term:

[ I(a) int left( sum_{k0}^{infty} frac{(-1)^k x^{2k 1} x^a}{(2k 1)!} right) frac{1}{ln x} dx ]

This simplifies to:

[ I(a) sum_{k0}^{infty} frac{(-1)^k}{(2k 1)!} int frac{x^{2k a 1}}{ln x} dx ]

By making a substitution (u ln x), we can solve the integral:

[ int frac{x^{2k a 1}}{ln x} dx int frac{e^u (2k a 1)}{u} du ]

This integral involves the exponential integral (mathrm{Ei}(u)), which is defined as:

[ mathrm{Ei}(u) int_{-u}^{infty} frac{e^{-t}}{t} dt ]

Thus, the integral can be expressed as:

[ I(a) -a sum_{k0}^{infty} frac{(-1)^k mathrm{Ei}((2k a 1) ln x)}{(2k 1)!} C ]

Setting (a 0), we find the solution for the original integral:

[ I - sum_{k0}^{infty} frac{(-1)^k mathrm{Ei}((2k 1) ln x)}{(2k 1)!} C ]

Conclusion

The integral (int frac{sin x}{ln x} dx) can be evaluated using the series expansion of the sine function and Feynman’s trick. Despite the complexity, this method allows us to express the solution in terms of the exponential integral function.

It is important to note that while the series representation provides a solution, the rate of convergence may be slow. Nonetheless, it offers a viable approach when traditional methods fail.