Evaluating a Special Integral: Techniques and Insights

While discussing integrals can sometimes be as complex as navigating through a dense forest, there are methods and tools that can make the journey smoother. In this article, we'll delve into the process of evaluating a specific integral using a combination of substitution and algebraic manipulation. This integral is not just a problem to solve; it's an opportunity to explore the beauty and power of integral calculus.

Introduction to the Integral

The integral we will evaluate is:

[ int_{sqrt{2}}^{infty} frac{dx}{x^{sqrt{2}} - x} ]

At first glance, this integral might seem daunting, but with the right approach, we can simplify and solve it. The key is to carefully manipulate the integrand and use appropriate substitutions.

Step-by-Step Solution

First, let's rewrite the integrand in a more manageable form:

[ int_{sqrt{2}}^{infty} frac{dx}{x^{sqrt{2}} - x} int_{sqrt{2}}^{infty} frac{dx}{x(x^{sqrt{2} - 1} - 1)} ]

Now, we make a substitution to simplify the integrand. Let:

[ y x^{1 - sqrt{2}} - 1 ]

Then, the differential (dy) can be calculated as:

[ dy (1 - sqrt{2})x^{-sqrt{2}} dx ]

Therefore, we have:

[ x^{-sqrt{2}} dx frac{dy}{1 - sqrt{2}} ]

Now, we need to change the limits of integration. When (x sqrt{2}),

[ y (sqrt{2})^{1 - sqrt{2}} - 1 ]

And as (x to infty),

[ y to 1 ]

So, the integral becomes:

[ int_{(sqrt{2})^{1 - sqrt{2}} - 1}^{1} frac{1 - sqrt{2}}{y} dy ]

This simplifies to:

[ (1 - sqrt{2}) int_{(sqrt{2})^{1 - sqrt{2}} - 1}^{1} frac{1}{y} dy ]

The integral of (frac{1}{y}) is (ln|y|), so we have:

[ (1 - sqrt{2}) [ln|y|]_{(sqrt{2})^{1 - sqrt{2}} - 1}^{1} ]

Evaluating the limits, we get:

[ (1 - sqrt{2}) (ln 1 - ln((sqrt{2})^{1 - sqrt{2}} - 1)) ]

Since (ln 1 0), this simplifies to:

[ (1 - sqrt{2}) (-ln((sqrt{2})^{1 - sqrt{2}} - 1)) ]

Rationalizing the denominator, we get:

[ -frac{1}{sqrt{2} - 1} ln((sqrt{2})^{1 - sqrt{2}} - 1) ]

Multiplying both the numerator and the denominator by (sqrt{2} 1), we get:

[ -frac{(sqrt{2} 1) ln((sqrt{2})^{1 - sqrt{2}} - 1)}{(2 - 1)} ]

This simplifies to:

[ -(sqrt{2} 1) ln((sqrt{2})^{1 - sqrt{2}} - 1) ]

Thus, the value of the integral is:

[ -frac{ln((sqrt{2})^{1 - sqrt{2}} - 1)}{sqrt{2} - 1} ]

Conclusion

Through careful manipulation and substitution, we have successfully evaluated the given integral. This process showcases the power and elegance of integral calculus and how it can be used to solve complex problems efficiently.

Remember, patience and a systematic approach are crucial when dealing with integrals. Whenever you face a complex integral, break it down into smaller, manageable parts, and use the tools and techniques at your disposal.