Solve The Integral: A Step-by-Step Guide

Hey guys! Today, we're diving deep into a fascinating integral equation that looks intimidating at first glance, but trust me, we'll break it down step by step. We're aiming to show that:

$$\frac{1}{2}\text{I}_1 + \text{I}_2 =\frac{\pi^3}{48}- \frac{\pi}{4} \text{arctanh}^2\left({\frac{1}{\sqrt{2}}}\right)$$

Where I₁ and I₂ are defined as definite integrals. Buckle up, it's going to be a fun ride!

Dissecting the Integrals: I₁ and I₂

Before we jump into the solution, let's take a closer look at our players: I₁ and I₂. These are definite integrals, which means we're looking for specific numerical values, not just functions. This involves a lot of strategic manipulation and clever substitutions.

I₁: The Arctangent and Logarithm Tango

Let's start with I₁:

$$\text{I}_1 = \int^1_{-1} \frac{\arctan\left({\frac{\sqrt{2}x}{1-x^2}}\right)\ln\left({\frac{1+x^4}{2}}\right)}{x+1} dx$$

This integral brings together a few interesting functions. We have the arctangent (arctan) which is the inverse tangent function, the natural logarithm (ln), and a rational function in the denominator. The interplay between these functions is what makes this integral challenging, but also quite beautiful.

Breaking Down the Components:

• Arctangent Term: The argument of the arctangent, $\frac{\sqrt{2}x}{1-x^2}$, hints at a possible trigonometric substitution. Remember the double angle formula for tangent? $tan(2θ) = \frac{2tan(θ)}{1-tan^2(θ)}$. This suggests that substituting $x = tan(θ)$ might simplify this part. Key concept here is trigonometric substitution, a powerful tool in our integration arsenal.
• Logarithm Term: The logarithm, $\ln\left({\frac{1+x^4}{2}}\right)$, looks a bit less friendly. The $x^4$ inside suggests that we might need to deal with some higher-order terms. However, let's hold our horses for now. Sometimes, these terms simplify nicely after a clever substitution or integration by parts. The form of the logarithm often suggests considering properties of logarithms or looking for ways to simplify the argument.
• Denominator: The $(x+1)$ in the denominator is a classic sign that we might need to consider partial fraction decomposition or look for ways to cancel it out. It’s often the little things that give us clues! The denominator's linear term hints at possible simplifications through algebraic manipulation or substitution.

Strategic Approaches for I₁:

1. Trigonometric Substitution: As hinted earlier, substituting $x = tan(θ)$ is a strong contender. This will likely simplify the arctangent term, but we need to see how it affects the rest of the integral.
2. Integration by Parts: Since we have a product of two functions (arctan and ln), integration by parts might be helpful. The key is to choose the 'u' and 'dv' strategically. Remember the formula: $\int u dv = uv - \int v du$. The success of integration by parts often depends on choosing appropriate 'u' and 'dv' functions.
3. Properties of Logarithms: We could try to split the logarithm using the property $\ln(\frac{a}{b}) = \ln(a) - \ln(b)$. This might lead to simpler integrals. Logarithmic properties are essential for simplifying complex logarithmic expressions within integrals.

I₂: The Arctanh Adventure

Now, let's turn our attention to I₂:

$$\text{I}_2 = \int^1_{-1} \frac{\text{arctanh}\left({\frac{\sqrt{2}x }{1+x^2}}\right)}{x+1} dx$$

This integral features the inverse hyperbolic tangent function, arctanh. For those less familiar, arctanh(x) is the inverse of the hyperbolic tangent function, tanh(x). It’s defined as: $\text{arctanh}(x) = \frac{1}{2}\ln(\frac{1+x}{1-x})$. Understanding the properties of inverse hyperbolic functions is crucial for handling integrals involving them.

Understanding Arctanh:

• Connection to Logarithms: The definition of arctanh as a logarithm is super important. It means we can often convert arctanh integrals into logarithmic integrals, which we might be more comfortable handling. This logarithmic representation provides a direct pathway for simplification and integration.
• Symmetry: The argument of the arctanh, $\frac{\sqrt{2}x }{1+x^2}$, is an odd function. This suggests that we might be able to exploit symmetry properties of definite integrals. Symmetry considerations can significantly simplify definite integral calculations, especially over symmetric intervals.

Strategic Approaches for I₂:

1. Logarithmic Conversion: The most obvious step is to replace arctanh with its logarithmic definition. This will transform the integral into a form involving logarithms, which we can then try to simplify.
2. Partial Fractions: Again, the $(x+1)$ in the denominator hints at partial fractions. We might need to combine this with the logarithmic form of arctanh to see how it plays out.
3. Symmetry: Let's keep an eye out for opportunities to use symmetry. If the integrand (the function inside the integral) has certain symmetry properties, we can simplify the integration limits or even show that the integral is zero.

The Grand Strategy: Combining I₁ and I₂

Our ultimate goal is to show that $\frac{1}{2}\text{I}_1 + \text{I}_2 =\frac{\pi^3}{48}- \frac{\pi}{4} \text{arctanh}^2\left({\frac{1}{\sqrt{2}}}\right)$. This means we can't just solve I₁ and I₂ independently. We need to think about how they might interact.

Key Considerations:

• Common Denominator: Both integrals have $(x+1)$ in the denominator. This is a strong indicator that we should try to combine them early on. Combining integrals with common denominators often simplifies the overall expression and reveals hidden cancellations or relationships.
• Target Expression: The right-hand side of our target equation involves $\pi^3$ and $\text{arctanh}^2(\frac{1}{\sqrt{2}})$. This suggests that we need to look for ways to generate these terms during our integration process. The presence of $\pi$ usually indicates trigonometric functions or definite integrals with specific limits, while the arctanh term suggests the involvement of inverse hyperbolic functions.

A Potential Roadmap:

1. Initial Simplifications: Start by applying basic simplification techniques to I₁ and I₂ individually. This might involve trigonometric substitutions for I₁, and logarithmic conversion for I₂.
2. Combine the Integrals: Once we have simplified forms, let's combine $\frac{1}{2}\text{I}_1$ and $\text{I}_2$ into a single integral. This is where the magic might happen!
3. Look for Cancellations: After combining, carefully examine the integrand for any terms that might cancel out or simplify. This is often the key to making progress.
4. Evaluate the Integral: Finally, use appropriate integration techniques (like substitution, integration by parts, or trigonometric identities) to evaluate the resulting integral.

Let's Get Our Hands Dirty: Solving I₁

Alright, let's start with I₁. Remember our strategic approach? We're going to try the trigonometric substitution $x = tan(θ)$.

Step 1: The Trigonometric Substitution

If $x = tan(θ)$, then $dx = sec^2(θ) dθ$. We also need to change the limits of integration. When $x = -1$, $θ = -\frac{\pi}{4}$, and when $x = 1$, $θ = \frac{\pi}{4}$.

Substituting these into I₁, we get:

$$\text{I}_1 = \int^{-\frac{\pi}{4}}_{\frac{\pi}{4}} \frac{\arctan\left({\frac{\sqrt{2}tan(θ)}{1-tan^2(θ)}}\right)\ln\left({\frac{1+tan^4(θ)}{2}}\right)}{tan(θ)+1} sec^2(θ) dθ$$

Step 2: Simplifying the Arctangent

Now, let's use the double angle formula for tangent: $tan(2θ) = \frac{2tan(θ)}{1-tan^2(θ)}$. Our arctangent term becomes:

$$\arctan\left({\frac{\sqrt{2}tan(θ)}{1-tan^2(θ)}}\right) = \arctan(\sqrt{2}tan(2θ))$$

So, I₁ now looks like:

$$\text{I}_1 = \int^{-\frac{\pi}{4}}_{\frac{\pi}{4}} \frac{\arctan(\sqrt{2}tan(2θ))\ln\left({\frac{1+tan^4(θ)}{2}}\right)}{tan(θ)+1} sec^2(θ) dθ$$

Step 3: Ugh... Still Complicated!

Okay, this looks a bit better, but it's still not super friendly. The $\arctan(\sqrt{2}tan(2θ))$ term is particularly annoying. The logarithm also hasn’t magically simplified. We might need a different approach or another clever trick. Don't worry, this is perfectly normal in the world of integration! Sometimes you need to try a few paths before you find the right one. It's crucial to remain flexible and adapt your strategy as needed. Persistence and a willingness to explore different techniques are key to success.

Let's pause here and rethink our strategy for I₁. Maybe the trigonometric substitution wasn't the silver bullet we hoped for. Perhaps we should explore integration by parts, focusing on the $\ln\left({\frac{1+x^4}{2}}\right)$ term. Or, we could try to simplify the logarithm first using its properties. Let's try something else.

Time for Plan B: Attacking I₂

Since I₁ is giving us a bit of a headache, let's switch gears and focus on I₂ for a while. Remember, the key to I₂ is the arctanh function.

Step 1: Logarithmic Conversion

The first thing we're going to do is replace arctanh with its logarithmic definition:

$$\text{arctanh}(x) = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$$

So, for our integral I₂:

$$\text{I}_2 = \int^1_{-1} \frac{\frac{1}{2}\ln\left(\frac{1+\frac{\sqrt{2}x }{1+x^2}}{1-\frac{\sqrt{2}x }{1+x^2}}\right)}{x+1} dx$$

Step 2: Simplifying the Logarithm

Let's simplify the argument of the logarithm inside the integral:

$$\frac{1+\frac{\sqrt{2}x }{1+x^2}}{1-\frac{\sqrt{2}x }{1+x^2}} = \frac{1+x^2+\sqrt{2}x}{1+x^2-\sqrt{2}x}$$

Now, I₂ looks like:

$$\text{I}_2 = \frac{1}{2}\int^1_{-1} \frac{\ln\left(\frac{1+x^2+\sqrt{2}x}{1+x^2-\sqrt{2}x}\right)}{x+1} dx$$

Step 3: Partial Fractions? Not So Fast!

We still have that pesky $(x+1)$ in the denominator, and the logarithm is still a bit intimidating. Partial fraction decomposition doesn't seem directly applicable here. Instead, let's see if we can rewrite the argument of the logarithm in a more insightful way.

Step 4: Completing the Square (Kind Of)

The expressions $1+x^2+\sqrt{2}x$ and $1+x^2-\sqrt{2}x$ look a bit like quadratic expressions. Let's try to "complete the square" (loosely speaking) to see if we can factor them. Remember, completing the square is a powerful technique for rewriting quadratic expressions and revealing hidden structures.

Consider:

$$x^2 + \sqrt{2}x + 1 = \left(x + \frac{1}{\sqrt{2}}\right)^2 + \frac{1}{2}$$

$$x^2 - \sqrt{2}x + 1 = \left(x - \frac{1}{\sqrt{2}}\right)^2 + \frac{1}{2}$$

Now our logarithm argument becomes:

$$\frac{\left(x + \frac{1}{\sqrt{2}}\right)^2 + \frac{1}{2}}{\left(x - \frac{1}{\sqrt{2}}\right)^2 + \frac{1}{2}}$$

I₂ is now:

$$\text{I}_2 = \frac{1}{2}\int^1_{-1} \frac{\ln\left(\frac{\left(x + \frac{1}{\sqrt{2}}\right)^2 + \frac{1}{2}}{\left(x - \frac{1}{\sqrt{2}}\right)^2 + \frac{1}{2}}\right)}{x+1} dx$$

Step 5: A Glimmer of Hope?

This looks…different. It's not immediately clear if this is better, but we've definitely changed the form of the integral. The squared terms and the fractions involving $\frac{1}{\sqrt{2}}$ suggest that we might be getting closer to our target expression involving $\text{arctanh}^2(\frac{1}{\sqrt{2}})$.

The next step might be to try a substitution that involves $x + \frac{1}{\sqrt{2}}$ or $x - \frac{1}{\sqrt{2}}$. Let’s keep pushing!

Back to I₁: A New Perspective

Before we dive deeper into I₂, let's revisit I₁. Sometimes, working on one part of a problem gives you insights into other parts. We struggled with the trigonometric substitution earlier, but maybe we can now see a different way forward.

Remember, I₁ was:

$$\text{I}_1 = \int^1_{-1} \frac{\arctan\left({\frac{\sqrt{2}x}{1-x^2}}\right)\ln\left({\frac{1+x^4}{2}}\right)}{x+1} dx$$

Key Insight: Double Angle Formula

The arctangent term, $\arctan\left({\frac{\sqrt{2}x}{1-x^2}}\right)$, is still begging for simplification. We know that $\frac{\sqrt{2}x}{1-x^2}$ looks a lot like the tangent double angle formula. But instead of substituting $x = tan(θ)$, let's directly use the identity:

$$\arctan\left({\frac{2x}{1-x^2}}\right) = 2\arctan(x)$$

We need a slight modification to apply this. Notice that our argument is $\frac{\sqrt{2}x}{1-x^2}$, not $\frac{2x}{1-x^2}$. So, this direct substitution isn't going to work perfectly here. But the idea is still good, and the arctangent term is still looking like it needs to be simplified.

Alternative Approach: Focus on the Logarithm

Maybe we've been too focused on the arctangent. What about the logarithm, $\ln\left({\frac{1+x^4}{2}}\right)$? Can we simplify this?

Logarithm Properties to the Rescue

Let's use the property $\ln(\frac{a}{b}) = \ln(a) - \ln(b)$:

$$\ln\left({\frac{1+x^4}{2}}\right) = \ln(1+x^4) - \ln(2)$$

This might seem like a small step, but it separates the constant term, $\ln(2)$, which could be helpful. Now I₁ becomes:

$$\text{I}_1 = \int^1_{-1} \frac{\arctan\left({\frac{\sqrt{2}x}{1-x^2}}\right)(\ln(1+x^4) - \ln(2))}{x+1} dx$$

We can split this into two integrals:

$$\text{I}_1 = \int^1_{-1} \frac{\arctan\left({\frac{\sqrt{2}x}{1-x^2}}\right)\ln(1+x^4)}{x+1} dx - \ln(2)\int^1_{-1} \frac{\arctan\left({\frac{\sqrt{2}x}{1-x^2}}\right)}{x+1} dx$$

Let's call these two integrals I₁ₐ and I₁ʙ:

$$\text{I}_{1a} = \int^1_{-1} \frac{\arctan\left({\frac{\sqrt{2}x}{1-x^2}}\right)\ln(1+x^4)}{x+1} dx$$

$$\text{I}_{1b} = \int^1_{-1} \frac{\arctan\left({\frac{\sqrt{2}x}{1-x^2}}\right)}{x+1} dx$$

So, $\text{I}_1 = \text{I}_{1a} - \ln(2)\text{I}_{1b}$. This decomposition might make things more manageable.

The Road Ahead: Combining the Pieces

We've made progress on both I₁ and I₂, but we're not there yet. We've explored trigonometric substitutions, logarithmic conversions, and properties, and even a bit of "completing the square." We've split I₁ into two smaller integrals, I₁ₐ and I₁ʙ.

Our next steps are likely to involve:

• Further Simplification of I₁ₐ and I₁ʙ: We need to find a way to tackle these integrals. Trigonometric substitutions might still be relevant, or perhaps integration by parts.
• Continuing to Work on I₂: We need to see where our "completing the square" approach leads us. A well-chosen substitution might be the key.
• Combining the Results: Remember, our ultimate goal is to compute $\frac{1}{2}\text{I}_1 + \text{I}_2$. We need to keep this in mind as we work on the individual integrals. Look for opportunities to combine terms or exploit cancellations.

This is a challenging problem, but we're making progress. The key is to keep exploring different techniques, be patient, and never give up! Stay tuned as we continue to unravel this integral equation mystery!

This is a long journey, guys, and we've only just scratched the surface. In the next installment, we'll dive deeper into solving I₁ₐ, I₁ʙ, and I₂, and finally bring all the pieces together to prove our initial equation. Keep your thinking caps on, and let's conquer this integral equation together!