Mixing
Integral involving Dirac delta $delta(ax-b)$
$begingroup$
I am trying to evaluate the integral
$$int_{-infty}^{infty} f(x) delta(ax-b) , dx$$
for $aneq0$.
From what I was taught, I would expect the answer to be $fbigl(frac{b}{a}bigr)$ since $acdotfrac{b}{a}-b=0$.
However, I can also do a change of variables $u=ax$. This makes the integral
$$frac{1}{a}int_{-infty}^{infty} f Bigl(frac{u}{a}Bigr) delta(u-b) , du.$$
I would expect the value of this expression to be $frac{1}{a} fbigl(frac{b}{a}bigr)$. How am I getting two different answers? Is one of my steps invalid? Wolfram|Alpha seems to agree with the second result. If this is correct, why do I have to make the substitution to find the correct answer?
Thank you
dirac-delta
asked Jan 24 at 22:43
MathIsFunMathIsFun
1915
$endgroup$
add a comment |
$begingroup$
I am trying to evaluate the integral
$$int_{-infty}^{infty} f(x) delta(ax-b) , dx$$
for $aneq0$.
From what I was taught, I would expect the answer to be $fbigl(frac{b}{a}bigr)$ since $acdotfrac{b}{a}-b=0$.
However, I can also do a change of variables $u=ax$. This makes the integral
$$frac{1}{a}int_{-infty}^{infty} f Bigl(frac{u}{a}Bigr) delta(u-b) , du.$$
I would expect the value of this expression to be $frac{1}{a} fbigl(frac{b}{a}bigr)$. How am I getting two different answers? Is one of my steps invalid? Wolfram|Alpha seems to agree with the second result. If this is correct, why do I have to make the substitution to find the correct answer?
Thank you
dirac-delta
asked Jan 24 at 22:43
MathIsFunMathIsFun
1915
$endgroup$
$begingroup$
There's no such rule which states that if we consider: $$I=int_{-infty}^{infty} f(x)delta(cdot)~dx$$ and if the function inside the Dirac delta function has a root at a point $x_1$ then the result is $I=f(x_1)$.
$endgroup$
– projectilemotion
Jan 24 at 22:57
add a comment |
$begingroup$
I am trying to evaluate the integral
$$int_{-infty}^{infty} f(x) delta(ax-b) , dx$$
for $aneq0$.
From what I was taught, I would expect the answer to be $fbigl(frac{b}{a}bigr)$ since $acdotfrac{b}{a}-b=0$.
However, I can also do a change of variables $u=ax$. This makes the integral
$$frac{1}{a}int_{-infty}^{infty} f Bigl(frac{u}{a}Bigr) delta(u-b) , du.$$
I would expect the value of this expression to be $frac{1}{a} fbigl(frac{b}{a}bigr)$. How am I getting two different answers? Is one of my steps invalid? Wolfram|Alpha seems to agree with the second result. If this is correct, why do I have to make the substitution to find the correct answer?
Thank you
dirac-delta
asked Jan 24 at 22:43
MathIsFunMathIsFun
1915
$endgroup$
I am trying to evaluate the integral
$$int_{-infty}^{infty} f(x) delta(ax-b) , dx$$
for $aneq0$.
From what I was taught, I would expect the answer to be $fbigl(frac{b}{a}bigr)$ since $acdotfrac{b}{a}-b=0$.
However, I can also do a change of variables $u=ax$. This makes the integral
$$frac{1}{a}int_{-infty}^{infty} f Bigl(frac{u}{a}Bigr) delta(u-b) , du.$$
I would expect the value of this expression to be $frac{1}{a} fbigl(frac{b}{a}bigr)$. How am I getting two different answers? Is one of my steps invalid? Wolfram|Alpha seems to agree with the second result. If this is correct, why do I have to make the substitution to find the correct answer?
Thank you
dirac-delta
dirac-delta
asked Jan 24 at 22:43
MathIsFunMathIsFun
1915
asked Jan 24 at 22:43
MathIsFunMathIsFun
1915
asked Jan 24 at 22:43
MathIsFunMathIsFun
1915
asked Jan 24 at 22:43
MathIsFunMathIsFun
1915
asked Jan 24 at 22:43
MathIsFunMathIsFun
1915
1915
$begingroup$
There's no such rule which states that if we consider: $$I=int_{-infty}^{infty} f(x)delta(cdot)~dx$$ and if the function inside the Dirac delta function has a root at a point $x_1$ then the result is $I=f(x_1)$.
$endgroup$
– projectilemotion
Jan 24 at 22:57
add a comment |
$begingroup$
There's no such rule which states that if we consider: $$I=int_{-infty}^{infty} f(x)delta(cdot)~dx$$ and if the function inside the Dirac delta function has a root at a point $x_1$ then the result is $I=f(x_1)$.
$endgroup$
– projectilemotion
Jan 24 at 22:57
$begingroup$
There's no such rule which states that if we consider: $$I=int_{-infty}^{infty} f(x)delta(cdot)~dx$$ and if the function inside the Dirac delta function has a root at a point $x_1$ then the result is $I=f(x_1)$.
$endgroup$
– projectilemotion
Jan 24 at 22:57
$begingroup$
There's no such rule which states that if we consider: $$I=int_{-infty}^{infty} f(x)delta(cdot)~dx$$ and if the function inside the Dirac delta function has a root at a point $x_1$ then the result is $I=f(x_1)$.
$endgroup$
– projectilemotion
Jan 24 at 22:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your original solution is indeed false. Because $delta$ isn't a function, you have to be a little bit more careful on what you can and cannot do. If you scale the "argument" you will indeed get this factor. Depending on how far into the theory you want to go you can show (or accept as a definition) that for $a neq 0$
$$int_{mathbb R} f(x)delta(ax)dx = frac{1}{|a|} int_{mathbb R} f(x/a)delta(x)dx.$$
(Check out the properties of the dirac delta distribution.)
answered Jan 24 at 22:55
flawrflawr
11.7k32546
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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
$begingroup$
Your original solution is indeed false. Because $delta$ isn't a function, you have to be a little bit more careful on what you can and cannot do. If you scale the "argument" you will indeed get this factor. Depending on how far into the theory you want to go you can show (or accept as a definition) that for $a neq 0$
$$int_{mathbb R} f(x)delta(ax)dx = frac{1}{|a|} int_{mathbb R} f(x/a)delta(x)dx.$$
(Check out the properties of the dirac delta distribution.)
answered Jan 24 at 22:55
flawrflawr
11.7k32546
$endgroup$
add a comment |
$begingroup$
Your original solution is indeed false. Because $delta$ isn't a function, you have to be a little bit more careful on what you can and cannot do. If you scale the "argument" you will indeed get this factor. Depending on how far into the theory you want to go you can show (or accept as a definition) that for $a neq 0$
$$int_{mathbb R} f(x)delta(ax)dx = frac{1}{|a|} int_{mathbb R} f(x/a)delta(x)dx.$$
(Check out the properties of the dirac delta distribution.)
answered Jan 24 at 22:55
flawrflawr
11.7k32546
$endgroup$
add a comment |
$begingroup$
Your original solution is indeed false. Because $delta$ isn't a function, you have to be a little bit more careful on what you can and cannot do. If you scale the "argument" you will indeed get this factor. Depending on how far into the theory you want to go you can show (or accept as a definition) that for $a neq 0$
$$int_{mathbb R} f(x)delta(ax)dx = frac{1}{|a|} int_{mathbb R} f(x/a)delta(x)dx.$$
(Check out the properties of the dirac delta distribution.)
answered Jan 24 at 22:55
flawrflawr
11.7k32546
$endgroup$
Your original solution is indeed false. Because $delta$ isn't a function, you have to be a little bit more careful on what you can and cannot do. If you scale the "argument" you will indeed get this factor. Depending on how far into the theory you want to go you can show (or accept as a definition) that for $a neq 0$
$$int_{mathbb R} f(x)delta(ax)dx = frac{1}{|a|} int_{mathbb R} f(x/a)delta(x)dx.$$
(Check out the properties of the dirac delta distribution.)
answered Jan 24 at 22:55
flawrflawr
11.7k32546
answered Jan 24 at 22:55
flawrflawr
11.7k32546
answered Jan 24 at 22:55
flawrflawr
11.7k32546
answered Jan 24 at 22:55
flawrflawr
11.7k32546
11.7k32546
add a comment |
add a comment |
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$begingroup$
There's no such rule which states that if we consider: $$I=int_{-infty}^{infty} f(x)delta(cdot)~dx$$ and if the function inside the Dirac delta function has a root at a point $x_1$ then the result is $I=f(x_1)$.
$endgroup$
– projectilemotion
Jan 24 at 22:57