Integrals of generally continuous functions: definition and example
$begingroup$
$f(x)$ is called "generally continuous functions" in $[a,b]$ if it is not continuous in a finite number of points of this interval.
Example: let $f(x) = dfrac{1}{|x-b|^alpha}$, with $alpha < 0$
To retrieve $displaystyle int_a^b f(x) dx$, I must retrieve $displaystyle int_a^{b+epsilon} f(x) dx$ with $epsilon < 0$. But
$$int_a^{b+epsilon} dfrac{1}{|x-b|^alpha} ; dx = int_a^{b+epsilon} dfrac{1}{(-x+b)^alpha} ; dx$$
I can't understand the last equality: how can I get $|x-b|^alpha$ from $(-x+b)^alpha$?
definite-integrals
asked Jan 9 at 14:31
user3204810user3204810
1947
$endgroup$
add a comment |
$begingroup$
$f(x)$ is called "generally continuous functions" in $[a,b]$ if it is not continuous in a finite number of points of this interval.
Example: let $f(x) = dfrac{1}{|x-b|^alpha}$, with $alpha < 0$
To retrieve $displaystyle int_a^b f(x) dx$, I must retrieve $displaystyle int_a^{b+epsilon} f(x) dx$ with $epsilon < 0$. But
$$int_a^{b+epsilon} dfrac{1}{|x-b|^alpha} ; dx = int_a^{b+epsilon} dfrac{1}{(-x+b)^alpha} ; dx$$
I can't understand the last equality: how can I get $|x-b|^alpha$ from $(-x+b)^alpha$?
definite-integrals
asked Jan 9 at 14:31
user3204810user3204810
1947
$endgroup$
add a comment |
$begingroup$
$f(x)$ is called "generally continuous functions" in $[a,b]$ if it is not continuous in a finite number of points of this interval.
Example: let $f(x) = dfrac{1}{|x-b|^alpha}$, with $alpha < 0$
To retrieve $displaystyle int_a^b f(x) dx$, I must retrieve $displaystyle int_a^{b+epsilon} f(x) dx$ with $epsilon < 0$. But
$$int_a^{b+epsilon} dfrac{1}{|x-b|^alpha} ; dx = int_a^{b+epsilon} dfrac{1}{(-x+b)^alpha} ; dx$$
I can't understand the last equality: how can I get $|x-b|^alpha$ from $(-x+b)^alpha$?
definite-integrals
asked Jan 9 at 14:31
user3204810user3204810
1947
$endgroup$
$f(x)$ is called "generally continuous functions" in $[a,b]$ if it is not continuous in a finite number of points of this interval.
Example: let $f(x) = dfrac{1}{|x-b|^alpha}$, with $alpha < 0$
To retrieve $displaystyle int_a^b f(x) dx$, I must retrieve $displaystyle int_a^{b+epsilon} f(x) dx$ with $epsilon < 0$. But
$$int_a^{b+epsilon} dfrac{1}{|x-b|^alpha} ; dx = int_a^{b+epsilon} dfrac{1}{(-x+b)^alpha} ; dx$$
I can't understand the last equality: how can I get $|x-b|^alpha$ from $(-x+b)^alpha$?
definite-integrals
definite-integrals
asked Jan 9 at 14:31
user3204810user3204810
1947
asked Jan 9 at 14:31
user3204810user3204810
1947
asked Jan 9 at 14:31
user3204810user3204810
1947
asked Jan 9 at 14:31
user3204810user3204810
1947
asked Jan 9 at 14:31
user3204810user3204810
1947
1947
add a comment |
add a comment |
1 Answer
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$begingroup$
If $x in [a, b+epsilon]$ then $|x-b|=b-x$ since $x < b$.
answered Jan 9 at 14:38
A. BailleulA. Bailleul
1587
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add a comment |
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$begingroup$
If $x in [a, b+epsilon]$ then $|x-b|=b-x$ since $x < b$.
answered Jan 9 at 14:38
A. BailleulA. Bailleul
1587
$endgroup$
add a comment |
$begingroup$
If $x in [a, b+epsilon]$ then $|x-b|=b-x$ since $x < b$.
answered Jan 9 at 14:38
A. BailleulA. Bailleul
1587
$endgroup$
add a comment |
$begingroup$
If $x in [a, b+epsilon]$ then $|x-b|=b-x$ since $x < b$.
answered Jan 9 at 14:38
A. BailleulA. Bailleul
1587
$endgroup$
If $x in [a, b+epsilon]$ then $|x-b|=b-x$ since $x < b$.
answered Jan 9 at 14:38
A. BailleulA. Bailleul
1587
answered Jan 9 at 14:38
A. BailleulA. Bailleul
1587
answered Jan 9 at 14:38
A. BailleulA. Bailleul
1587
answered Jan 9 at 14:38
A. BailleulA. Bailleul
1587
1587
add a comment |
add a comment |
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