Proving Integrals with Floor Functions and Radical Fractions: A Comprehensive Guide

When dealing with integrals that involve floor functions and radical fractions, we need to carefully analyze the behavior of these functions over specific intervals. This article explores how to evaluate such integrals, focusing on a detailed method involving the breakdown of the integral into subintervals.

Understanding Floor Functions and Radical Fractions

Floor functions, denoted by lfloor x rfloor, represent the greatest integer less than or equal to x. For instance, lfloor 5.9 rfloor 5. These functions can cause jumps at integer values, and they make the integral evaluation more complex when combined with radical fractions like x/sqrt{2} and x/sqrt{3}.

Similarly, radical fractions create additional points where the value of the integrand changes. These points are crucial for breaking down the integral into manageable parts.

Method for Integration

Identify Key Points: Determine the points at which the floor functions and radical fractions cause a jump in the integrand. For this example, these points are x 1, 2, 3, 4, sqrt{2}, 2sqrt{2}, 3sqrt{2}, sqrt{3}, 2sqrt{3}, 3sqrt{3}, 4, 2sqrt{3}, 3sqrt{2}, 5. Define Subintervals: Split the original interval into smaller subintervals based on these key points. In our case, we have the subintervals I_1 [0, 1], I_2 [1, sqrt{2}], I_3 [sqrt{2}, sqrt{3}], I_4 [sqrt{3}, 2], ... , I_{10} [3sqrt{2}, 5]. Increase the Integral: Evaluate the integral within each subinterval, considering the value of the floor functions within that interval. For instance, in the interval I_1 [0, 1], the floor functions are zero, so the integral simplifies to a constant value.

Examples of Integration

Interval I_1 [0, 1]:

displaystyle int_{I_1} -1^{lfloor x rfloor lfloor x/sqrt{2} rfloor lfloor x/sqrt{3} rfloor} dx int_0^1 -1^{0 0 0} dx 1.

Interval I_2 [1, sqrt{2}]:

displaystyle int_{I_2} -1^{lfloor x rfloor lfloor x/sqrt{2} rfloor lfloor x/sqrt{3} rfloor} dx int_1^{sqrt{2}} -1^{1 0 0} dx 1 - sqrt{2}.

Interval I_3 [sqrt{2}, sqrt{3}]:

displaystyle int_{I_3} -1^{lfloor x rfloor lfloor x/sqrt{2} rfloor lfloor x/sqrt{3} rfloor} dx int_{sqrt{2}}^{sqrt{3}} -1^{1 1 0} dx sqrt{3} - sqrt{2}.

Interval I_4 [sqrt{3}, 2]:

displaystyle int_{I_4} -1^{lfloor x rfloor lfloor x/sqrt{2} rfloor lfloor x/sqrt{3} rfloor} dx int_{sqrt{3}}^2 -1^{1 1 1} dx sqrt{3} - 2.

Interval I_5 [2, 2sqrt{2}]:

displaystyle int_{I_5} -1^{lfloor x rfloor lfloor x/sqrt{2} rfloor lfloor x/sqrt{3} rfloor} dx int_2^{2sqrt{2}} -1^{2 1 1} dx 2sqrt{2} - 2.

Interval I_6 [2sqrt{2}, 3]:

displaystyle int_{I_6} -1^{lfloor x rfloor lfloor x/sqrt{2} rfloor lfloor x/sqrt{3} rfloor} dx int_{2sqrt{2}}^3 -1^{2 2 1} dx 2sqrt{2} - 3.

Interval I_7 [3, 2sqrt{3}]:

displaystyle int_{I_7} -1^{lfloor x rfloor lfloor x/sqrt{2} rfloor lfloor x/sqrt{3} rfloor} dx int_3^{2sqrt{3}} -1^{3 2 1} dx 2sqrt{3} - 3.

Interval I_8 [2sqrt{3}, 4]:

displaystyle int_{I_8} -1^{lfloor x rfloor lfloor x/sqrt{2} rfloor lfloor x/sqrt{3} rfloor} dx int_{2sqrt{3}}^4 -1^{3 2 2} dx 2sqrt{3} - 4.

Interval I_9 [4, 3sqrt{2}]:

displaystyle int_{I_9} -1^{lfloor x rfloor lfloor x/sqrt{2} rfloor lfloor x/sqrt{3} rfloor} dx int_4^{3sqrt{2}} -1^{4 2 2} dx 3sqrt{2} - 4.

Interval I_{10} [3sqrt{2}, 5]:

displaystyle int_{I_{10}} -1^{lfloor x rfloor lfloor x/sqrt{2} rfloor lfloor x/sqrt{3} rfloor} dx int_{3sqrt{2}}^5 -1^{4 3 2} dx 3sqrt{2} - 5.

Final Calculation

Since the intervals I_1 I_2 ... I_{10} form a pairwise disjoint union of the interval [0, 5], we can sum the integrals of each subinterval to get the final answer:

displaystyle int_0^5 -1^{lfloor x rfloor lfloor x/sqrt{2} rfloor lfloor x/sqrt{3} rfloor} dx sum_{k1}^{10} int_{I_k} -1^{lfloor x rfloor lfloor x/sqrt{2} rfloor lfloor x/sqrt{3} rfloor} dx boxed{-21 6sqrt{3} 8sqrt{2}}.

This comprehensive step-by-step approach ensures a clear and accurate evaluation of the integral, making it a valuable tool for solving similar problems in mathematics.