7
$\begingroup$

I am asked to solve $\int_0^\frac{\pi}{4}\tan x\, \mathrm{d}x$.

This is what I did:

$$ \begin{align} &\int_0^\frac{\pi}{4}\tan x \,\mathrm{d}x&\\ = {} &\int_0^\frac{\pi}{4}\frac{\sin x}{\cos x}\, \mathrm{d}x&\\ = {} &\int_0^\frac{\pi}{4}\sin x\frac{-1}{-\cos x} \,\mathrm{d}x&\\ = {} &-\int_0^\frac{\pi}{4}\varphi'(x)\frac{1}{\varphi(x)} \,\mathrm{d}x &\text{where $\varphi(x):=-\cos x$} \\ = {} &-\int_{\varphi(0)}^{\varphi(\frac{\pi}{4})} \frac{1}{z} \,\mathrm{d}z \\ = {} &-\int_{-1}^{-\frac{1}{\sqrt{2}}} \frac{1}{z}\, \mathrm{d}z \\ = {} &\Big[\ln z\Big]_{-1}^{-\frac{1}{\sqrt{2}}} \\ = {} &\ln \left(-\frac{1}{\sqrt{2}}\right) - \ln (-1) \end{align} $$

And this is where I am stuck, because the solution is definitely not a complex number. I know the correct answer (it's $-\ln\left(\frac{1}{\sqrt{2}}\right)$, my problem is that I don't know where I made a mistake.

$\endgroup$
  • 3
    $\begingroup$ A primitive of $1/x$ is $\ln(|x|)$. $\endgroup$ – Michael Hoppe Jun 8 '18 at 20:36
  • 2
    $\begingroup$ you've forgotten negative sign at the pre-last line, and l$n|x|$ $\endgroup$ – haqnatural Jun 8 '18 at 20:37
16
$\begingroup$

It turns out that $\displaystyle\int\frac1z\,\mathrm dz=\log|z|$, not $\log z$.

$\endgroup$
8
$\begingroup$

You have lost a $-$ sign, so you should actually get $$\ln\left(-1\right)-\ln\left(-\frac{1}{\sqrt{2}}\right)$$ And now, you can use the fact that $$\ln(a)-\ln(b)=\ln\left(\frac{a}{b}\right)$$ to reach the desired result: $$\ln\left(-1\right)-\ln\left(-\frac{1}{\sqrt{2}}\right)=\ln\left(\frac{-1}{-\frac{1}{\sqrt{2}}}\right)=\ln\left(\frac{1}{\frac{1}{\sqrt{2}}}\right)=\frac{\ln 2}{2}$$ It works, because the logaritm of a negative number (the main branch) $-x$ (where $x>0$) is $\ln(-x)=\ln(x)+i \pi$, and in your case, the $i\pi$'s will cancel each other out. They would not cancel out if you were trying to compute, for example $\int_{-2}^{2} \frac{1}{x} \mathrm{d}x$. In this case, the complex result can tell you that you did something wrong.

But you can avoid the mess with the complex numbers by simply saying that $\int \frac{1}{x} \mathrm{d}x=\ln|x|$, as José Carlos Santos mentioned it already.

$\endgroup$
  • $\begingroup$ What do you mean by $\ln(-1)$? $\endgroup$ – Najib Idrissi Jun 8 '18 at 21:08
  • $\begingroup$ @NajibIdrissi What do you mean? I think you know the answer. It's either a complex number or just "a formal thing" which will lead you to the desired result. $\endgroup$ – Botond Jun 8 '18 at 21:15
  • $\begingroup$ The main branch of the logarithm is not defined on negative numbers... $\endgroup$ – Najib Idrissi Jun 8 '18 at 22:00
  • 3
    $\begingroup$ @NajibIdrissi The main branch of $\log$ is usually defined such that its imaginary part is in the interval $(-\pi,\pi]$ (see e.g. wikipedia). Therefore, $\log(-1)$ is perfectly well-defined and equal to $i\pi$. More generally, the main branch of $\log$ is well-defined on negative numbers and equal to $\log(-x)=\log x+i\pi$. $\endgroup$ – AccidentalFourierTransform Jun 9 '18 at 0:45
  • 2
    $\begingroup$ @NajibIdrissi the main branch isn't continuous there, but it's perfectly well defined. $\endgroup$ – Brevan Ellefsen Jun 9 '18 at 1:36
6
$\begingroup$

You have $$ \int_a^b\frac{1}{z}\,dz $$ with $a<0$ and $b<0$. The antiderivative to use is $\ln(-z)$, which is a function defined on $(-\infty,0)$.

You can avoid the problem by writing instead $$ \int_0^{\pi/4}(-\sin x)\frac{-1}{\cos x}\,dx= \int_0^{\pi/4}\varphi'(x)\frac{-1}{\varphi(x)}\,dx $$ where $\varphi(x)=\cos x$. For $x=0$ we have $\varphi(x)=1$; for $x=\pi/4$ we have $\varphi(x)=1/\sqrt{2}$, so the integral becomes $$ \int_1^{1/\sqrt{2}}-\frac{1}{z}\,dz= \int_{1/\sqrt{2}}^1 \frac{1}{z}\,dz= \Bigl[\ln z\Bigr]_{1/\sqrt{2}}^1=-\ln\frac{1}{\sqrt{2}}=\ln\sqrt{2}= \frac{1}{2}\ln2 $$

$\endgroup$

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.