<p>Rewrite, by separating variables:
<span class="math-container">$$
\frac{1}{Q} d Q= -\frac{1}{P} dP
$$</span>
Integrating both parts gives:
<span class="math-container">$$
\ln(Q) = - \ln(P) + C = \ln(1/P) + C, \qquad C \in \mathbb{R}
$$</span>
Taking exponents on both sides gives that there is  a <span class="math-container">$A  = e^C &gt; 0$</span> such that:
<span class="math-container">$$
Q = -\frac{A}{P}
$$</span></p>
<p>Double check for correctness. We have <span class="math-container">$\frac{dQ}{dP} = -\frac{A}{P^2}$</span>, so:
<span class="math-container">$$
\frac{dQ}{dP} \frac{P}{Q} = -\frac{A}{P^2}\frac{P}{-\frac{A}{P}} = -1.
$$</span></p>

CC BY-SA 4.0
2024-09-23T14:21:23.940Z
2024-09-23T14:21:23.940Z
tdm
<p>There is mistake already in the step 2. In this problem <span class="math-container">$Q(P)$</span> so you need to take integral wrt <span class="math-container">$P$</span> so the equations should look like;</p>
<p><span class="math-container">$$\int (P dQ) dP=− \int (Q dP) dP$$</span></p>

CC BY-SA 4.0
2024-09-22T23:15:26.200Z
But can't I write the equations as $\int PQ'(P)\ dP=\int Q(P)\ dP$? This makes the integration variable with respect to P on both sides.
vbxr
CC BY-SA 4.0
2024-09-22T10:56:10.500Z
2024-09-22T10:56:10.500Z
1muflon1
<p>I was trying to solve <span class="math-container">$E_D=\dfrac{dQ}{dP}\cdot\dfrac{P}{Q}=-1$</span>, assuming <span class="math-container">$Q\neq0$</span>.</p>
<p><span class="math-container">$$
\begin{align*}
P\ dQ&amp;=-Q\ dP\\
\int P\ dQ&amp;=-\int Q\ dP\\
\end{align*}
$$</span></p>
<p>Integration by parts yields:</p>
<p><span class="math-container">$$
\begin{align*}
PQ-\int Q\ dP&amp;=-\int Q\ dP\\
PQ&amp;=0\\
P&amp;=0
\end{align*}
$$</span></p>
<p>But this isn't right for many reasons. First, plugging <span class="math-container">$P=0$</span> into the original equation for <span class="math-container">$E_D$</span> yields 0 (since it's safe to assume <span class="math-container">$dQ\neq0$</span> and <span class="math-container">$dP\neq0$</span>). Second, when there's unit elasticity, this should maximize revenue, but if <span class="math-container">$P=0$</span>, this means revenue is 0.</p>
<p>What have I done wrong in these calculations?</p>

CC BY-SA 4.0
2024-09-22T10:50:30.680Z
Welcome to the site.  Are you assuming that elasticity of demand is constant at $-1$ over the whole demand curve, or just that it equals $-1$ at one point?  Presumably the former, otherwise the integration wouldn't make sense.
Adam Bailey
CC BY-SA 4.0
2024-09-22T23:14:39.597Z
@AdamBailey I think that might be the reason why, although that doesn't explain why plugging in $P=0$ wouldn't make $E_D=-1$ (just purely from a math point of view, no microeconomics perspective). I was assuming the latter. What would happen if I assumed the former?
vbxr
<microeconomics><elasticity>
Elasticity of demand calculation yields unit elasticity at $0. What's wrong?
vbxr
# Elasticity of demand calculation yields unit elasticity at $0. What's wrong?

(Asked by: username_0 on 2024-09-22)

I was trying to solve $E\_D=\dfrac{dQ}{dP}\cdot\dfrac{P}{Q}=-1$, assuming $Q\neq0$.

$$
\begin{align\*}
P\ dQ&=-Q\ dP\\
\int P\ dQ&=-\int Q\ dP\\
\end{align\*}
$$

Integration by parts yields:

$$
\begin{align\*}
PQ-\int Q\ dP&=-\int Q\ dP\\
PQ&=0\\
P&=0
\end{align\*}
$$

But this isn't right for many reasons. First, plugging $P=0$ into the original equation for $E\_D$ yields 0 (since it's safe to assume $dQ\neq0$ and $dP\neq0$). Second, when there's unit elasticity, this should maximize revenue, but if $P=0$, this means revenue is 0.

What have I done wrong in these calculations?

**Tags:** microeconomics, elasticity

> **Comments:**
> **username_1**: Welcome to the site.  Are you assuming that elasticity of demand is constant at $-1$ over the whole demand curve, or just that it equals $-1$ at one point?  Presumably the former, otherwise the integration wouldn't make sense.  
> **username_0**: @AdamBailey I think that might be the reason why, although that doesn't explain why plugging in $P=0$ wouldn't make $E_D=-1$ (just purely from a math point of view, no microeconomics perspective). I was assuming the latter. What would happen if I assumed the former?  

---

## 2 answers

### Answer by username_2 on 2024-09-23. Score: 2 | Accepted answer

Rewrite, by separating variables:
$$
\frac{1}{Q} d Q= -\frac{1}{P} dP
$$
Integrating both parts gives:
$$
\ln(Q) = - \ln(P) + C = \ln(1/P) + C, \qquad C \in \mathbb{R}
$$
Taking exponents on both sides gives that there is a $A = e^C > 0$ such that:
$$
Q = -\frac{A}{P}
$$

Double check for correctness. We have $\frac{dQ}{dP} = -\frac{A}{P^2}$, so:
$$
\frac{dQ}{dP} \frac{P}{Q} = -\frac{A}{P^2}\frac{P}{-\frac{A}{P}} = -1.
$$


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### Answer by username_3 on 2024-09-22. Score: 0

There is mistake already in the step 2. In this problem $Q(P)$ so you need to take integral wrt $P$ so the equations should look like;

$$\int (P dQ) dP=− \int (Q dP) dP$$

> **Comments:**
> **username_0**: But can't I write the equations as $\int PQ'(P)\ dP=\int Q(P)\ dP$? This makes the integration variable with respect to P on both sides.  

---

