A slight modification of Beta function The Beta function is defined as follows: for real values $a,b&gt;0$
$$B(a,b) = \int_0^1 w^{a-1}(1-w)^{b-1} dw.$$

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 $c&gt;0$. 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!! :)
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:

$$\boxed{ B(a, b)\, _1F_1(a;a+b;-c)}$$
According to Maple, it is

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

Where $M$ is the Whittaker M function.