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I know how to integrate the Gamma function using various gamma formula. But, I was wondering if the following type of problems also can be solved using the gamma integral:

$$\int_0^\infty \frac 1 {y^m}\exp(-y) \, dy$$

Please tell me the method that we can use to do the above type of integral where the value of $m$ can be any positive integer. Thanks in advance.

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$$ \int_0^\infty \frac 1 {y^m} \exp(-y) \, dy = \int_0^\infty y^n \exp(-y) \, dy \text{ where } n = -m. $$ The latter integral has a finite value if $n>-1$ and is $+\infty$ if $n\le-1.$

Therefore the former integral has a finite value if $m<1$ and is $+\infty$ if $m\ge 1.$

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    $\begingroup$ Does that mean for values of $m$ larger than$1$, the above integral will always be $\+infty$. $\endgroup$ – userNoOne Jan 9 '18 at 3:05
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    $\begingroup$ @RAHUlJHa : Yes. An easy way to see that is this: $$ \int_0^\infty \frac 1 {y^m} e^{-y} \, dy \ge \int_0^1 \frac 1 {y^m} e^{-y} \, dy \ge \int_0^1 \frac 1 {y^m} \cdot \frac 1 e \, dy = +\infty. $$ $\endgroup$ – Michael Hardy Jan 9 '18 at 4:13
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The problem happens because of the lower bound equal to $0$. $$\int_\epsilon^\infty \frac 1 {y^m} \exp(-y) \, dy =\Gamma (1-m,\epsilon )\qquad \text{if}\qquad \Im(\epsilon )\neq 0\lor \epsilon >0$$ If $\epsilon=0$, as Michael Hardy already answered, $$\int_0^\infty \frac 1 {y^m} \exp(-y) \, dy =\Gamma (1-m)\qquad \text{if}\qquad \Re(m)<1$$ Close to $m=1$, the asymptotics is $$\Gamma (1-m)=-\frac{1}{m-1}-\gamma -\frac{6 \gamma ^2+\pi ^2}{12} (m-1)+O\left((m-1)^2\right)$$

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