Evaluating Complex Integrals Using Logarithmic and Floor Functions

The problem at hand involves evaluating a complex integral by making strategic substitutions and utilizing properties of both the logarithmic and floor functions. This approach not only simplifies the integral but also provides a clear and structured way to determine its value.

Introduction

Let us consider the integral

[ int_0^1 frac{1}{lfloor 1 - log_2{t} rfloor} , dt. ]

Initially, the integral seems challenging due to the floor function and the logarithmic term. However, by employing a strategic substitution, we can significantly simplify the problem.

Alternative Substitution: t 1 - x

By defining the substitution ( t 1 - x ), we transform the integral into a more manageable form:

[ int_0^1 frac{1}{lfloor 1 - log_2{t} rfloor} , dt int_0^1 frac{1}{lfloor 1 - log_2{(1 - x)} rfloor} , dx. ]

Properties of Floor Function

The floor function, denoted by ( lfloor y rfloor ), yields the greatest integer less than or equal to ( y ). This property makes it possible to analyze the behavior of the integrand more systematically.

By examining the condition ( lfloor 1 - log_2{t} rfloor n ), we can derive the bounds for ( t ). Specifically,

[ n leq 1 - log_2{t}

implies

[ frac{1}{2^n} leq t

This translates the integral into a sum of simpler integrals:

[ int_0^1 frac{1}{lfloor 1 - log_2{t} rfloor} , dt sum_{n1}^{infty} int_{1/2^n}^{1/2^{n-1}} frac{1}{n} , dt. ]

Evaluation of the Integral

Each term in the sum can be evaluated directly:

[ int_{1/2^n}^{1/2^{n-1}} frac{1}{n} , dt frac{1}{n} left( frac{1}{2^{n-1}} - frac{1}{2^n} right) frac{1}{n cdot 2^n}. ]

Therefore, the integral becomes:

[ sum_{n1}^{infty} frac{1}{n cdot 2^n}. ]

Utilizing the Geometric Series

To evaluate this series, we use the geometric series:

[ frac{1}{1 - t} sum_{k0}^{infty} t^k. ]

By integrating both sides of this equation from ( 0 ) to ( x ), we obtain:

[ -ln(1 - t) sum_{k1}^{infty} frac{t^k}{k}. ]

Substituting ( t frac{1}{2} ) into this series, we find:

[ -lnleft(1 - frac{1}{2}right) sum_{k1}^{infty} frac{left(frac{1}{2}right)^k}{k} sum_{n1}^{infty} frac{1}{n cdot 2^n}. ]

Thus, the integral evaluates to:

[ int_0^1 frac{1}{lfloor 1 - log_2{t} rfloor} , dt -lnleft( frac{1}{2} right) ln(2). ]

Conclusion

In conclusion, the integral can be effectively evaluated by making a strategic substitution and utilizing the properties of the floor and logarithmic functions. This approach not only simplifies the problem but also provides a clear path to its solution, demonstrating the power of mathematical techniques in solving complex integrals.