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Carlo Beenakker
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$$L_{n,m}(t)=\int \frac{x^n}{(1+x)^m} e^{-xt}\,dx=\Gamma (n+1) U(n+1,-m+n+2,t),$$ with $U$ the confluent hypergeometric function.

More explicit expressions, in terms of an incomplete Gamma function, can be given for definite values of $n,m$, for example, $$\{L_{1,1},L_{1,2},\ldots L_{1,5}\}=\left\{e^t \Gamma (-1,t),2 e^t \Gamma (-2,t),6 e^t \Gamma (-3,t),24 e^t \Gamma (-4,t),120 e^t \Gamma (-5,t)\right\}$$ $$\{L_{2,2},L_{2,3},\ldots L_{2,5}\}=\left\{\frac{1}{t}-e^t (t+2) \Gamma (0,t)+1,\frac{1}{t^2}-\frac{2}{t}+e^t (t+3) \Gamma (0,t)-1,\frac{t (t (t+3)-2)+2}{t^3}-e^t (t+4) \Gamma (0,t),e^t (t+5) \Gamma (0,t)-\frac{(t-1) (t (t (t+5)+2)+6)}{t^4}\right\}.$$

$$L_{n,m}(t)=\int \frac{x^n}{(1+x)^m} e^{-xt}\,dx=\Gamma (n+1) U(n+1,-m+n+2,t),$$ with $U$ the confluent hypergeometric function.

More explicit expressions can be given for definite values of $n,m$, for example, $$\{L_{1,1},L_{1,2},\ldots L_{1,5}\}=\left\{e^t \Gamma (-1,t),2 e^t \Gamma (-2,t),6 e^t \Gamma (-3,t),24 e^t \Gamma (-4,t),120 e^t \Gamma (-5,t)\right\}$$ $$\{L_{2,2},L_{2,3},\ldots L_{2,5}\}=\left\{\frac{1}{t}-e^t (t+2) \Gamma (0,t)+1,\frac{1}{t^2}-\frac{2}{t}+e^t (t+3) \Gamma (0,t)-1,\frac{t (t (t+3)-2)+2}{t^3}-e^t (t+4) \Gamma (0,t),e^t (t+5) \Gamma (0,t)-\frac{(t-1) (t (t (t+5)+2)+6)}{t^4}\right\}.$$

$$L_{n,m}(t)=\int \frac{x^n}{(1+x)^m} e^{-xt}\,dx=\Gamma (n+1) U(n+1,-m+n+2,t),$$ with $U$ the confluent hypergeometric function.

More explicit expressions, in terms of an incomplete Gamma function, can be given for definite values of $n,m$, for example, $$\{L_{1,1},L_{1,2},\ldots L_{1,5}\}=\left\{e^t \Gamma (-1,t),2 e^t \Gamma (-2,t),6 e^t \Gamma (-3,t),24 e^t \Gamma (-4,t),120 e^t \Gamma (-5,t)\right\}$$ $$\{L_{2,2},L_{2,3},\ldots L_{2,5}\}=\left\{\frac{1}{t}-e^t (t+2) \Gamma (0,t)+1,\frac{1}{t^2}-\frac{2}{t}+e^t (t+3) \Gamma (0,t)-1,\frac{t (t (t+3)-2)+2}{t^3}-e^t (t+4) \Gamma (0,t),e^t (t+5) \Gamma (0,t)-\frac{(t-1) (t (t (t+5)+2)+6)}{t^4}\right\}.$$
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Carlo Beenakker
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$$\int \frac{x^n}{(1+x)^m} e^{-xt}\,dx=\Gamma (n+1) U(n+1,-m+n+2,t),$$$$L_{n,m}(t)=\int \frac{x^n}{(1+x)^m} e^{-xt}\,dx=\Gamma (n+1) U(n+1,-m+n+2,t),$$ with $U$ the confluent hypergeometric function.

More explicit expressions can be given for definite values of $n,m$, for example, $$\{L_{1,1},L_{1,2},\ldots L_{1,5}\}=\left\{e^t \Gamma (-1,t),2 e^t \Gamma (-2,t),6 e^t \Gamma (-3,t),24 e^t \Gamma (-4,t),120 e^t \Gamma (-5,t)\right\}$$ $$\{L_{2,2},L_{2,3},\ldots L_{2,5}\}=\left\{\frac{1}{t}-e^t (t+2) \Gamma (0,t)+1,\frac{1}{t^2}-\frac{2}{t}+e^t (t+3) \Gamma (0,t)-1,\frac{t (t (t+3)-2)+2}{t^3}-e^t (t+4) \Gamma (0,t),e^t (t+5) \Gamma (0,t)-\frac{(t-1) (t (t (t+5)+2)+6)}{t^4}\right\}.$$

$$\int \frac{x^n}{(1+x)^m} e^{-xt}\,dx=\Gamma (n+1) U(n+1,-m+n+2,t),$$ with $U$ the confluent hypergeometric function.

$$L_{n,m}(t)=\int \frac{x^n}{(1+x)^m} e^{-xt}\,dx=\Gamma (n+1) U(n+1,-m+n+2,t),$$ with $U$ the confluent hypergeometric function.

More explicit expressions can be given for definite values of $n,m$, for example, $$\{L_{1,1},L_{1,2},\ldots L_{1,5}\}=\left\{e^t \Gamma (-1,t),2 e^t \Gamma (-2,t),6 e^t \Gamma (-3,t),24 e^t \Gamma (-4,t),120 e^t \Gamma (-5,t)\right\}$$ $$\{L_{2,2},L_{2,3},\ldots L_{2,5}\}=\left\{\frac{1}{t}-e^t (t+2) \Gamma (0,t)+1,\frac{1}{t^2}-\frac{2}{t}+e^t (t+3) \Gamma (0,t)-1,\frac{t (t (t+3)-2)+2}{t^3}-e^t (t+4) \Gamma (0,t),e^t (t+5) \Gamma (0,t)-\frac{(t-1) (t (t (t+5)+2)+6)}{t^4}\right\}.$$
Source Link
Carlo Beenakker
  • 188.1k
  • 18
  • 448
  • 651

$$\int \frac{x^n}{(1+x)^m} e^{-xt}\,dx=\Gamma (n+1) U(n+1,-m+n+2,t),$$ with $U$ the confluent hypergeometric function.