Reverse engineering $int m'(H) HdH $
$begingroup$
I have the following inequality for a positive, strictly decreasing, continuous, differentiable function $m:R_{++} rightarrow R_{++}$:
$$ - frac{m(H_0) - m(H_1)}{log(H_0) - log_(H_1)} > 1, forall H_1, H_0$$
where $H_1 > H_0$. Define $x = log(H)$, and take the limit as $x_1 = log(H_1)$ goes to $x_0 = log(H_0)$ to obtain:
$$ - lim_{x_1 rightarrow x_0} frac{m(e^{x_0}) - m(e^{x_1})}{x_0 - x_1} > 1$$
$$ -m'(e^{x_0})e^{x_0} > 1$$
$$ - m'(H_0)H_0 > 1$$
Now, say I want to integrate the function on the Left-Hand-Side
$$F(H)= int -m'(H)HdH$$
I should expect that I could reverse engineer the process by substituting $H = e^x$ in the integral, but since $dH = e^x dx$ I get:
$$tilde F(x) = F(e^x) = int -m'(e^x)e^{2x}dx$$
And so $tilde F(H) neq -int m(e^x)e^x dx $. Why doesn't this work as I expected it to?
calculus integration limits derivatives
edited Dec 21 '18 at 9:19
Alex Vong
1,340819
asked Dec 20 '18 at 18:54
Francisco AndrésFrancisco Andrés
556
$endgroup$
add a comment |
$begingroup$
I have the following inequality for a positive, strictly decreasing, continuous, differentiable function $m:R_{++} rightarrow R_{++}$:
$$ - frac{m(H_0) - m(H_1)}{log(H_0) - log_(H_1)} > 1, forall H_1, H_0$$
where $H_1 > H_0$. Define $x = log(H)$, and take the limit as $x_1 = log(H_1)$ goes to $x_0 = log(H_0)$ to obtain:
$$ - lim_{x_1 rightarrow x_0} frac{m(e^{x_0}) - m(e^{x_1})}{x_0 - x_1} > 1$$
$$ -m'(e^{x_0})e^{x_0} > 1$$
$$ - m'(H_0)H_0 > 1$$
Now, say I want to integrate the function on the Left-Hand-Side
$$F(H)= int -m'(H)HdH$$
I should expect that I could reverse engineer the process by substituting $H = e^x$ in the integral, but since $dH = e^x dx$ I get:
$$tilde F(x) = F(e^x) = int -m'(e^x)e^{2x}dx$$
And so $tilde F(H) neq -int m(e^x)e^x dx $. Why doesn't this work as I expected it to?
calculus integration limits derivatives
edited Dec 21 '18 at 9:19
Alex Vong
1,340819
asked Dec 20 '18 at 18:54
Francisco AndrésFrancisco Andrés
556
$endgroup$
add a comment |
$begingroup$
I have the following inequality for a positive, strictly decreasing, continuous, differentiable function $m:R_{++} rightarrow R_{++}$:
$$ - frac{m(H_0) - m(H_1)}{log(H_0) - log_(H_1)} > 1, forall H_1, H_0$$
where $H_1 > H_0$. Define $x = log(H)$, and take the limit as $x_1 = log(H_1)$ goes to $x_0 = log(H_0)$ to obtain:
$$ - lim_{x_1 rightarrow x_0} frac{m(e^{x_0}) - m(e^{x_1})}{x_0 - x_1} > 1$$
$$ -m'(e^{x_0})e^{x_0} > 1$$
$$ - m'(H_0)H_0 > 1$$
Now, say I want to integrate the function on the Left-Hand-Side
$$F(H)= int -m'(H)HdH$$
I should expect that I could reverse engineer the process by substituting $H = e^x$ in the integral, but since $dH = e^x dx$ I get:
$$tilde F(x) = F(e^x) = int -m'(e^x)e^{2x}dx$$
And so $tilde F(H) neq -int m(e^x)e^x dx $. Why doesn't this work as I expected it to?
calculus integration limits derivatives
edited Dec 21 '18 at 9:19
Alex Vong
1,340819
asked Dec 20 '18 at 18:54
Francisco AndrésFrancisco Andrés
556
$endgroup$
I have the following inequality for a positive, strictly decreasing, continuous, differentiable function $m:R_{++} rightarrow R_{++}$:
$$ - frac{m(H_0) - m(H_1)}{log(H_0) - log_(H_1)} > 1, forall H_1, H_0$$
where $H_1 > H_0$. Define $x = log(H)$, and take the limit as $x_1 = log(H_1)$ goes to $x_0 = log(H_0)$ to obtain:
$$ - lim_{x_1 rightarrow x_0} frac{m(e^{x_0}) - m(e^{x_1})}{x_0 - x_1} > 1$$
$$ -m'(e^{x_0})e^{x_0} > 1$$
$$ - m'(H_0)H_0 > 1$$
Now, say I want to integrate the function on the Left-Hand-Side
$$F(H)= int -m'(H)HdH$$
I should expect that I could reverse engineer the process by substituting $H = e^x$ in the integral, but since $dH = e^x dx$ I get:
$$tilde F(x) = F(e^x) = int -m'(e^x)e^{2x}dx$$
And so $tilde F(H) neq -int m(e^x)e^x dx $. Why doesn't this work as I expected it to?
calculus integration limits derivatives
calculus integration limits derivatives
edited Dec 21 '18 at 9:19
Alex Vong
1,340819
asked Dec 20 '18 at 18:54
Francisco AndrésFrancisco Andrés
556
edited Dec 21 '18 at 9:19
Alex Vong
1,340819
asked Dec 20 '18 at 18:54
Francisco AndrésFrancisco Andrés
556
edited Dec 21 '18 at 9:19
Alex Vong
1,340819
edited Dec 21 '18 at 9:19
Alex Vong
1,340819
edited Dec 21 '18 at 9:19
Alex Vong
1,340819
1,340819
asked Dec 20 '18 at 18:54
Francisco AndrésFrancisco Andrés
556
asked Dec 20 '18 at 18:54
Francisco AndrésFrancisco Andrés
556
asked Dec 20 '18 at 18:54
Francisco AndrésFrancisco Andrés
556
556
add a comment |
add a comment |
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