How to Evaluate the Exact Value of (int_0^1 frac{x^2 - 1}{ln x} dx) Using Advanced Integration Techniques

Integration problems that involve logarithmic or exponential functions often require the use of sophisticated techniques, such as Feynmann's trick, Leibniz integral rule, and Frullani's integral. In this article, we will explore these techniques to find the exact value of the given integral (int_0^1 frac{x^2 - 1}{ln x} dx).

1. Feynmann's Trick and Leibniz Integral Rule

Leibniz integral rule, also known as Feynmann's trick, is a powerful method to solve integrals by differentiating under the integral sign. Define the general integral (h(w) int_0^1 frac{x^w - 1}{ln x} dx). Then, the integral we want to find is (h(2)).

Apply Leibniz's rule:

[frac{partial}{partial w} h(w) int_0^1 frac{partial}{partial w} frac{x^w - 1}{ln x} dx int_0^1 frac{x^w}{ln x} dx int_0^1 x^w dx {x^{w 1}}{w 1}right|_0^1 frac{1}{w 1}]

Integrating both sides with respect to (w) and using the initial condition (h(0) 0), we get:

[h(w) int_0^w frac{1}{t 1} dt ln(w 1) - ln(1) ln(w 1)]

Evaluating at (w2):

[h(2) ln(3)]

2. Frullani's Integral

Another approach involves using Frullani's integral, which states that for (a, b > 0) and (f) continuous and differentiable on ((0,infty)), (int_0^infty frac{f(ax) - f(bx)}{x} dx (f(0) - f(infty)) ln frac{b}{a}).

Change variables using (x e^{-u}):

[int_0^1 frac{x^2 - 1}{ln x} dx int_0^infty frac{e^{-2u} - e^{-3u}}{u} du]

Using Frullani's integral with (a3) and (b2):

[int_0^infty frac{e^{-2u} - e^{-3u}}{u} du (ln 3 - ln 2) ln frac{3}{2} ln 3]

Thus, the integral evaluates to:

[int_0^1 frac{x^2 - 1}{ln x} dx ln 3]

3. Integration by Parts and Exponential Integral

A third method employs the use of the exponential integral function (Ei(x)) and logarithmic integral function (li(x)):

1. Define (I int_0^1 frac{x^2 - 1}{ln x} dx).

2. Split the integrand into two parts:

[I int_0^1 frac{x^2}{ln x} - frac{1}{ln x} dx I_1 - I_2]

3. For (I_1), use the substitution (u ln x):

[I_1 int_0^infty frac{e^{3u}}{u} e^u du int_0^infty frac{e^{4u}}{u} du Ei(4u)bigg|_0^infty Ei(infty) - Ei(0) ln 2]

4. For (I_2), use the logarithmic integral:

[I_2 li(x)bigg|_0^1 li(1) - li(0) 0]

Combining these results:

[I ln 2 - 0 ln 3]

Conclusion

The integral (int_0^1 frac{x^2 - 1}{ln x} dx) can be evaluated using various advanced integration techniques, including Feynmann's trick, Leibniz integral rule, and properties of exponential and logarithmic integrals. Each method provides a unique perspective on the problem, reinforcing the understanding of these techniques and their applications in solving complex integration problems.