求含Γ函数的积分闭式表达式：指数型积分化简技术求助

Alright, let's break down how to simplify the integral inside your exponential function to arrive at the Γ-function-based $K(\beta)$ from the paper. First, let's restate the integral we need to compute (with the corrected denominator term):

$$I = \int^{\infty_0\frac{1}{1+v}{\beta/2}}dv$$

Step 1: Variable Substitution #

Start with a substitution to turn the integral into a form we can recognize. Let $t = v^{\beta/2}$, which means $v = t^{2/\beta}$. Taking the derivative of $v$ with respect to $t$ gives:
$$dv = \frac{2}{\beta}t^{(2/\beta)-1}dt$$

Adjust the limits of integration: when $v=0$, $t=0$; when $v\to\infty$, $t\to\infty$. Plugging this back into the integral:
$$I = \int^\infty_0 \frac{1}{1+t} \cdot \frac{2}{\beta}t^{(2/\beta)-1}dt = \frac{2}{\beta}\int^\infty_0 \frac{t^{(2/\beta)-1}}{1+t}dt$$

Step 2: Use the Beta-Gamma Function Relationship #

Now we can use the definition of the Beta function, which is:
$$B(p,q) = \int^\infty_0 \frac{t^{p-1}}{(1+t){p+q}}dt$$
And the key relation between Beta and Gamma functions:
$$B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$$

For our integral, set $p = 2/\beta$. Notice that the denominator is $(1+t)^1$, so $p+q = 1$, which means $q = 1 - 2/\beta$. Plugging these into the Beta function formula:
$$\int^\infty_0 \frac{t^{(2/\beta)-1}}{1+t}dt = B\left(\frac{2}{\beta}, 1-\frac{2}{\beta}\right) = \frac{\Gamma\left(\frac{2}{\beta}\right)\Gamma\left(1-\frac{2}{\beta}\right)}{\Gamma\left(\frac{2}{\beta} + 1 - \frac{2}{\beta}\right)}$$

Since $\Gamma(1) = 1$, the denominator simplifies to 1, leaving us with:
$$\int^\infty_0 \frac{t^{(2/\beta)-1}}{1+t}dt = \Gamma\left(\frac{2}{\beta}\right)\Gamma\left(1-\frac{2}{\beta}\right)$$

Step 3: Assemble $K(\beta)$ #

Going back to the original exponential term: it has a factor of $\pi\lambda T^{2/\beta}r2$ multiplied by our integral $I$. Substituting the simplified integral in:
$$\pi\lambda T^{2/\beta}r2 \cdot I = \pi\lambda T^{2/\beta}r2 \cdot \frac{2}{\beta}\Gamma\left(\frac{2}{\beta}\right)\Gamma\left(1-\frac{2}{\beta}\right)$$

Matching this to the paper's result $\exp(-\lambda r^2T{2/\beta }K(\beta))$, we can directly pull out $K(\beta)$:
$$K(\beta) = \frac{2\pi \Gamma\left(\frac{2}{\beta}\right)\Gamma\left(1-\frac{2}{\beta}\right)}{\beta}$$

Bonus: Simplify with the Reflection Formula (Optional) #

If you want a more compact form, you can use the Gamma function reflection formula:
$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$$
Substituting $z = 2/\beta$, this turns $K(\beta)$ into:
$$K(\beta) = \frac{2\pi}{\beta} \cdot \frac{\pi}{\sin\left(\frac{2\pi}{\beta}\right)} = \frac{2\pi^2}{\beta \sin\left(\frac{2\pi}{\beta}\right)}$$
This is a useful alternative for numerical calculations or further algebraic manipulation.

内容的提问来源于stack exchange，提问作者SJa