Evaluate $\int-x^{1-n}e^{xt}\ dx$

Question

I have to evaluate $$\large\displaystyle\int-x^{1-n}e^{xt}\ dx$$ with respect to x but I am not sure how. I have tried integration by parts but this gets very complicated, is there an easier way? Thank you!

Answer 1

$$I(x)=\int-x^{1-n}\cdot\exp(xt)\,\mathrm dx$$ Let $u=-xt$ $$I(x)=\int\left(\frac {-u}t\right)^{1-n}\exp(-u)\,\mathrm du$$ $$I(x)=(-t)^{n-1}\int u^{1-n}\exp(-u)\,\mathrm du$$ The lower incomplete gamma function $\gamma(a,x)$ is defined by $$\gamma(a,x)=\int_0^xt^{a-1}\exp(-t)\,\mathrm dt$$ If $a=2-n$ then the above integral is equal to the integral in our problem, up to a constant. Therefore $$I(x)=(-t)^{n-1}\gamma(2-n, -xt)+c_1$$ However the lower incomplete gamma function is very unknown to most people, however there is another function called the incomplete gamma function, which is more known, and defined as the following: $$\Gamma(a,x)=\int_x^\infty t^{a-1}\exp(-t)\,\mathrm dt$$ By definition, the following is true, where $\Gamma(x)$ is the gamma function $$\gamma(a,x)+\Gamma(a,x)=\Gamma(a)$$ $$\gamma(a,x)=\Gamma(a)-\Gamma(a,x)$$ Therefore we can say that $$I(x)=(-t)^{n-1}\left(\Gamma(2-n)-\Gamma(2-n, -xt)\right)+c_1$$ Which reduces to $$\boxed{\displaystyle I(x)=-(-t)^{n-1}\Gamma(2-n, -xt)+c_2}$$