Evaluating $int_{-infty}^{infty} xf(x)delta(x-a) dx$.
1
Evaluate
$$int_{-infty}^{infty} xf(x)delta(x-a) dx.$$
I suspect that I could create another function, let's say $g(x)=xf(x)$ and perform the integral which would just give $g(a)=af(a)$ but I'm not too sure. Thanks!
integration special-functions
edited Nov 27 at 2:42
Shaun
8,642113680
asked Nov 27 at 2:32
killerownage2006
61
add a comment |
1
Evaluate
$$int_{-infty}^{infty} xf(x)delta(x-a) dx.$$
I suspect that I could create another function, let's say $g(x)=xf(x)$ and perform the integral which would just give $g(a)=af(a)$ but I'm not too sure. Thanks!
integration special-functions
edited Nov 27 at 2:42
Shaun
8,642113680
asked Nov 27 at 2:32
killerownage2006
61
Is this function $delta(x-a)$ any type of special function? I rarely see $delta$ used as a name for a function if not, and such context would be absolutely important. (Granted names are arbitrary but I generally just tend to see $f, g, h$ etc.)
– Eevee Trainer
Nov 27 at 2:33
1
take a look here for a similar question. Yes, the answer is $a f(a)$
– Masacroso
Nov 27 at 2:35
Yes it is the special function area 1 width zero. So it just selects the value at a point, like a sampler. It is used in study of signals and systems with LaPlace transforms. Commonly known as Delta Function. It is first derivative of step function. Such a trivial integral would not be online.
– John L Winters
Nov 27 at 2:41
Subtle point: One has to assume that $f(a)$ exists (because, for instance, $f$ might be defined a.e. except at $a$, otherwise).
– Shaun
Nov 27 at 2:46
add a comment |
1
1
1
Evaluate
$$int_{-infty}^{infty} xf(x)delta(x-a) dx.$$
I suspect that I could create another function, let's say $g(x)=xf(x)$ and perform the integral which would just give $g(a)=af(a)$ but I'm not too sure. Thanks!
integration special-functions
edited Nov 27 at 2:42
Shaun
8,642113680
asked Nov 27 at 2:32
killerownage2006
61
Evaluate
$$int_{-infty}^{infty} xf(x)delta(x-a) dx.$$
I suspect that I could create another function, let's say $g(x)=xf(x)$ and perform the integral which would just give $g(a)=af(a)$ but I'm not too sure. Thanks!
integration special-functions
integration special-functions
edited Nov 27 at 2:42
Shaun
8,642113680
asked Nov 27 at 2:32
killerownage2006
61
edited Nov 27 at 2:42
Shaun
8,642113680
asked Nov 27 at 2:32
killerownage2006
61
edited Nov 27 at 2:42
Shaun
8,642113680
edited Nov 27 at 2:42
Shaun
8,642113680
edited Nov 27 at 2:42
Shaun
8,642113680
8,642113680
asked Nov 27 at 2:32
killerownage2006
61
asked Nov 27 at 2:32
killerownage2006
61
asked Nov 27 at 2:32
killerownage2006
61
61
Is this function $delta(x-a)$ any type of special function? I rarely see $delta$ used as a name for a function if not, and such context would be absolutely important. (Granted names are arbitrary but I generally just tend to see $f, g, h$ etc.)
– Eevee Trainer
Nov 27 at 2:33
1
take a look here for a similar question. Yes, the answer is $a f(a)$
– Masacroso
Nov 27 at 2:35
Yes it is the special function area 1 width zero. So it just selects the value at a point, like a sampler. It is used in study of signals and systems with LaPlace transforms. Commonly known as Delta Function. It is first derivative of step function. Such a trivial integral would not be online.
– John L Winters
Nov 27 at 2:41
Subtle point: One has to assume that $f(a)$ exists (because, for instance, $f$ might be defined a.e. except at $a$, otherwise).
– Shaun
Nov 27 at 2:46
add a comment |
Is this function $delta(x-a)$ any type of special function? I rarely see $delta$ used as a name for a function if not, and such context would be absolutely important. (Granted names are arbitrary but I generally just tend to see $f, g, h$ etc.)
– Eevee Trainer
Nov 27 at 2:33
1
take a look here for a similar question. Yes, the answer is $a f(a)$
– Masacroso
Nov 27 at 2:35
Yes it is the special function area 1 width zero. So it just selects the value at a point, like a sampler. It is used in study of signals and systems with LaPlace transforms. Commonly known as Delta Function. It is first derivative of step function. Such a trivial integral would not be online.
– John L Winters
Nov 27 at 2:41
Subtle point: One has to assume that $f(a)$ exists (because, for instance, $f$ might be defined a.e. except at $a$, otherwise).
– Shaun
Nov 27 at 2:46
Is this function $delta(x-a)$ any type of special function? I rarely see $delta$ used as a name for a function if not, and such context would be absolutely important. (Granted names are arbitrary but I generally just tend to see $f, g, h$ etc.)
– Eevee Trainer
Nov 27 at 2:33
Is this function $delta(x-a)$ any type of special function? I rarely see $delta$ used as a name for a function if not, and such context would be absolutely important. (Granted names are arbitrary but I generally just tend to see $f, g, h$ etc.)
– Eevee Trainer
Nov 27 at 2:33
1
1
take a look here for a similar question. Yes, the answer is $a f(a)$
– Masacroso
Nov 27 at 2:35
take a look here for a similar question. Yes, the answer is $a f(a)$
– Masacroso
Nov 27 at 2:35
Yes it is the special function area 1 width zero. So it just selects the value at a point, like a sampler. It is used in study of signals and systems with LaPlace transforms. Commonly known as Delta Function. It is first derivative of step function. Such a trivial integral would not be online.
– John L Winters
Nov 27 at 2:41
Yes it is the special function area 1 width zero. So it just selects the value at a point, like a sampler. It is used in study of signals and systems with LaPlace transforms. Commonly known as Delta Function. It is first derivative of step function. Such a trivial integral would not be online.
– John L Winters
Nov 27 at 2:41
Subtle point: One has to assume that $f(a)$ exists (because, for instance, $f$ might be defined a.e. except at $a$, otherwise).
– Shaun
Nov 27 at 2:46
Subtle point: One has to assume that $f(a)$ exists (because, for instance, $f$ might be defined a.e. except at $a$, otherwise).
– Shaun
Nov 27 at 2:46
add a comment |
1 Answer
1
active
oldest
votes
3
The general rule in integrating over a delta function is:
$$intlimits_{-infty}^infty g(x) delta(x-a) dx = g(a)$$
Here your $g(x)$ is $x f(x)$.
answered Nov 27 at 2:44
David G. Stork
9,74921232
Okay thank you, I just wanted to confirm if I could do this.
– killerownage2006
Nov 27 at 3:06
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
The general rule in integrating over a delta function is:
$$intlimits_{-infty}^infty g(x) delta(x-a) dx = g(a)$$
Here your $g(x)$ is $x f(x)$.
answered Nov 27 at 2:44
David G. Stork
9,74921232
Okay thank you, I just wanted to confirm if I could do this.
– killerownage2006
Nov 27 at 3:06
add a comment |
3
The general rule in integrating over a delta function is:
$$intlimits_{-infty}^infty g(x) delta(x-a) dx = g(a)$$
Here your $g(x)$ is $x f(x)$.
answered Nov 27 at 2:44
David G. Stork
9,74921232
Okay thank you, I just wanted to confirm if I could do this.
– killerownage2006
Nov 27 at 3:06
add a comment |
3
3
3
The general rule in integrating over a delta function is:
$$intlimits_{-infty}^infty g(x) delta(x-a) dx = g(a)$$
Here your $g(x)$ is $x f(x)$.
answered Nov 27 at 2:44
David G. Stork
9,74921232
The general rule in integrating over a delta function is:
$$intlimits_{-infty}^infty g(x) delta(x-a) dx = g(a)$$
Here your $g(x)$ is $x f(x)$.
answered Nov 27 at 2:44
David G. Stork
9,74921232
answered Nov 27 at 2:44
David G. Stork
9,74921232
answered Nov 27 at 2:44
David G. Stork
9,74921232
answered Nov 27 at 2:44
David G. Stork
9,74921232
9,74921232
Okay thank you, I just wanted to confirm if I could do this.
– killerownage2006
Nov 27 at 3:06
add a comment |
Okay thank you, I just wanted to confirm if I could do this.
– killerownage2006
Nov 27 at 3:06
Okay thank you, I just wanted to confirm if I could do this.
– killerownage2006
Nov 27 at 3:06
Okay thank you, I just wanted to confirm if I could do this.
– killerownage2006
Nov 27 at 3:06
add a comment |
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Is this function $delta(x-a)$ any type of special function? I rarely see $delta$ used as a name for a function if not, and such context would be absolutely important. (Granted names are arbitrary but I generally just tend to see $f, g, h$ etc.)
– Eevee Trainer
Nov 27 at 2:33
1
take a look here for a similar question. Yes, the answer is $a f(a)$
– Masacroso
Nov 27 at 2:35
Yes it is the special function area 1 width zero. So it just selects the value at a point, like a sampler. It is used in study of signals and systems with LaPlace transforms. Commonly known as Delta Function. It is first derivative of step function. Such a trivial integral would not be online.
– John L Winters
Nov 27 at 2:41
Subtle point: One has to assume that $f(a)$ exists (because, for instance, $f$ might be defined a.e. except at $a$, otherwise).
– Shaun
Nov 27 at 2:46