4u integration qn (1 Viewer)

synthesisFR · Jul 11, 2023

lemme re try it one last time

Masaken · Jul 11, 2023

guys I think I got it
can someone check tho it looks sus

synthesisFR · Jul 11, 2023

Masaken said:
guys I think I got it
can someone check tho it looks sus
View attachment 38939

i think thats good

howcanibesmarter · Jul 11, 2023

Masaken said:
guys I think I got it
can someone check tho it looks sus

looks good to me

Trebla · Jul 11, 2023

Just a remark. Generally speaking, definite integrals where one of the points is undefined (x = 0 in this case) does not really appear in the HSC exam itself - if it ever does then they make it very clear you are taking the limiting value rather than evaluating it at that undefined end point.

tywebb · Jul 12, 2023

Here is a quicker way:

$\textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)\ \!dx=\frac{i}{2}[\text{Li}_2(e^{2ix})]_0^\pi=0\text{ where Li}_2\text{ is the dilogarithm function}$

$\text{However}$

$\begin{aligned}\textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)dx&=\textstyle\int_0^\pi(\ln2+\ln(-i)+ix+\ln(\sin x))\ \!dx\\&=\textstyle\pi\ln2+\pi\cdot\frac{-\pi i}{2}+i\frac{\pi^2}{2}+\int_0^\pi\ln(\sin x)\ \!dx\\\textstyle\therefore \int_0^\pi\ln(\sin x)\ \!dx&=-\pi\ln2\end{aligned}$

$\textstyle\text{and by symmetry}\int_0^{\frac{\pi}{2}}\ln(\sin x)\ \!dx=-\frac{\pi}{2}\ln2$

synthesisFR · Jul 12, 2023

tywebb said:
Here is a quicker way:

$\textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)\ \!dx=\frac{i}{2}[\text{Li}_2(e^{2ix})]_0^\pi=0\text{ where Li}_2\text{ is the dilogarithm function}$

$\text{However}$

$\begin{aligned}\textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)dx&=\textstyle\int_0^\pi(\ln2+\ln(-i)+ix+\ln(\sin x))\ \!dx\\&=\textstyle\pi\ln2+\pi\cdot\frac{-\pi i}{2}+i\frac{\pi^2}{2}+\int_0^\pi\ln(\sin x)\ \!dx\\\textstyle\therefore \int_0^\pi\ln(\sin x)\ \!dx&=-\pi\ln2\end{aligned}$

$\textstyle\text{and by symmetry}\int_0^{\frac{\pi}{2}}\ln(\sin x)\ \!dx=-\frac{\pi}{2}\ln2$

We are in high school

Average Boreduser · Jul 12, 2023

synthesisFR said:
We are in high school

skill issue bud.

tywebb · Aug 3, 2023

tywebb said:
Here is a quicker way:

$\textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)\ \!dx=\frac{i}{2}[\text{Li}_2(e^{2ix})]_0^\pi=0\text{ where Li}_2\text{ is the dilogarithm function}$

$\text{However}$

$\begin{aligned}\textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)dx&=\textstyle\int_0^\pi(\ln2+\ln(-i)+ix+\ln(\sin x))\ \!dx\\&=\textstyle\pi\ln2+\pi\cdot\frac{-\pi i}{2}+i\frac{\pi^2}{2}+\int_0^\pi\ln(\sin x)\ \!dx\\\textstyle\therefore \int_0^\pi\ln(\sin x)\ \!dx&=-\pi\ln2\end{aligned}$

$\textstyle\text{and by symmetry}\int_0^{\frac{\pi}{2}}\ln(\sin x)\ \!dx=-\frac{\pi}{2}\ln2$

So there's this dude from Harvard University, Lars Ahlfors

He is like the coolest dude ever for complex numbers

- like the GOAT of complex numbers

who wrote the book Complex Analysis - probably the one which has been used more than any other for that subject.

My very short solution actually came from his book.

My version is somewhat less verbose than his, but essentially the same.

4u integration qn (1 Viewer)

synthesisFR

しゃけ

Masaken

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synthesisFR

しゃけ

howcanibesmarter

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Trebla

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tywebb

dangerman

synthesisFR

しゃけ

Average Boreduser

𝙱𝚊𝚗𝚗𝚎𝚍

tywebb

dangerman

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