Proof: Definite Integrals & Gamma Function Relation

Hey guys! Today, we're diving into a fascinating problem that intertwines definite integrals, the Gamma function, and a bit of clever manipulation to prove the following identity:

$$\left(\int_{0}^{\infty}x^{-1/4}e^{-x}dx\right)^4\left(\int_{0}^{1}\frac{dt}{\sqrt{1-t^4}}\right)^2=\frac{\pi^3}{8}$$

Buckle up; it's going to be a fun ride!

Part 1: Tackling the Gamma Function Integral

Let's start with the integral involving the exponential function. This integral is directly related to the Gamma function, a special function that extends the factorial function to complex numbers. The Gamma function is defined as:

$$\Gamma(z) = \int_{0}^{\infty} t^{z-1}e^{-t} dt$$

Our integral looks very similar to this. Specifically, we have:

$$\int_{0}^{\infty}x^{-1/4}e^{-x}dx$$

Comparing this with the definition of the Gamma function, we can see that $z - 1 = -1/4$, which implies $z = 3/4$. Therefore, our integral is simply $\Gamma(3/4)$:

$$\int_{0}^{\infty}x^{-1/4}e^{-x}dx = \Gamma\left(\frac{3}{4}\right)$$

So, the left-hand side of our target equation now contains $\Gamma(3/4)^4$. This is a great start! The Gamma function has some amazing properties, and we'll leverage those soon.

Digging Deeper into the Gamma Function

The Gamma function is a cornerstone in many areas of mathematics and physics. Its definition as an integral allows us to work with factorials of non-integer and even complex numbers. One of the most crucial properties we'll use is the reflection formula:

$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$$

This formula connects the Gamma function at $z$ with its value at $1-z$, and it's perfect for simplifying expressions involving Gamma functions with fractional arguments. In our case, we have $\Gamma(3/4)$, so let's see how the reflection formula can help us. We can pair it with $\Gamma(1 - 3/4) = \Gamma(1/4)$:

$$\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{1}{4}\right) = \frac{\pi}{\sin(\frac{3\pi}{4})} = \frac{\pi}{\sin(\frac{\pi}{4})} = \frac{\pi}{\frac{\sqrt{2}}{2}} = \pi\sqrt{2}$$

This is a significant simplification. Now we know the product of $\Gamma(3/4)$ and $\Gamma(1/4)$. This will be useful when we want to express $\Gamma(3/4)$ in terms of $\pi$ and $\Gamma(1/4)$.

Part 2: Unraveling the Elliptic Integral

Now, let's turn our attention to the second integral:

$$\int_{0}^{1}\frac{dt}{\sqrt{1-t^4}}$$

This integral is an example of an elliptic integral. Elliptic integrals are a class of integrals that arise in various geometric problems, such as finding the arc length of an ellipse (hence the name). They generally don't have elementary closed-form solutions, but they can be expressed in terms of special functions or other integrals. In this particular case, we can relate it back to the Gamma function!

To tackle this, we'll use a substitution. Let $t^4 = u$, so $4t^3 dt = du$, which means $dt = \frac{du}{4t^3} = \frac{du}{4u^{3/4}}$. Substituting this into our integral, we get:

$$\int_{0}^{1}\frac{dt}{\sqrt{1-t^4}} = \int_{0}^{1} \frac{1}{\sqrt{1-u}} \cdot \frac{du}{4u^{3/4}} = \frac{1}{4} \int_{0}^{1} u^{-3/4} (1-u)^{-1/2} du$$

Connecting to the Beta Function

The integral we now have looks very similar to the Beta function, which is defined as:

$$B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

Comparing our integral with the Beta function, we identify $x - 1 = -3/4$ and $y - 1 = -1/2$. Thus, $x = 1/4$ and $y = 1/2$. Therefore, our integral can be expressed as:

$$\frac{1}{4} \int_{0}^{1} u^{-3/4} (1-u)^{-1/2} du = \frac{1}{4} B\left(\frac{1}{4}, \frac{1}{2}\right) = \frac{1}{4} \frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{\Gamma(\frac{3}{4})}$$

We know that $\Gamma(1/2) = \sqrt{\pi}$. Therefore:

$$\int_{0}^{1}\frac{dt}{\sqrt{1-t^4}} = \frac{1}{4} \frac{\Gamma(\frac{1}{4})\sqrt{\pi}}{\Gamma(\frac{3}{4})}$$

Part 3: Putting it All Together

Now we have expressions for both integrals in terms of Gamma functions and $\pi$. Let's substitute these back into the original equation we want to prove:

$$\left(\int_{0}^{\infty}x^{-1/4}e^{-x}dx\right)^4\left(\int_{0}^{1}\frac{dt}{\sqrt{1-t^4}}\right)^2=\frac{\pi^3}{8}$$

Substituting our results:

$$\left[\Gamma\left(\frac{3}{4}\right)\right]^4 \left[\frac{1}{4} \frac{\Gamma(\frac{1}{4})\sqrt{\pi}}{\Gamma(\frac{3}{4})}\right]^2 = \frac{\pi^3}{8}$$

Let's simplify this expression:

$$\Gamma\left(\frac{3}{4}\right)^4 \cdot \frac{1}{16} \cdot \frac{\Gamma\left(\frac{1}{4}\right)^2 \pi}{\Gamma\left(\frac{3}{4}\right)^2} = \frac{\pi^3}{8}$$

$$\frac{1}{16} \Gamma\left(\frac{3}{4}\right)^2 \Gamma\left(\frac{1}{4}\right)^2 \pi = \frac{\pi^3}{8}$$

Now, remember the reflection formula? We know that $\Gamma(3/4)\Gamma(1/4) = \pi\sqrt{2}$. Let's substitute that in:

$$\frac{1}{16} (\pi\sqrt{2})^2 \pi = \frac{\pi^3}{8}$$

$$\frac{1}{16} (2\pi^2) \pi = \frac{\pi^3}{8}$$

$$\frac{2\pi^3}{16} = \frac{\pi^3}{8}$$

$$\frac{\pi^3}{8} = \frac{\pi^3}{8}$$

And there you have it! We've successfully proven the identity. The key was recognizing the integrals as related to the Gamma and Beta functions and then using the reflection formula to simplify the expression. Isn't it awesome how different areas of math connect like this?

Conclusion: A Symphony of Mathematical Concepts

Proving this identity was a beautiful demonstration of how different mathematical concepts harmonize. The Gamma function, with its elegant definition and powerful properties, played a starring role. The Beta function provided a crucial link between the elliptic integral and the Gamma function. And the reflection formula acted as the bridge that allowed us to simplify the expression and arrive at the final result.

This journey underscores the importance of recognizing patterns and connections in mathematics. By understanding the fundamental properties of special functions and mastering techniques like substitution, we can unlock the secrets hidden within complex-looking integrals and reveal the elegant truths that lie beneath the surface. Keep exploring, keep questioning, and keep discovering the beauty of mathematics!