Is it possible to use Newton's method on a function with an integral?

Question

So I have a root solving problem where I want to find x, but my equation is: $$ 0=1-p+c_2\int_0^Xt^{\alpha-1}e^{-t/2}dt $$ where $0\le p\le1, c_2\ge2$, $1\le\alpha$. Is it possible to use Newton's method on equation containing an integral? If so, how would one calculate the derivative, or would the just use the standard derivative formula $\frac{f(x+h)-f(x)}{h}$ for h small?

Answer 1

In other words, you want to find the zero of function $$f(x)=\Gamma \left(\alpha ,\frac{x}{2}\right)-k \qquad \text{where} \qquad k=\Gamma (\alpha )-\frac{ (\alpha -1)}{c \,2^{\alpha }}$$ with,as said in comments, $$f'(x)=-2^{-\alpha } e^{-x/2} x^{\alpha -1}$$

The only problem would be to find a suitable $x_0$ to start iterations.

Answer 2

Let

$$ f(x) = c_2 \int_0^x t^{\alpha -1} \operatorname{e}^{-\frac{t}{2}}\operatorname d t + 1 - p $$

We note that $f$ is defined for $x \ge 0 $, moreover $$ f(0) = 1 -p \ge 0$$ If we differentiate, we obtain

$$ f'(x) = x^{\alpha-1}\operatorname{e}^{-\frac{x}{2}}\ge 0 \qquad \text{ for all $x\ge 0$ } $$

Since $f' > 0 $, the function $f$ is in increasing.

Conclusion: the only possible zero is $x=0$, if $p=1$.