I have integral of $\dfrac{1}{t^2 + 1}$ and integral of $\dfrac{t}{t^2 + 1}$ whose output is $\arctan(t)$ and $\dfrac12\ln(t^2 + 1)$ respectively.

Are there any similar unexpected results when we take integral? A link to such list would be enough for me. I'm asking this because while solving heat equations in differential equations I always face similar things. Thus I need to be prepared when I face something similar where the result of the integral is tangent, sine, cosine, permutation, cot, power function, etc.

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    $\begingroup$ What is "unexpected" about this? $\endgroup$ – user61527 Jun 14 '14 at 20:00
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    $\begingroup$ @SanathDevalapurkar, not elementary for me. $\endgroup$ – ilhan Jun 14 '14 at 20:06
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    $\begingroup$ @ilhan $$\int\dfrac{dt}{1+t^2}=C+\arctan{t}$$ $\endgroup$ – user122283 Jun 14 '14 at 20:09
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    $\begingroup$ @ilhan Don't feel bad, I don't see a connection between the integrals either. $\endgroup$ – Git Gud Jun 14 '14 at 20:11
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    $\begingroup$ Apparently, everything is easy for Sanath... nothing is unexpected. @Sanath you are lying to yourself and everyone if this is "elementary" as there is nothing elementary about these two objects except the simple computation but this has nothing to do with the computation and everything to do with intuition. $\endgroup$ – Squirtle Jun 15 '14 at 1:45

I wouldn't say that either of these results is unexpected, but I agree that inverse $\tan$ wouldn't spring to mind to somebody who only knew the power rule for integration.

Let's say we've got $\int \frac{1}{\color{red}{1+t^2}}\color{green}{dt}$.

Let $t=\tan(\theta) \iff \color{purple}{\theta=\arctan(t)}$.

Now, $\underbrace{\frac{dt}{d\theta}=\sec^2(\theta)}_{\textrm{try differentiating tan by the quotient rule!}} \iff dt=\color{green}{\sec^2(\theta)d\theta}$

Then the integral becomes $$\int \frac{1}{\color{red}{1+\tan^2(\theta)}}\color{green}{\sec^2(\theta) d\theta}=\require{cancel} \int\frac{1}{{\cancel{\color{red}{\sec^2(\theta)}}}}\cancel{\color{green}{\sec^2(\theta)}}\color{green}{d\theta}=\int 1d\theta=\color{purple}{\theta}+C=\color{purple}{\arctan(t)}+C.$$

For the second integral, use the substitution $u=t^2+1.$ I'll leave you to prove the second result!

By the way, in line with the whole 'unexpected integrals topic', for me, at least: $$\int \frac{1}{x}dx =\ln|x|+C$$ (unintuitive because $\frac{1}{x}$ doesn't obey the power rule like every other power does and $\frac{1}{x}$ doesn't seem to be related to $\ln$ in any way).

A couple more, following your pattern of unintuitiveness, are:

  • $\int \frac{1}{\sqrt{1-x^2}}dx=\arcsin(x)+C$
  • $\int \frac{1}{\sqrt{1+x^2}}dx=\mathrm{arsinh}(x)+C$
  • $\int \frac{1}{\sqrt{x^2-1}}dx=\mathrm{arcosh}(x)+C$
  • $\int \frac{1}{1-x^2}dx=\frac{1}{2}\ln \left| \frac{x+1}{x-1}\right|+C$ (for $x \neq \pm1$).
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    $\begingroup$ I guess 'unintuitive' is inherently subjective, but I'm not sure I'd call $\int \frac{1}{1-x^2} dx$ unintuitive, since you can get the result using partial fractions and some log rules. $\endgroup$ – Strants Jun 14 '14 at 22:37

There is a great deal of similarity between the two results:

$\begin{align} &\displaystyle \int \dfrac{t}{t^2 + 1} dt = \dfrac{\ln (t^2 + 1)}{2} + C = \dfrac{\ln(t-i)(t + i)}{2} + C &= \boxed{\dfrac{\ln(t - i) + \ln(t + i)}{2} + C}\\ &\displaystyle \int \dfrac{1}{t^2 + 1} dt = \arctan t + C = \dfrac{1}{2i}\ln\dfrac{t-i}{t+i} + C & = \boxed{\dfrac{\ln(t - i) - \ln(t + i)}{2i} + C} \end{align}$

Tell me those aren't similar!


Let me illustrate the case you give at the beginning of the post. Each function will be developed as an infinite series built at the origin. So, $$\dfrac{1}{t^2 + 1}=1-t^2+t^4-t^6+t^8-t^{10}+t^{12}+O\left(t^{13}\right)$$ from which we can derive (just multiply each term by $t$) $$\dfrac{t}{t^2 + 1}=t-t^3+t^5-t^7+t^9-t^{11}+t^{13}+O\left(t^{14}\right)$$ and the functions "look" similar. But now, let us integrate them and get $$\int \dfrac{dt}{t^2 + 1}=t-\frac{t^3}{3}+\frac{t^5}{5}-\frac{t^7}{7}+\frac{t^9}{9}-\frac{t^{11}}{11}+\frac{t^{ 13}}{13}+O\left(t^{14}\right)$$ $$\int \dfrac{t ~~dt}{t^2 + 1}=\frac{t^2}{2}-\frac{t^4}{4}+\frac{t^6}{6}-\frac{t^8}{8}+\frac{t^{10}}{10}-\frac{t^{12 }}{12}+\frac{t^{14}}{14}+O\left(t^{15}\right)$$ After integration, they really don't look any more at all to each other. So, there is no reason for the integrals to look similar.

However, you could find some similarities in the following integrals $$I_n=\int \dfrac{t^{2n}}{t^2 + 1}~~dt$$ $$J_n=\int \dfrac{t^{2n-1}}{t^2 + 1}~~dt$$ since $$I_0=\tan ^{-1}(t)$$ $$I_1=t-\tan ^{-1}(t)$$ $$I_2=\frac{t^3}{3}-t+\tan ^{-1}(t)$$ $$I_3=\frac{t^5}{5}-\frac{t^3}{3}+t-\tan ^{-1}(t)$$ and $$J_1=\frac{1}{2} \log \left(t^2+1\right)$$ $$J_2=\frac{t^2}{2}-\frac{1}{2} \log \left(t^2+1\right)$$ $$J_3=\frac{1}{4} \left(\left(t^2-1\right)^2+2 \log \left(t^2+1\right)\right)$$


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