Special case of this general integral
$begingroup$
We know that the integral of,
$$ int frac{1}{x} dx = ln x$$
ignoring any integration constants.
Consider the integral of some more `general' function,
$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx $$
i.e. in the $y=z=0$ case we recover the original integral ($1/x$)
Now, if I put the second integral into WolframAlpha, it spits out,
$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx = ln left(sqrt{x^2+y^2+z^2} + xright)$$.
I naively expected that in the $y=z=0$ case, this answer would revert to $ln x$, but instead we get $ln 2x$
What gives?
integration
asked Dec 4 '18 at 18:25
user1887919user1887919
225112
$endgroup$
add a comment |
$begingroup$
We know that the integral of,
$$ int frac{1}{x} dx = ln x$$
ignoring any integration constants.
Consider the integral of some more `general' function,
$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx $$
i.e. in the $y=z=0$ case we recover the original integral ($1/x$)
Now, if I put the second integral into WolframAlpha, it spits out,
$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx = ln left(sqrt{x^2+y^2+z^2} + xright)$$.
I naively expected that in the $y=z=0$ case, this answer would revert to $ln x$, but instead we get $ln 2x$
What gives?
integration
asked Dec 4 '18 at 18:25
user1887919user1887919
225112
$endgroup$
4
$begingroup$
Technically it's $intfrac{1}{x}dx=ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,,x>0$.
$endgroup$
– J.G.
Dec 4 '18 at 18:30
add a comment |
$begingroup$
We know that the integral of,
$$ int frac{1}{x} dx = ln x$$
ignoring any integration constants.
Consider the integral of some more `general' function,
$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx $$
i.e. in the $y=z=0$ case we recover the original integral ($1/x$)
Now, if I put the second integral into WolframAlpha, it spits out,
$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx = ln left(sqrt{x^2+y^2+z^2} + xright)$$.
I naively expected that in the $y=z=0$ case, this answer would revert to $ln x$, but instead we get $ln 2x$
What gives?
integration
asked Dec 4 '18 at 18:25
user1887919user1887919
225112
$endgroup$
We know that the integral of,
$$ int frac{1}{x} dx = ln x$$
ignoring any integration constants.
Consider the integral of some more `general' function,
$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx $$
i.e. in the $y=z=0$ case we recover the original integral ($1/x$)
Now, if I put the second integral into WolframAlpha, it spits out,
$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx = ln left(sqrt{x^2+y^2+z^2} + xright)$$.
I naively expected that in the $y=z=0$ case, this answer would revert to $ln x$, but instead we get $ln 2x$
What gives?
integration
integration
asked Dec 4 '18 at 18:25
user1887919user1887919
225112
asked Dec 4 '18 at 18:25
user1887919user1887919
225112
asked Dec 4 '18 at 18:25
user1887919user1887919
225112
asked Dec 4 '18 at 18:25
user1887919user1887919
225112
asked Dec 4 '18 at 18:25
user1887919user1887919
225112
225112
4
$begingroup$
Technically it's $intfrac{1}{x}dx=ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,,x>0$.
$endgroup$
– J.G.
Dec 4 '18 at 18:30
add a comment |
4
$begingroup$
Technically it's $intfrac{1}{x}dx=ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,,x>0$.
$endgroup$
– J.G.
Dec 4 '18 at 18:30
4
4
$begingroup$
Technically it's $intfrac{1}{x}dx=ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,,x>0$.
$endgroup$
– J.G.
Dec 4 '18 at 18:30
$begingroup$
Technically it's $intfrac{1}{x}dx=ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,,x>0$.
$endgroup$
– J.G.
Dec 4 '18 at 18:30
add a comment |
1 Answer
1
active
oldest
votes
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Note that
$$ln 2x=ln x +ln 2=ln x +C$$
answered Dec 4 '18 at 18:27
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
$endgroup$
– user1887919
Dec 4 '18 at 18:30
$begingroup$
@user1887919 No since the constant cancels out, let try also with some numerical example.
$endgroup$
– gimusi
Dec 4 '18 at 18:31
$begingroup$
@user1887919 You are welcome! Refer also to that related OP, Bye
$endgroup$
– gimusi
Dec 4 '18 at 18:36
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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$begingroup$
Note that
$$ln 2x=ln x +ln 2=ln x +C$$
answered Dec 4 '18 at 18:27
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
$endgroup$
– user1887919
Dec 4 '18 at 18:30
$begingroup$
@user1887919 No since the constant cancels out, let try also with some numerical example.
$endgroup$
– gimusi
Dec 4 '18 at 18:31
$begingroup$
@user1887919 You are welcome! Refer also to that related OP, Bye
$endgroup$
– gimusi
Dec 4 '18 at 18:36
add a comment |
$begingroup$
Note that
$$ln 2x=ln x +ln 2=ln x +C$$
answered Dec 4 '18 at 18:27
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
$endgroup$
– user1887919
Dec 4 '18 at 18:30
$begingroup$
@user1887919 No since the constant cancels out, let try also with some numerical example.
$endgroup$
– gimusi
Dec 4 '18 at 18:31
$begingroup$
@user1887919 You are welcome! Refer also to that related OP, Bye
$endgroup$
– gimusi
Dec 4 '18 at 18:36
add a comment |
$begingroup$
Note that
$$ln 2x=ln x +ln 2=ln x +C$$
answered Dec 4 '18 at 18:27
gimusigimusi
92.9k84494
$endgroup$
Note that
$$ln 2x=ln x +ln 2=ln x +C$$
answered Dec 4 '18 at 18:27
gimusigimusi
92.9k84494
answered Dec 4 '18 at 18:27
gimusigimusi
92.9k84494
answered Dec 4 '18 at 18:27
gimusigimusi
92.9k84494
answered Dec 4 '18 at 18:27
gimusigimusi
92.9k84494
92.9k84494
$begingroup$
Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
$endgroup$
– user1887919
Dec 4 '18 at 18:30
$begingroup$
@user1887919 No since the constant cancels out, let try also with some numerical example.
$endgroup$
– gimusi
Dec 4 '18 at 18:31
$begingroup$
@user1887919 You are welcome! Refer also to that related OP, Bye
$endgroup$
– gimusi
Dec 4 '18 at 18:36
add a comment |
$begingroup$
Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
$endgroup$
– user1887919
Dec 4 '18 at 18:30
$begingroup$
@user1887919 No since the constant cancels out, let try also with some numerical example.
$endgroup$
– gimusi
Dec 4 '18 at 18:31
$begingroup$
@user1887919 You are welcome! Refer also to that related OP, Bye
$endgroup$
– gimusi
Dec 4 '18 at 18:36
$begingroup$
Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
$endgroup$
– user1887919
Dec 4 '18 at 18:30
$begingroup$
Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
$endgroup$
– user1887919
Dec 4 '18 at 18:30
$begingroup$
@user1887919 No since the constant cancels out, let try also with some numerical example.
$endgroup$
– gimusi
Dec 4 '18 at 18:31
$begingroup$
@user1887919 No since the constant cancels out, let try also with some numerical example.
$endgroup$
– gimusi
Dec 4 '18 at 18:31
$begingroup$
@user1887919 You are welcome! Refer also to that related OP, Bye
$endgroup$
– gimusi
Dec 4 '18 at 18:36
$begingroup$
@user1887919 You are welcome! Refer also to that related OP, Bye
$endgroup$
– gimusi
Dec 4 '18 at 18:36
add a comment |
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$begingroup$
Technically it's $intfrac{1}{x}dx=ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,,x>0$.
$endgroup$
– J.G.
Dec 4 '18 at 18:30