Antiderivative of $x^ae^\left( bx \right)$

Question

I need to know the primitive function (Antiderivative) of this function: $$f(x)=x^ae^\left( bx \right)$$ where $a$ and $b$ is a positive constants please how could I find the primitive of function ? is there any technique concerning this types?

Thanks in advance

Answer 1

This is a standard integration question which can be solved by using reduction formulas (just a fancy application of making repeated integration by parts). Main idea is you use integration by parts, which will ultimately required differentiating $x^a$ reducing the index by 1, then you generalize the rule, call it $I_0$, repeatedly apply until you get to $I_0$. Other similar integrals which make use of this are: $e^{x} \sin(x)$

Answer 2

The fundamental theorem of calculus states

$$\frac{d}{dx}\left(\int_{g(x)}^{\infty} h(t)\,dt\right)=-h(g(x))g'(x).\tag{1}$$

The definition of the Incomplete gamma function is $$\Gamma(a,x) = \int_x^{\infty} t^{a-1}\,e^{-t}\, dt.\tag{2}$$

Combining these two gives $$\frac{d}{dx}\Gamma(a+1,-bx)\stackrel{(2)}=\frac{d}{dx}\left(\int_{-bx}^{\infty} t^{a}\,e^{-t}\, dt\right)\stackrel{(1)}=-(-bx)^ae^{bx}(-b)=b(-b)^{a}x^ae^{bx}.\tag{3}$$

Therefore \begin{align} \int x^a e^{bx}\,dx&=\int b(-b)^{a}x^a e^{bx}b^{-1}(-b)^{-a}\,dx\\&\stackrel{(3)}= b^{-1}(-b)^{-a}\int\left(\frac{d}{dx}\Gamma(a+1,-bx)\right)\,dx\\&= b^{-1}(-b)^{-a}\Gamma(a+1,-bx)+C. \end{align}

Answer 3

Note that if $a\in\mathbb{N}$, then $$f(x)=x^ae^\left( bx \right)$$ can be written as $$f(x)=\frac{\partial^a}{\partial b^a}e^{bx}.$$ So you can first find an an antiderivative of $e^{bx}$ and then differentiate it $a$ times w.r.t $b$. Otherwise you need the incomplete gamma function as said by others.