How to evaluate this Integral using Beta and Gamma Functions
Published June 19, 2025, 8:09 a.m. by james
Introduction
We are given the definite integral:
$$I=\int_0^1 \frac{\sqrt{x}}{\sqrt{1+x^2} + \sqrt{1-x^2}} \, dx$$
This document details the step-by-step evaluation of this integral using properties of rationalization, substitution, and specialized functions like the Beta and Gamma functions.
Step 1: Rationalize the Denominator
Rationalizing the denominator of the integrand via conjugates and simplifying gives us:
$$I = \int_0^1 \frac{\sqrt{x} (\sqrt{1+x^2} - \sqrt{1-x^2})}{(1+x^2) - (1-x^2)} \, dx = \int_0^1 \frac{\sqrt{1+x^2} - \sqrt{1-x^2}}{2x^{3/2}} \, dx$$
Step 2: Apply Substitution
Next, we make the substitution $t = x^2$. This implies $dt = 2x \, dx$, so $dx = \frac{dt}{2\sqrt{t}} = \frac{1}{2}t^{-1/2} \, dt$. The limits of integration remain from $0$ to $1$. Substituting, we split the integral into two terms:
$$I = \frac{1}{4} \int_0^1 t^{-5/4} (1+t)^{1/2} \, dt - \frac{1}{4} \int_0^1 t^{-5/4} (1 - t)^{1/2} \, dt$$
Step 3: Transform the First Integral
For the first term, we use the interval-inverting substitution replacing $t$ with $t^{-1}$. If $u = t^{-1}$, then $du = -t^{-2} \, dt$. When t=0, u=infinity. When t=1, u=1.
$$\int_0^1 t^{-5/4} (1+t)^{1/2} \, dt = \int_{\infty}^1 (u^{-1})^{-5/4} (1+u^{-1})^{1/2} (-u^{-2} \, du)$$
$$= \int_1^{\infty} u^{5/4} \left(\frac{u+1}{u}\right)^{1/2} u^{-2} \, du = \int_1^{\infty} u^{5/4} \frac{(u+1)^{1/2}}{u^{1/2}} u^{-2} \, du$$
$$= \int_1^{\infty} u^{5/4 - 1/2 - 2} (u+1)^{1/2} \, du = \int_1^{\infty} u^{5/4 - 2/4 - 8/4} (u+1)^{1/2} \, du = \int_1^{\infty} u^{-5/4} (u+1)^{1/2} \, du$$
Therefore, it follows from averaging the integrals (having the same integrands) that:
$$\int_0^1 t^{-5/4} (1+t)^{1/2} \, dt = \frac{1}{2} \int_0^{\infty} t^{-5/4} (t + 1)^{1/2} \, dt$$
Step 4: Express in terms of Beta Functions
Thus, we have:
$$I = \frac{1}{4} \cdot \frac{1}{2} \int_0^{\infty} t^{-5/4} (1+t)^{1/2} \, dt - \frac{1}{4} \int_0^1 t^{-5/4} (1 - t)^{1/2} \, dt$$
We can rewrite the exponents to fit the Beta function definition $$B(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt$$ and $$B(x,y) = \int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}} dt$$.
$$I = \frac{1}{8} \int_0^{\infty} \frac{t^{-\frac{1}{4} - 1}}{(1+t)^{\frac{1}{4} - \frac{1}{2}}} \, dt - \frac{1}{4} \int_0^1 t^{-\frac{1}{4} - 1} (1 - t)^{\frac{3}{2} - 1} \, dt$$
So the first integral term is of the form $B(x,y)$ where $x-1 = -5/4 \Rightarrow x = -1/4$, and $x+y = -1/2 \Rightarrow y = -1/2 - (-1/4) = -1/4$. The second integral term is of the form $B(x,y)$ where $x-1 = -5/4 \Rightarrow x = -1/4$, and $y-1 = 1/2 \Rightarrow y = 3/2$.
Each of the remaining integrals is a special case of the Beta function. Expressing the results in terms of the Gamma function using $$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$:
$$I = \frac{1}{8} \cdot B\left(-\frac{1}{4}, -\frac{1}{4}\right) - \frac{1}{4} \cdot B\left(-\frac{1}{4}, \frac{3}{2}\right)$$
$$I = \frac{1}{8} \cdot \frac{\Gamma\left(-\frac{1}{4}\right) \, \Gamma\left(-\frac{1}{4}\right)}{\Gamma\left(-\frac{1}{4} - \frac{1}{4}\right)} - \frac{1}{4} \cdot \frac{\Gamma\left(-\frac{1}{4}\right) \, \Gamma\left(\frac{3}{2}\right)}{\Gamma\left(-\frac{1}{4} + \frac{3}{2}\right)}$$
$$I = \frac{1}{8} \cdot \frac{\Gamma^2\left(-\frac{1}{4}\right)}{\Gamma\left(-\frac{1}{2}\right)} - \frac{1}{4} \cdot \frac{\Gamma\left(-\frac{1}{4}\right) \, \Gamma\left(\frac{3}{2}\right)}{\Gamma\left(\frac{5}{4}\right)}$$
Step 5: Apply Gamma Function Properties
Using the property $\Gamma(z+1) = z\Gamma(z)$ and $\Gamma(1/2) = \sqrt{\pi}$:
$\Gamma(-1/4) = \frac{\Gamma(3/4)}{-1/4} = -4\Gamma(3/4)$
$\Gamma(-1/2) = \frac{\Gamma(1/2)}{-1/2} = -2\sqrt{\pi}$
$\Gamma(3/2) = \frac{1}{2}\Gamma(1/2) = \frac{1}{2}\sqrt{\pi}$
$\Gamma(5/4) = \frac{1}{4}\Gamma(1/4)$
Substitute these values back into the expression for $I$:
$$I = \frac{1}{8} \cdot \frac{(-4 \, \Gamma(\frac{3}{4}))^2}{-2\sqrt{\pi}} - \frac{1}{4} \cdot \frac{(-4 \cdot \Gamma(\frac{3}{4})) \cdot (\frac{1}{2} \sqrt{\pi})}{\frac{1}{4} \, \Gamma(\frac{1}{4})}$$
$$I = \frac{1}{8} \cdot \frac{16 \, \Gamma^2(\frac{3}{4})}{-2\sqrt{\pi}} - \frac{1}{4} \cdot \frac{-2\sqrt{\pi} \, \Gamma(\frac{3}{4})}{\frac{1}{4} \, \Gamma(\frac{1}{4})}$$
$$I = -\frac{\Gamma^2(\frac{3}{4})}{\sqrt{\pi}} + \frac{2\sqrt{\pi} \, \Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}$$
Step 6: Apply Euler’s Reflection Formula
Finally, we use Euler’s reflection formula $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$. For $z = 1/4$:
$$\Gamma\left(\frac{1}{4}\right) \Gamma\left(1 - \frac{1}{4}\right) = \Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{3}{4}\right) = \frac{\pi}{\sin\left(\frac{\pi}{4}\right)} = \frac{\pi}{\frac{1}{\sqrt{2}}} = \pi\sqrt{2}$$
So, $\Gamma\left(\frac{1}{4}\right) = \frac{\pi\sqrt{2}}{\Gamma(\frac{3}{4})}$. Substitute this into the expression for $I$:
$$I = -\frac{\Gamma^2(\frac{3}{4})}{\sqrt{\pi}} + \frac{2\sqrt{\pi} \, \Gamma(\frac{3}{4})}{\frac{\pi\sqrt{2}}{\Gamma(\frac{3}{4})}}$$
$$I = -\frac{\Gamma^2(\frac{3}{4})}{\sqrt{\pi}} + \frac{2\sqrt{\pi} \, \Gamma^2(\frac{3}{4})}{\pi\sqrt{2}}$$
To combine the terms, we simplify the second term: $$\frac{2\sqrt{\pi}}{\pi\sqrt{2}} = \frac{\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{\pi}}{\sqrt{\pi} \cdot \sqrt{\pi} \cdot \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{\pi}}$$.
So,
$$I = -\frac{\Gamma^2(\frac{3}{4})}{\sqrt{\pi}} + \frac{\sqrt{2}}{\sqrt{\pi}} \, \Gamma^2\left(\frac{3}{4}\right)$$
$$I = \frac{\sqrt{2} - 1}{\sqrt{\pi}} \, \Gamma^2\left(\frac{3}{4}\right)$$
Conclusion
The final evaluated integral is:
$$\boxed{\frac{\sqrt{2} - 1}{\sqrt{\pi}} \, \Gamma^2\left(\frac{3}{4}\right)}$$
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