Why is this integral (that looks complicated) so easy to evaluate?
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4
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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
edited Nov 25 at 10:31
José Carlos Santos
145k22115214
asked Nov 25 at 10:24
hephaes
1627
add a comment |
up vote
4
down vote
favorite
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
edited Nov 25 at 10:31
José Carlos Santos
145k22115214
asked Nov 25 at 10:24
hephaes
1627
2
If all else fails, the integral is quite easily solved using integration by parts.
– MisterRiemann
Nov 25 at 10:26
The solution holds only as long as $t<1$.
– Alecos Papadopoulos
Nov 26 at 1:57
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
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
edited Nov 25 at 10:31
José Carlos Santos
145k22115214
asked Nov 25 at 10:24
hephaes
1627
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
edited Nov 25 at 10:31
José Carlos Santos
145k22115214
asked Nov 25 at 10:24
hephaes
1627
edited Nov 25 at 10:31
José Carlos Santos
145k22115214
asked Nov 25 at 10:24
hephaes
1627
edited Nov 25 at 10:31
José Carlos Santos
145k22115214
edited Nov 25 at 10:31
José Carlos Santos
145k22115214
edited Nov 25 at 10:31
José Carlos Santos
145k22115214
145k22115214
asked Nov 25 at 10:24
hephaes
1627
asked Nov 25 at 10:24
hephaes
1627
asked Nov 25 at 10:24
hephaes
1627
1627
2
If all else fails, the integral is quite easily solved using integration by parts.
– MisterRiemann
Nov 25 at 10:26
The solution holds only as long as $t<1$.
– Alecos Papadopoulos
Nov 26 at 1:57
add a comment |
2
If all else fails, the integral is quite easily solved using integration by parts.
– MisterRiemann
Nov 25 at 10:26
The solution holds only as long as $t<1$.
– Alecos Papadopoulos
Nov 26 at 1:57
2
2
If all else fails, the integral is quite easily solved using integration by parts.
– MisterRiemann
Nov 25 at 10:26
If all else fails, the integral is quite easily solved using integration by parts.
– MisterRiemann
Nov 25 at 10:26
The solution holds only as long as $t<1$.
– Alecos Papadopoulos
Nov 26 at 1:57
The solution holds only as long as $t<1$.
– Alecos Papadopoulos
Nov 26 at 1:57
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
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}
answered Nov 25 at 10:30
José Carlos Santos
145k22115214
How can the mean of exponential rv be negative
– Neymar
Nov 25 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 at 10:33
By the rule of lhopitals
– Neymar
Nov 25 at 10:34
@Neymar I see. Thanks.
– hephaes
Nov 25 at 10:35
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
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}
answered Nov 25 at 10:30
José Carlos Santos
145k22115214
How can the mean of exponential rv be negative
– Neymar
Nov 25 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 at 10:33
By the rule of lhopitals
– Neymar
Nov 25 at 10:34
@Neymar I see. Thanks.
– hephaes
Nov 25 at 10:35
add a comment |
up vote
3
down vote
accepted
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}
answered Nov 25 at 10:30
José Carlos Santos
145k22115214
How can the mean of exponential rv be negative
– Neymar
Nov 25 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 at 10:33
By the rule of lhopitals
– Neymar
Nov 25 at 10:34
@Neymar I see. Thanks.
– hephaes
Nov 25 at 10:35
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
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}
answered Nov 25 at 10:30
José Carlos Santos
145k22115214
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}
answered Nov 25 at 10:30
José Carlos Santos
145k22115214
answered Nov 25 at 10:30
José Carlos Santos
145k22115214
answered Nov 25 at 10:30
José Carlos Santos
145k22115214
answered Nov 25 at 10:30
José Carlos Santos
145k22115214
145k22115214
How can the mean of exponential rv be negative
– Neymar
Nov 25 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 at 10:33
By the rule of lhopitals
– Neymar
Nov 25 at 10:34
@Neymar I see. Thanks.
– hephaes
Nov 25 at 10:35
add a comment |
How can the mean of exponential rv be negative
– Neymar
Nov 25 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 at 10:33
By the rule of lhopitals
– Neymar
Nov 25 at 10:34
@Neymar I see. Thanks.
– hephaes
Nov 25 at 10:35
How can the mean of exponential rv be negative
– Neymar
Nov 25 at 10:33
How can the mean of exponential rv be negative
– Neymar
Nov 25 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 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 at 10:33
By the rule of lhopitals
– Neymar
Nov 25 at 10:34
By the rule of lhopitals
– Neymar
Nov 25 at 10:34
@Neymar I see. Thanks.
– hephaes
Nov 25 at 10:35
@Neymar I see. Thanks.
– hephaes
Nov 25 at 10:35
add a comment |
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If all else fails, the integral is quite easily solved using integration by parts.
– MisterRiemann
Nov 25 at 10:26
The solution holds only as long as $t<1$.
– Alecos Papadopoulos
Nov 26 at 1:57