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I just managed to catch up to an anime called Boku No hero, and to my surprise they showed an integral in ep 80-81 I think. Where the main character tried to evaluate it and got $\frac{107}{12}$ but the answer was $\frac{107}{28}$. I'm just curious how would anyone evaluate this? So the integral was

$$I =\int^{\log(1+\sqrt{2})}_0\left(\frac{e^x-e^{-x}}{2}\right)^3\left(\frac{e^x-e^{-x}}{2}\right)^{11}dx$$enter image description here

This would be the same as integrating $$I=\int^{\log(1+\sqrt{2})}_0\sinh^{14}xdx$$ But I'm not sure how to go on from here. Is there any hidden trick that I fail to see?

user8700908
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    There are probably a few typos ; $\int_0^{\ln(1+\sqrt{2})} \sinh(x)^3 \cosh(x)^{11} \ dx = \frac{107}{28}$, courtesy of Wolfram Alpha, but that's the closest I can get. – D. Thomine Mar 09 '20 at 15:06
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    Fun fact: When one of my friends reached this point of the anime, he stopped and computed the integral before continuing watching. – Qi Zhu Mar 09 '20 at 15:06
  • Or it could be $\int_0^{\ln(1+\sqrt{2})} \cosh(x)^{3} \sinh(x)^{11}; dx = \frac{13}{84}$. – Robert Israel Mar 09 '20 at 15:08
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    The integral cannot be the two mentioned numbers. The integrand $\sinh(x)^{14} \in [0,1]$ for $x \in [0,\log(1+\sqrt{2})]$, this means the integral is bounded from above by $\log(1+\sqrt{2}) \sim 0.8814 < 1$ – achille hui Mar 09 '20 at 15:15
  • Might be a typo in the anime as @D.Thomine mentioned. – user8700908 Mar 09 '20 at 15:16
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    As far as interesting math in anime is concerned, there is a cool one appearing near the end of Assassination Classroom as well (season 2 episode 12, as far as googling could tell me). That's more geometric, though, asking about the volume of the region inside a unit cube that is closer to the center than to any of the vertices. The presented solution is elegant as well. – Arthur Mar 09 '20 at 15:18
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    According to this video, there is a typo in the animation. In the manga, the 2nd factor has a $+$ sign instead of minus sign. So the correct integral is $\int_0^{\log(1+\sqrt{2})} \sinh^3 x \cosh^{11}x dx$ – achille hui Mar 09 '20 at 15:24
  • @Arthur Yeah, i've seen that one. The solution was elegant indeed. I saw some intimidating integrals on Magic Kaito ep 1 as well. – user8700908 Mar 09 '20 at 15:25
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    That integral has terms $$\dfrac{e^x-\hat{e}^{!x}}2$$ (a hat / circumflex over the $e$ like $\displaystyle\hat e^{!x}$). What is $\hat e$? It's clearly not $e^{-x}$. – emacs drives me nuts Mar 09 '20 at 16:06
  • Probably another typo. Clearly $e^{\jhat x}$ makes no sense – user8700908 Mar 09 '20 at 20:24
  • Does this anime got anymore math in it? Is it worth watching regardless of the math? – Zacky Mar 12 '20 at 14:57
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    @Zacky, unfortunatley this is the only scene with math in it. I haven’t read the manga so I can’t say for sure. But YES, this anime is the a superb anime (IMO) It has a good character development, stunning art, good plot and it has two awesome movies out already. If you’re into anime you should really watch this. – user8700908 Mar 12 '20 at 22:41

1 Answers1

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It seems there are many typos in this adaptation of manga into anime. In Chapter 167 of manga (in both english and chinese translation), the main character and his classmates have been asked to evaluate the integral below:

Klak! an integral
(image extracted from reahheroacademia.com )

As one can see, in the second factor of the integrand, there is a '+' instead of '-' between the two exponentials. The correct integral to evaluate is

$$\mathcal{I} = \int_0^{\log(1+\sqrt{2})} \sinh^3 x\cosh^{11} x dx$$

If one zoom in above picture, one will notice the color of the blackboard is not a solid gray. It is a pattern obtained by scattering white dots over black background. It is hard to recognize the '-' in those $e^{-x}$ appear in the integrand. This may explain why in the anime, it is mistakenly to be $\hat{e}^x$.

Finally in the manga, unlike the numbers $\frac{107}{12}$ and $\frac{107}{24}$ quoted in question, the main character has evaluated the integral to $\frac{107}{14}$ and the correct answer in manga $\frac{107}{28}$ is the correct one in real world.

Let's back to the math and change variable to $u = \sinh(x)^2$.

When $x$ changes from $0$ to $\log(1+\sqrt{2})$, $u$ increases from $0$ monotonically to $1$. Since $\cosh^2x = \sinh^2x + 1$ and $\frac{du}{dx} = 2\sinh x\cosh x$, we can evaluate the integral as:

$$\begin{align} \mathcal{I} &= \frac12\int_0^{\log(1+\sqrt{2})} (\sinh^2 x)(\cosh^2 x)^5 (2\sinh x\cosh x)dx\\ &= \frac12\int_0^1 u (u+1)^5 du\\ &= \frac12\int_0^1 (u+1)^6 - (u+1)^5 du\\ &= \frac12\left[\frac{(u+1)^7}{7}-\frac{(u+1)^6}{6}\right]_0^1\\ &= \frac12\left[\frac{128-1}{7} - \frac{64-1}{6}\right]\\ &= \frac{107}{28}\end{align}$$

achille hui
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