Does there exist a $b$ such that the integral exists?

Question

I came across the following integral while working on my thesis,

$$ \int_{0}^{\infty} x^{b-1} (1-x)^{z_1+1} dx $$ where $z_1 \in \mathbb{C}$ and $z_1 = a - wi$ where $a > 0$. Does there exists a $b$ such that the above integral converges?

I was trying on wolfram alpha, and it mentions something about hypergeometric function, but I'm not familiar with the concept.

Answer 1

tl;dr: No. For $a>0$, there is no such $b$.

First, as $x \rightarrow \infty$, your integrand is approaching $x^{b-1}x^{a+1-w\mathrm{i}} = x^{a+b-w\mathrm{i}}$. By comparison and the $p$-test, this cannot converge if $a+b \geq -1$, so convergence of your integral requires $a+b < -1$.

Second, for $x \rightarrow 0$, your integrand is approaching $x^{b-1}\cdot 1$. By comparison and the $p$-test, this cannot converge if $b-1 \leq -1$, so convergence of your integral requires $b > 0$.

Third, for $x \rightarrow 1$, your integrand is approaching $1 \cdot (1-x)^{a+1-w\mathrm{i}}$. As in the second case, this cannot converge if $a+1 \leq -1$, so convergence of your integral requires $a > -2$.

These constraints give a triangle in the $a,b$-plane, with vertices $(-2,0)$, $(-1,0)$, and $(-2,1)$. As long as $(a,b)$ is a point in the interior of that triangle, your integral converges.

Your integral is related to the Beta function. In fact, $\int_0^1$ of your integrand is a value of the Beta function. Today, I'm too lazy to find the change of variable (some variant of $x \mapsto 1/x$ or $x \mapsto 1-1/x$) to turn the $\int_1^\infty$ integral into a value of Beta.

---

Since OP apparently wants a CAS to provide the integral and conditions ... Mathematica 11.2 gives

Assuming[{Element[a|b|w, Reals]},
  FullSimplify[
    Integrate[
      x^(b-1)(1-x)^(a-wI+1), 
      {x,0,Infinity}, 
      GenerateConditions -> True
    ]
  ]
]

(* Result:  
ConditionalExpression[
  Gamma[2+a-Iw] (
    Gamma[b]/Gamma[2+a+b-Iw] + (-1)^(1+a-Iw)Gamma[-1-a-b+Iw]/Gamma[1-b]
  ),
  b > 0 && a > -2 && a+b < -1
]
*)

And these are the same conditions on $a$ and $b$ identified above.