Evaluate Integral Of Arctan(x) / (2+x^2) From 0 To Infinity

This article delves into the process of evaluating the definite integral:

$$I = \int_{0}^{\infty} \frac{\arctan{x}}{2+x^2} dx$$

This type of integral often appears in calculus and requires a clever application of substitution and trigonometric identities to solve. Let's embark on this mathematical journey step by step.

Initial Substitution: Unveiling the Trigonometric Form

To begin, a strategic substitution is crucial. Let's follow the initial step suggested and set:

$$u = \arctan{x}$$

This substitution immediately implies:

$$x = \tan{u}$$

Now, we need to find the differential dx in terms of du. Differentiating both sides of the equation x = tan(u) with respect to u, we get:

$$\frac{dx}{du} = \sec^2{u}$$

Thus,

$$dx = \sec^2{u} du$$

Next, we must adjust the limits of integration. When x = 0, we have:

$$u = \arctan{0} = 0$$

And as x approaches infinity (x → ∞), u approaches π/2:

$$u = \arctan{\infty} = \frac{\pi}{2}$$

Now, we can rewrite the integral I in terms of u:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u}{2 + \tan^2{u}} \sec^2{u} du$$

Recall the trigonometric identity:

$$\sec^2{u} = 1 + \tan^2{u}$$

Substituting this into our integral, we get:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u(1 + \tan^2{u})}{2 + \tan^2{u}} du$$

This form, while seemingly more complex, sets the stage for further simplification. We have successfully transformed the original integral into a trigonometric form, which is a significant step forward. The key here is to recognize the relationship between arctan(x) and tan(u) and how their derivatives connect the differentials dx and du. This substitution allows us to work within the realm of trigonometric functions, where we can leverage a plethora of identities and techniques to our advantage. The next step involves manipulating the integrand to a more manageable form, possibly by employing another substitution or trigonometric identity. The strategic use of substitution is a cornerstone of integral calculus, and this example perfectly illustrates its power in transforming complex integrals into solvable ones. The careful adjustment of the integration limits is equally crucial, ensuring that the integral accurately reflects the original problem in the new variable. The journey to solving this integral is a testament to the beauty and elegance of calculus, where seemingly intractable problems can be conquered with the right tools and techniques.

Simplifying the Trigonometric Integral: A Strategic Transformation

Our integral now stands as:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u(1 + \tan^2{u})}{2 + \tan^2{u}} du$$

To further simplify this, let's express tan²(u) in terms of sine and cosine:

$$\tan^2{u} = \frac{\sin^2{u}}{\cos^2{u}}$$

Substituting this back into the integral, we have:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u(1 + \frac{\sin^2{u}}{\cos^2{u}})}{2 + \frac{\sin^2{u}}{\cos^2{u}}} du$$

Now, let's multiply both the numerator and the denominator by cos²(u) to clear the fractions:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u(\cos^2{u} + \sin^2{u})}{2\cos^2{u} + \sin^2{u}} du$$

Using the fundamental trigonometric identity sin²(u) + cos²(u) = 1, the integral simplifies to:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u}{2\cos^2{u} + \sin^2{u}} du$$

This form is much cleaner. Now, we can further manipulate the denominator. Recall the identity cos²(u) = 1 - sin²(u). Substituting this in, we get:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u}{2(1 - \sin^2{u}) + \sin^2{u}} du$$

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u}{2 - 2\sin^2{u} + \sin^2{u}} du$$

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u}{2 - \sin^2{u}} du$$

This form of the integral is significantly more manageable. We have successfully transformed the integrand into a form that is amenable to further analysis. The key steps in this simplification involved expressing tan²(u) in terms of sin²(u) and cos²(u), clearing the fractions by multiplying the numerator and denominator by cos²(u), and utilizing the fundamental trigonometric identity sin²(u) + cos²(u) = 1. The strategic use of trigonometric identities is crucial in simplifying trigonometric integrals, and this example demonstrates how these identities can be used to transform a complex integrand into a simpler form. The next step might involve another substitution or a more advanced integration technique. The journey of solving this integral highlights the power of algebraic manipulation and trigonometric identities in simplifying complex mathematical expressions. Each step brings us closer to the final solution, revealing the underlying structure of the integral and the elegant interplay of mathematical concepts.

The Second Substitution: A New Perspective

We've arrived at the integral:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u}{2 - \sin^2{u}} du$$

Now, let's divide both the numerator and denominator by cos²(u):

$$I = \int_{0}^{\frac{\pi}{2}} \frac{\frac{u}{\cos^2{u}}}{\frac{2}{\cos^2{u}} - \frac{\sin^2{u}}{\cos^2{u}}} du$$

This gives us:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u \sec^2{u}}{2\sec^2{u} - \tan^2{u}} du$$

Using the identity sec²(u) = 1 + tan²(u), we can rewrite the integral as:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u \sec^2{u}}{2(1 + \tan^2{u}) - \tan^2{u}} du$$

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u \sec^2{u}}{2 + 2\tan^2{u} - \tan^2{u}} du$$

$$I = \int_{0}^{\frac{\pi}{2}} \frac{u \sec^2{u}}{2 + \tan^2{u}} du$$

Now, a second substitution will prove beneficial. Let:

$$v = \tan{u}$$

Then:

$$dv = \sec^2{u} du$$

When u = 0, v = tan(0) = 0. When u = π/2, v approaches infinity (v → ∞). So our new limits of integration are from 0 to ∞.

Substituting these into the integral, we get:

$$I = \int_{0}^{\infty} \frac{\arctan{v}}{2 + v^2} dv$$

Notice something remarkable? We've arrived back at our original integral! This might seem circular, but it's a crucial step. We will use this seemingly redundant transformation to our advantage shortly. The division by cos²(u) and the subsequent substitution v = tan(u) might appear to bring us back to square one, but they have subtly altered the form of the integral in a way that will allow us to exploit a symmetry. This is a common technique in advanced integration, where seemingly redundant steps can reveal hidden relationships and lead to a solution. The key here is to recognize that the limits of integration have changed back to their original form, suggesting a possible connection between the original integral and the transformed one. This connection will be the key to unlocking the final solution. The beauty of this approach lies in its ability to leverage the structure of the integral itself to find a solution, demonstrating the power of mathematical manipulation and insight.

Exploiting Symmetry: The Final Solution

We've shown that:

$$I = \int_{0}^{\infty} \frac{\arctan{x}}{2+x^2} dx = \int_{0}^{\frac{\pi}{2}} \frac{u}{2-\sin^2{u}} du = \int_{0}^{\infty} \frac{\arctan{v}}{2 + v^2} dv$$

This might seem like we're going in circles, but let's reconsider the integral in the v variable. Since v is just a dummy variable, we can replace it with x:

$$I = \int_{0}^{\infty} \frac{\arctan{x}}{2 + x^2} dx$$

Now, let's consider a different approach. We can also use integration by parts on the original integral. Let:

$$u = \arctan{x}$$

$$dv = \frac{1}{2 + x^2} dx$$

Then:

$$du = \frac{1}{1 + x^2} dx$$

To find v, we integrate dv:

$$v = \int \frac{1}{2 + x^2} dx = \frac{1}{\sqrt{2}} \arctan{\frac{x}{\sqrt{2}}}$$

Now, applying integration by parts:

$$\int u dv = uv - \int v du$$

$$I = \left[ \arctan{x} \cdot \frac{1}{\sqrt{2}} \arctan{\frac{x}{\sqrt{2}}} \right]_{0}^{\infty} - \int_{0}^{\infty} \frac{1}{\sqrt{2}} \arctan{\frac{x}{\sqrt{2}}} \cdot \frac{1}{1 + x^2} dx$$

Let's evaluate the first term:

$$\lim_{x \to \infty} \arctan{x} \cdot \frac{1}{\sqrt{2}} \arctan{\frac{x}{\sqrt{2}}} = \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\pi}{2} = \frac{\pi^2}{4\sqrt{2}}$$

At x = 0, the term is 0. So the first term evaluates to:

$$\frac{\pi^2}{4\sqrt{2}}$$

Now we have:

$$I = \frac{\pi^2}{4\sqrt{2}} - \frac{1}{\sqrt{2}} \int_{0}^{\infty} \frac{\arctan{\frac{x}{\sqrt{2}}}}{1 + x^2} dx$$

Let's make one final substitution: Let y = x/√2, so x = √2y and dx = √2 dy. The limits of integration remain 0 and ∞.

$$I = \frac{\pi^2}{4\sqrt{2}} - \frac{1}{\sqrt{2}} \int_{0}^{\infty} \frac{\arctan{y}}{1 + 2y^2} \sqrt{2} dy$$

$$I = \frac{\pi^2}{4\sqrt{2}} - \int_{0}^{\infty} \frac{\arctan{y}}{1 + 2y^2} dy$$

$$I = \frac{\pi^2}{4\sqrt{2}} - I$$

$$2I = \frac{\pi^2}{8}$$

$$I = \frac{\pi^2}{8\sqrt{2}}$$

Finally,

$$I = \frac{\pi^2}{8\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\pi^2\sqrt{2}}{16}$$

Thus, the integral evaluates to:

$$I = \frac{\pi^2}{8\sqrt{2}}$$

This completes our evaluation of the integral. The key to solving this integral was a combination of strategic substitutions, trigonometric identities, and integration by parts. Each step built upon the previous one, ultimately leading us to the final solution. The process highlights the interconnectedness of different mathematical concepts and the power of combining multiple techniques to solve a single problem. The journey through this integral is a testament to the beauty and elegance of calculus, where perseverance and strategic thinking can unlock the solutions to seemingly complex problems.

Find the value of the definite integral: ∫[0 to ∞] (arctan(x) / (2 + x²)) dx.

Evaluate Integral of arctan(x) / (2+x^2) from 0 to Infinity