Evaluating the Integral of sin(ln x)/ln x from 0 to 1

The integral in question is quite elegant and requires the application of several mathematical techniques to achieve a solution. This article will guide you through the step-by-step process of evaluating the integral [I int_{0}^{1} frac{sin(ln x)}{ln x} dx]

Using Substitution

We start with the integral:

[I int_{0}^{1} frac{sin(ln x)}{ln x} dx]

To simplify this integral, we use the substitution:

[x e^{-t}] This substitution leads to the following transformations:

When (x 0), (t to infty) When (x 1), (t 0) The differential (dx -e^{-t} dt)

Substituting these into the original integral:

[I int_{infty}^{0} frac{sin(-t)}{-t} e^{-t} (-dt)]

Since (sin(-t) -sin(t)), we can rewrite the integral as:

[I int_{infty}^{0} frac{-sin(t)}{-t} e^{-t} (-dt) int_{infty}^{0} frac{sin(t)}{t} e^{-t} dt]

Reversing the limits of integration:

[I int_{0}^{infty} frac{sin(t)}{t} e^{-t} dt]

This form of the integral is a well-known result in integral calculus:

[int_{0}^{infty} frac{sin(at)}{t} e^{-at} dt tan^{-1}left( frac{1}{a} right)]

For (a 1):

[I tan^{-1}(1) frac{pi}{4}]

The Laplace Transform Approach

We can also solve the integral using the concept of the Laplace transform. Let's define:

[I int_{0}^{1}frac{sin(ln x)}{ln x} dx]

Expressing the integral in terms of the Laplace transform, we have:

[I mathcal{L}{ln x | x 1}]

Let (g(x) frac{sin x}{x}) and (f(x) x g(x) sin x). The Laplace transform of (f(x)) is given by:

[mathcal{L}{sin x} frac{1}{1 p^2}]

Therefore, we have:

[mathcal{L}{g(x)} -arctan(p)]

Applying this to the definite integral:

[mathcal{L}{g(x)} -arctan(0) frac{pi}{2}]

Thus:

[I -arctan(1) frac{pi}{4}]

Complex Analysis Approach

Another method to evaluate the integral involves complex analysis. Let:

[mathcal{I} int_0^1 frac{sin(ln x)}{ln x} dx]

Using the expansion of (sin(ln x)) in terms of exponentials, we get:

[sin(ln x) frac{e^{iln x} - e^{-iln x}}{2i}]

Substituting this into the integral:

[mathcal{I} frac{1}{2} int_0^1 int_{-1}^{1} e^{it ln x} dt dx]

Letting the inner integral over (t) be a function of (x), we get:

[mathcal{I} frac{1}{2} int_0^1 int_{-1}^{1} x^{it} dt dx]

Evaluating the inner integral:

[int_{-1}^{1} x^{it} dt left. frac{x^{it 1}}{it 1} right|_{-1}^{1}]

Therefore:

[mathcal{I} frac{1}{2} int_0^1 left( frac{x^{i-1}}{i-1} - frac{x^{i 1}}{i 1} right) dx]

This can be simplified further:

[mathcal{I} frac{1}{2} int_{-1}^{1} frac{1}{1-t^2} dt arctan(1) - arctan(-1) frac{pi}{4}]

Conclusion

The integral (int_{0}^{1} frac{sin(ln x)}{ln x} dx frac{pi}{4}) can be effectively evaluated using various methods, including substitution, Laplace transform, and complex analysis. Each method provides a unique perspective on solving this integral, highlighting the beauty and power of mathematical techniques.