Why is this integral (that looks complicated) so easy to evaluate?












4














I encounter the following integral during Statistics lecture and the professor quickly solved it as if he assumed we know that this integral is easy so solve:



$$ int_{0}^{infty} xe^{-(1-t)x} dx = frac{1}{(1-t)^2}$$



I suspect that this involves a probability distribution but I do not know which. Why is this integral so easy to solve? I know I must have been missing something.






integration probability-theory probability-distributions improper-integrals







share|cite|improve this question




















  • 2




    If all else fails, the integral is quite easily solved using integration by parts.
    – MisterRiemann
    Nov 25 '18 at 10:26










  • The solution holds only as long as $t<1$.
    – Alecos Papadopoulos
    Nov 26 '18 at 1:57


















4














I encounter the following integral during Statistics lecture and the professor quickly solved it as if he assumed we know that this integral is easy so solve:



$$ int_{0}^{infty} xe^{-(1-t)x} dx = frac{1}{(1-t)^2}$$



I suspect that this involves a probability distribution but I do not know which. Why is this integral so easy to solve? I know I must have been missing something.






integration probability-theory probability-distributions improper-integrals







share|cite|improve this question




















  • 2




    If all else fails, the integral is quite easily solved using integration by parts.
    – MisterRiemann
    Nov 25 '18 at 10:26










  • The solution holds only as long as $t<1$.
    – Alecos Papadopoulos
    Nov 26 '18 at 1:57
















4












4








4







I encounter the following integral during Statistics lecture and the professor quickly solved it as if he assumed we know that this integral is easy so solve:



$$ int_{0}^{infty} xe^{-(1-t)x} dx = frac{1}{(1-t)^2}$$



I suspect that this involves a probability distribution but I do not know which. Why is this integral so easy to solve? I know I must have been missing something.






integration probability-theory probability-distributions improper-integrals







share|cite|improve this question















I encounter the following integral during Statistics lecture and the professor quickly solved it as if he assumed we know that this integral is easy so solve:



$$ int_{0}^{infty} xe^{-(1-t)x} dx = frac{1}{(1-t)^2}$$



I suspect that this involves a probability distribution but I do not know which. Why is this integral so easy to solve? I know I must have been missing something.






integration probability-theory probability-distributions improper-integrals




integration probability-theory probability-distributions improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 10:31









José Carlos Santos

150k22121221




150k22121221










asked Nov 25 '18 at 10:24









hephaes

1729




1729








  • 2




    If all else fails, the integral is quite easily solved using integration by parts.
    – MisterRiemann
    Nov 25 '18 at 10:26










  • The solution holds only as long as $t<1$.
    – Alecos Papadopoulos
    Nov 26 '18 at 1:57
















  • 2




    If all else fails, the integral is quite easily solved using integration by parts.
    – MisterRiemann
    Nov 25 '18 at 10:26










  • The solution holds only as long as $t<1$.
    – Alecos Papadopoulos
    Nov 26 '18 at 1:57










2




2




If all else fails, the integral is quite easily solved using integration by parts.
– MisterRiemann
Nov 25 '18 at 10:26




If all else fails, the integral is quite easily solved using integration by parts.
– MisterRiemann
Nov 25 '18 at 10:26












The solution holds only as long as $t<1$.
– Alecos Papadopoulos
Nov 26 '18 at 1:57






The solution holds only as long as $t<1$.
– Alecos Papadopoulos
Nov 26 '18 at 1:57












1 Answer
1






active

oldest

votes


















3














For each $lambda<0$,begin{align}int_0^infty xe^{lambda x},mathrm dx&=left[xfrac{e^{lambda x}}lambdaright]_{x=0}^{x=infty}-int_0^inftyfrac{e^{lambda x}}lambda,mathrm dx\&=-left[frac{e^{lambda x}}{lambda^2}right]_{x=0}^{x=infty}\&=frac1{lambda^2}.end{align}






share|cite|improve this answer





















  • How can the mean of exponential rv be negative
    – Neymar
    Nov 25 '18 at 10:33










  • Thank you for the answer! May I also know why can we be sure $xe^{-lambda x} $ always converges to some number as $x$ tends to infinity?
    – hephaes
    Nov 25 '18 at 10:33












  • By the rule of lhopitals
    – Neymar
    Nov 25 '18 at 10:34










  • @Neymar I see. Thanks.
    – hephaes
    Nov 25 '18 at 10:35











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3














For each $lambda<0$,begin{align}int_0^infty xe^{lambda x},mathrm dx&=left[xfrac{e^{lambda x}}lambdaright]_{x=0}^{x=infty}-int_0^inftyfrac{e^{lambda x}}lambda,mathrm dx\&=-left[frac{e^{lambda x}}{lambda^2}right]_{x=0}^{x=infty}\&=frac1{lambda^2}.end{align}






share|cite|improve this answer





















  • How can the mean of exponential rv be negative
    – Neymar
    Nov 25 '18 at 10:33










  • Thank you for the answer! May I also know why can we be sure $xe^{-lambda x} $ always converges to some number as $x$ tends to infinity?
    – hephaes
    Nov 25 '18 at 10:33












  • By the rule of lhopitals
    – Neymar
    Nov 25 '18 at 10:34










  • @Neymar I see. Thanks.
    – hephaes
    Nov 25 '18 at 10:35
















3














For each $lambda<0$,begin{align}int_0^infty xe^{lambda x},mathrm dx&=left[xfrac{e^{lambda x}}lambdaright]_{x=0}^{x=infty}-int_0^inftyfrac{e^{lambda x}}lambda,mathrm dx\&=-left[frac{e^{lambda x}}{lambda^2}right]_{x=0}^{x=infty}\&=frac1{lambda^2}.end{align}






share|cite|improve this answer





















  • How can the mean of exponential rv be negative
    – Neymar
    Nov 25 '18 at 10:33










  • Thank you for the answer! May I also know why can we be sure $xe^{-lambda x} $ always converges to some number as $x$ tends to infinity?
    – hephaes
    Nov 25 '18 at 10:33












  • By the rule of lhopitals
    – Neymar
    Nov 25 '18 at 10:34










  • @Neymar I see. Thanks.
    – hephaes
    Nov 25 '18 at 10:35














3












3








3






For each $lambda<0$,begin{align}int_0^infty xe^{lambda x},mathrm dx&=left[xfrac{e^{lambda x}}lambdaright]_{x=0}^{x=infty}-int_0^inftyfrac{e^{lambda x}}lambda,mathrm dx\&=-left[frac{e^{lambda x}}{lambda^2}right]_{x=0}^{x=infty}\&=frac1{lambda^2}.end{align}






share|cite|improve this answer












For each $lambda<0$,begin{align}int_0^infty xe^{lambda x},mathrm dx&=left[xfrac{e^{lambda x}}lambdaright]_{x=0}^{x=infty}-int_0^inftyfrac{e^{lambda x}}lambda,mathrm dx\&=-left[frac{e^{lambda x}}{lambda^2}right]_{x=0}^{x=infty}\&=frac1{lambda^2}.end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 '18 at 10:30









José Carlos Santos

150k22121221




150k22121221












  • How can the mean of exponential rv be negative
    – Neymar
    Nov 25 '18 at 10:33










  • Thank you for the answer! May I also know why can we be sure $xe^{-lambda x} $ always converges to some number as $x$ tends to infinity?
    – hephaes
    Nov 25 '18 at 10:33












  • By the rule of lhopitals
    – Neymar
    Nov 25 '18 at 10:34










  • @Neymar I see. Thanks.
    – hephaes
    Nov 25 '18 at 10:35


















  • How can the mean of exponential rv be negative
    – Neymar
    Nov 25 '18 at 10:33










  • Thank you for the answer! May I also know why can we be sure $xe^{-lambda x} $ always converges to some number as $x$ tends to infinity?
    – hephaes
    Nov 25 '18 at 10:33












  • By the rule of lhopitals
    – Neymar
    Nov 25 '18 at 10:34










  • @Neymar I see. Thanks.
    – hephaes
    Nov 25 '18 at 10:35
















How can the mean of exponential rv be negative
– Neymar
Nov 25 '18 at 10:33




How can the mean of exponential rv be negative
– Neymar
Nov 25 '18 at 10:33












Thank you for the answer! May I also know why can we be sure $xe^{-lambda x} $ always converges to some number as $x$ tends to infinity?
– hephaes
Nov 25 '18 at 10:33






Thank you for the answer! May I also know why can we be sure $xe^{-lambda x} $ always converges to some number as $x$ tends to infinity?
– hephaes
Nov 25 '18 at 10:33














By the rule of lhopitals
– Neymar
Nov 25 '18 at 10:34




By the rule of lhopitals
– Neymar
Nov 25 '18 at 10:34












@Neymar I see. Thanks.
– hephaes
Nov 25 '18 at 10:35




@Neymar I see. Thanks.
– hephaes
Nov 25 '18 at 10:35


















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