0
$\begingroup$

I was trying to evalute the integral $$\int \frac{1}{x^2+1} \,dx$$ by partial fractions.

$$\frac{1}{x^2+1} = \frac{1}{2i}\left(\frac{1}{x-i} - \frac{1}{x+i}\right)$$

Therefore, \begin{aligned} \int \frac{1}{x^2+1} \,dx &= \frac{1}{2i} \int \left(\frac{1}{x-i} - \frac{1}{x+i}\right) \,dx \\ &= \frac{1}{2i} (\ln(x-i) - \ln(x+i)) \\ &= \arctan\left(\frac{1}{x}\right) +C \end{aligned}

Because $$ x - i = \exp\left(\sqrt{x^2+1}+i\arctan{\frac{1}{x}}\right)$$ and $$ x + i = \exp\left(\sqrt{x^2+1}-i\arctan{\frac{1}{x}}\right).$$

This differs from what you get from trigonometric substitution, which is where I'm having difficulties in finding my error.

In the last step, I'm assuming that $\ln(z)$ works as you would expect for complex numbers.

$\endgroup$
4
  • 2
    $\begingroup$ You have swapped $+$ and $-$ in $\exp$. Thus the answer is $-\arctan \frac{1}{x} + C$ which is the same as $\arctan x + C$ $\endgroup$
    – uranix
    Jul 31 '20 at 16:45
  • $\begingroup$ Also it should be $x - i = \exp(\log \sqrt{x^2+1} - i \arctan(1/x))$ (for $x > 0$). $\endgroup$ Jul 31 '20 at 16:50
  • $\begingroup$ How is $\arctan(1/x) = - \arctan(x)$ $\endgroup$
    – Roskiller
    Jul 31 '20 at 16:53
  • 2
    $\begingroup$ No, $\arctan(x)+\arctan(1/x)=\operatorname{sgn}(x)\pi/2$ for all $x\neq 0$, and you absorb the $\pm\pi/2$ into $C$. $\endgroup$ Jul 31 '20 at 17:24
0
$\begingroup$

In the your fifth equation you will actually get $$I=-\tan^{-1}(1/x)+C$$ and due to the identity: $$\tan^{-1} x+\tan^{-1} (1/x)= \pi/2$$ We get $$I=\pi/2+\tan^{-1} x+ D$$ Uther wise the standard formula gives $$I=\tan^{-1} x+E$$ These two results differ only by a constant, which is normal in indefinite integration.

$\endgroup$

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.