3
$\begingroup$

$\int\frac{15}{x}dx$ would be $$15\int\frac{1}{x}dx = 5\ln|x|+c$$

This seems like a silly question but I'm feeling exceptionally dense today. Why would you apply the logarithm rule, why wouldn't raising $x$ to a $-1$ exponent work?

$\endgroup$
  • 1
    $\begingroup$ Your solution is correct. $\int x^tdx=\frac{1}{t+1}x^{t+1} +C$ is valid for $t\neq -1$, so doesn't apply here. $\endgroup$ – vadim123 Jun 2 '13 at 4:06
  • $\begingroup$ @vadim123 Ok, i see, thanks! $\endgroup$ – Ellen Jun 2 '13 at 4:10
  • 1
    $\begingroup$ As to "why," if you increment the power of $x^{-1}$ you should find something isn't right since you'll be left with $x^0=1$, whose derivative is not $x^{-1}$ (and which will actually give you 0 area for any bounds). $\endgroup$ – Zen Jun 2 '13 at 4:21
  • 1
    $\begingroup$ Also if you try to take the antiderivative of $x^{-1}$ using the power rule, you would obtain $\dfrac{x^0}{0}$. This is undefined, since the expression divides by zero. $\endgroup$ – Adriano Jun 2 '13 at 4:44
5
$\begingroup$

Let $a$ be a fixed positive number. We know by integration that $$\int_1^a \frac{dx}{x}=\ln a-\ln 1=\ln a.\tag{1}$$

We try to reconcile this with the ordinary formula for the integral of a power. Note that if $w\ne 1$, then $$\int_1^a x^{-w}\,dx= \frac{1}{1-w}\left(a^{1-w}-1^{1-w}\right).\tag{2}$$ There is an obvious problem in using this when $w=1$, because of the division by $0$. So instead let us take the limit as $w$ approaches $1$ of the expression at the right side of (2). So, letting $h=1-w$, we want to find $$\lim_{h\to 0}\frac{a^h -1}{h}.\tag{3}$$ We recognize the limit (3) as the derivative of the function $a^t$ with respect to $t$, at $t=0$. Simce $a^t=e^{(\ln a)t}$, the derivative at $t=0$ is $\ln a$.

This is the same as the answer (1) that we obtained by using the standard formula for $\int \frac{dx}{x}$. So this formula can be thought of as a limiting case of the 'usual" formula (2).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.