A slight modification of Beta function

Question

The Beta function is defined as follows: for real values $a,b>0$ $$B(a,b) = \int_0^1 w^{a-1}(1-w)^{b-1} dw.$$

It is known that $$B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},$$ where $\Gamma$ is the Gamma function.

Is there a way to find a closed expression or bound for the following integral? $$I:=\int_0^1 w^{a-1}(1-w)^{b-1} e^{-cw} dw,$$ where $c>0$. Of course, one has $$|I| \leq B(a,b),$$ but I would like to keep an expression in terms of the exponential of $c$. ($c$ will be some $x^2$ which should then be an integrable function on $\mathbb{R}$, that is why).

Any ideas or suggestions? Thanks a lot!! :)

Answer 1

Use Taylor series for the exponential and get

$$\int_0^1 w^{a-1}(1-w)^{b-1}\sum_{k = 0}^{+\infty}\frac{(-c)^k}{k!}w^k\ \text{d}w$$

Which becomes

$$\sum_{k = 0}^{+\infty}\frac{(-c)^k}{k!}\int_0^1 w^{a-1+k}(1-w)^{b-1}\ \text{d}w$$

The integral can be computed easily, and gives

$$\frac{\Gamma(b)\Gamma(a+k)}{\Gamma(a+b+k)}$$

which is nothing but the Beta

$$B(b, a, k)$$

hence in the end you have

$$\sum_{k = 0}^{+\infty}\frac{(-c)^k}{k!}\ B(b, a, k)$$

Further Manipulations

Since $B(a, b, k)$ is expressed as we wrote above, you can take out the first Gamma term which does not depend on $k$ and get a familiar result:

$$\Gamma(b)\sum_{k = 0}^{+\infty}\frac{(-c)^k}{k!}\frac{\Gamma(a+k)}{\Gamma(a+b+k)}$$

The series can be summed into The so called HyperGeometric Function:

$$\sum_{k = 0}^{+\infty}\frac{(-c)^k}{k!}\frac{\Gamma(a+k)}{\Gamma(a+b+k)} = \frac{\Gamma (a) \, _1F_1(a;a+b;-c)}{\Gamma (a+b)}$$

Hence the final result is:

$$\Gamma(b)\frac{\Gamma (a) \, _1F_1(a;a+b;-c)}{\Gamma (a+b)}$$

We again recognize a Beta, so finally:

$$\boxed{ B(a, b)\, _1F_1(a;a+b;-c)}$$

Answer 2

Since $c>0$, $\mathrm{e}^{-c w}$ is convex for $w \in [0,1]$. So replace $\mathrm{e}^{-c w}$ with its average value of the line through its endpoints on the interval, $\dfrac{1}{2}(\mathrm{e}^{0}+\mathrm{e}^{-c})$. That is, $$ I < \frac{1}{2}(1+\mathrm{e}^{-c}) B(a,b) \text{.} $$

Answer 3

According to Maple, it is

$$ \eqalign{ \Gamma \left( a \right) &{{\rm e}^{-c/2}}{c}^{-a/2-1-b/2}\Gamma \left( b \right) \cr&\frac{\left( \left( 1+a \right) \left( b+a \right) { {M}_{a/2-b/2+1,\,b/2+a/2+1/2}\left(c\right)}+b{{ M}_{a/2-b/2,\, b/2+a/2+1/2}\left(c\right)} \left( c+a+b \right) \right) }{\Gamma \left( b+a+2 \right) }} $$

Where $M$ is the Whittaker M function.

Answer 4

Using the following integral defintion of the confluent hypergeometric function \begin{equation} {}_{1}\mathrm{F}_{1}(\alpha ; \beta ;x) = \frac{\Gamma(\beta)}{\Gamma(\alpha)\Gamma(\beta - \alpha)} \int\limits_{0}^{1} (1-z)^{\beta - \alpha -1} z^{\alpha -1} \mathrm{e}^{xz} dz \end{equation} for $\mathrm{Re}\,\alpha \gt \mathrm{Re}\,\beta \gt 0$, we have $a=\alpha \,,\, b = \beta - \alpha \,,\, -c=x$ \begin{align} I &= \frac{\Gamma(b)\Gamma(a)}{\Gamma(a+b)}\,{}_{1}\mathrm{F}_{1}(a;a+b;-c) \\ &= \mathrm{B}(a,b)\,{}_{1}\mathrm{F}_{1}(a;a+b;-c) \end{align}