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I tried integrating this integral below by substituting $x = \sqrt{i}\,t$,

\begin{align} \int \frac{\mathrm d x}{1+x^4} & =\sqrt{i} \int \frac{\mathrm d t}{1-t^4}\\ & =\sqrt{i}\left[\int \frac{\mathrm d t}{1-t^2}+\int \frac{\mathrm d t}{1+t^2}\right] \\ & =\sqrt{i}\left[\frac{1}{2} \ln \frac{|1+t|}{|1-t|}+\tan ^{-1} t\right]+c \\ & =\sqrt{i}\left[\frac{1}{2} \ln \frac{\left|\sqrt{i}+x\right|}{\left|\sqrt{i}-x\right|}+\tan ^{-1} \frac{x}{\sqrt{i}}\right]+c\end{align}

Now, I’m stuck here. How do I convert it back to real-valued functions? I tried searching for identities similar to $\tan ^{-1} x=\dfrac{i}{2} \ln\dfrac{|1-ix|}{|1+ix|}$(this helped me to integrate $\displaystyle\int \frac{\mathrm d x}{1+x^2}$ in a similar way to this one).

On a broader note, how do I convert such complex functions that arise in real integrals back to real functions?

RobPratt
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Kirusanth
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    I don't think you're allowed to make a complex substitution in an indefinite integral, because from what I understand is that a complex substitution is essentially the same as integrating along the right contour, but for that there need to be bounds of integration. So to fix your solution you must write the indefinite integral as the definite integral $$\int_0^{x} \frac{1}{1+t^4} dt = F(x)+1 = \int \frac{1}{1+x^4} dx + 1$$ from here you can make the complex substitution and ensure you get a real answer. (Maybe this is just me missing out some standard simplification, my CA is really bad) – Kraken Mar 15 '25 at 14:05
  • @Kraken, can you please elaborate your answer in a comment, in more easy-to-understand terms? I'm sorry. But I know only a little about complex analysis. I read about contour integration only last week ... – Kirusanth Mar 15 '25 at 15:31
  • Do you want help with this integral in general or do you want to solve this integral by making a complex substitution? – Random Math Enthusiast Mar 15 '25 at 16:54
  • @RandomMathEnthusiast By a complex substituion, indeed. The general way is already there on MSE, https://math.stackexchange.com/questions/333611/evaluating-int-frac1x41-dx – Kirusanth Mar 16 '25 at 04:51

1 Answers1

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You have forgotten a factor of $2:$

$$\begin{align}\int \frac{\mathrm d x}{1+x^4} & =\sqrt i\int \frac{\mathrm d t}{1-t^4}\\ & =\frac{\sqrt{i}}{\color{green}2}\left[\int \frac{\mathrm d t}{1-t^2}+\int \frac{\mathrm d t}{1+t^2}\right]\\&=\frac{\sqrt i}2\left[\frac12\ln\left(\frac{\sqrt i+x}{\sqrt i-x}\right)+\tan^{-1}\frac{x}{\sqrt i}\right]+C\end{align}$$

To convert these complex-valued functions back to real-valued functions, the following identities will come in handy

$$\ln(z)=\ln|z|+i(2n\pi+\operatorname{arg}z)$$ $$\implies\ln(x+yi)=\frac12\ln\left(x^2+y^2\right)+i\tan^{-1}\frac{y}x+2n\pi i$$ $$\tan^{-1}=\frac{i}2\ln\left(\frac{1-ix}{1+ix}\right)$$ $$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right)\forall\,xy<1$$ $$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)\forall\,xy>-1$$

Now, there seems to be an apparent ambiguity in the choice of $\sqrt i$, as it can be both $\pm\frac{i+1}{\sqrt 2}$, i.e., $e^\frac{\pi\iota}4$ and $e^\frac{5\pi\imath}4$. But notice that both these values of $\sqrt i$ yield the same anti-derivative as the function $\displaystyle f(t)=\frac{t}4\ln\left(\frac{t-x}{t+x}\right)+\frac{t}2\tan^{-1}\frac{x}t$ is even. So, let $\sqrt i=\frac{i+1}{\sqrt2}$, then the given integral $\mathcal I$ becomes

$$\begin{align}\mathcal I&=\frac{\sqrt i}4\ln\left(\frac{\sqrt i+x}{\sqrt i-x}\right)+\frac{i\sqrt i}2\ln\left(\frac{1-\sqrt ix}{1+\sqrt ix}\right)+C\\&=\frac{i+1}{4\sqrt2}\ln\left(\frac{x+\frac1{\sqrt2}+\frac{i}{\sqrt2}}{-x+\frac1{\sqrt2}+\frac{i}{\sqrt2}}\right)+\frac{i-1}{4\sqrt2}\ln\left(\frac{1-\frac{x}{\sqrt2}-\frac{ix}{\sqrt2}}{1+\frac{x}{\sqrt2}+\frac{ix}{\sqrt2}}\right)+C\\&=\frac{i+1}{4\sqrt2}\left[\frac12\ln\left|x^2+\sqrt2x+1\right|+i\tan^{-1}\frac1{\sqrt2x+1}-\frac12\ln\left|x^2-\sqrt2x+1\right|+i\tan^{-1}\frac1{\sqrt2x-1}\right]+\frac{i-1}{4\sqrt2}\left[\frac12\ln\left|x^2-\sqrt2x+1\right|+i\tan^{-1}\frac{x}{x-\sqrt2}-\frac12\ln\left|x^2+\sqrt2x+1\right|-i\tan^{-1}\frac{x}{x+\sqrt2}\right]+C\\&\overset{*}=\frac1{4\sqrt2}\ln\left|\frac{x^2+\sqrt2x+1}{x^2-\sqrt2x+1}\right|-\frac1{2\sqrt2}\tan^{-1}\frac{\sqrt2x}{x^2-1}+C\end{align}$$

$*:$ $\tan^{-1}\frac{x}{x-\sqrt2}-\tan^{-1}\frac{x}{x+\sqrt2}=\tan^{-1}\frac1{\sqrt2x+1}+\tan^{-1}\frac{x}{\sqrt x-1}=\tan^{-1}\frac{\sqrt2x}{x^2-1}$

Poonguzhali Annadurai
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  • It would be nice if you edit and make your answer quite more elaborate, please... Especially those steps that follow that styled uppercase I. – Kirusanth Mar 16 '25 at 04:58
  • @Kirusanth I have added more details in my answer. – Poonguzhali Annadurai Mar 16 '25 at 11:18
  • Just one more query, have you ignored the complex parts at at the end? If so, why do we actually get complex parts in a real integral, why don't they become identically zero by themselves at the end? BTW, really thanks for your patience to edit your answer for my request... – Kirusanth Mar 16 '25 at 13:42
  • I think there is a sign error for one of the $itan^{-1}()$ in the step before the last one. I can't edit it as it asks for a minimum of 6 characters to be changed. – Kirusanth Mar 16 '25 at 13:54
  • @Kirusanth No, I haven’t ignored them; on simplifying them, the imaginary parts cancel out; just that I’ve not included those simplifications. And, are you talking about the term $\tan^{-1}\frac1{\sqrt2x-1}$? If yes, I changed its sign from $-$ to $+$ since it was $-\tan^{-1}\frac1{1-\sqrt2x}$ initially. – Poonguzhali Annadurai Mar 16 '25 at 14:15
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    Oh okkk. Really, really thanks for your help! Amazing answer... – Kirusanth Mar 16 '25 at 14:29