#integrals

N-gated Hacker Newsngate
2025-11-29

Oh, the sheer thrill of evaluating integrals! 🙄 Let's all pretend single-handedly invented while ignoring the guy named who did it decades earlier. 🤦‍♂️ Clearly, contour integration is for those who didn't get the memo on Feynman's party trick. 🎉
zackyzz.github.io/feynman.html

2025-11-05

have you ever needed a symbol for three snakes sharing the same hula hoop?

well, unicode has got you covered!

U+2230 : ∰

#math
#calculus
#integrals

Nom__XD, 飲ん,诺穆 (They/Them)🐈‍⬛nomxd
2025-08-25

I could probably find this table somewhere, but it was very enjoyable doing it myself, as a friend once told me. drive.google.com/file/d/1Qv7AZ 🐾

Nom__XD, 飲ん,诺穆 (They/Them)🐈‍⬛nomxd
2025-08-20

Did I told you I am obsessing about integrals? Well, I am, and I'm posting some of them here shorturl.at/TfwGx Within my random thoughts folder for the lack of a better name. I've gained a few points on math skills lately.shorturl.at/TfwGx 🐾

Pustam | पुस्तम | পুস্তম🇳🇵pustam_egr@mathstodon.xyz
2025-05-14

Integrals of inverse functions!

Proof without words (see image; credit: Jonathan Steinbuch, CC BY-SA 3.0, via Wikimedia Commons)...

For any montonic and invertible function \(f(x)\) in the interval \([a,b]\):
\[\displaystyle\int_a^bf(x)~ \mathrm dx+\int_{f(a)=c}^{f(b)=d}f^{-1}(x)~\mathrm dx=b\cdot f(b)-a\cdot f(a)=bd-ac\]

If \(F\) is an antiderivative of \(f\), then the antiderivatives of \(f^{-1}\) are:
\[\boxed{\displaystyle\int f^{-1}(y)~\mathrm dy=yf^{-1}(y)-F\circ f^{-1}(y)+C}\]
where \(C\) is an arbitrary constant (of integration), and \(\circ\) is the composition operator (function composition).

For example:
\[\begin{align*}\displaystyle\int \sin^{-1}(y) \, \mathrm dy &= y\sin^{-1}(y) - (-\cos(\sin^{-1}(y)))+C\\ &=y\sin^{-1}(y)+\sqrt{1-y^2}+C\end{align*}\]

\[\displaystyle\int \ln(y) \, dy = y\ln(y)-\exp(\ln(y)) + C= y\ln(y)-y + C.\]

#Function #InverseFunction #InverseFunctions #Functions #Integral #Integrals #Antiderivative #Integration #Calculus #FunctionComposition #CompositeFunction)

Emmanuel José Garcíavanclides88@mathstodon.xyz
2025-03-29

The USM, the Dominican method 🇩🇴, has relegated Euler substitutions to mere historical relics. Modern integration has a new name.

Method draft: drive.google.com/file/d/12DayP
#math #calculus #integrals #method #euler #halfangleapproach #symmetrymatters

2025-03-19

It's really fucking aggravating when you try to compute an integral, get to a solution, and then when you look at the instructor's solution, you find him using a formula that WAS NEVER INTRODUCED BEFORE JUST NOW!

🤬

#math #calculus #integrals #fuck

Emmanuel José Garcíavanclides88@mathstodon.xyz
2025-03-01

The USM 𝐢𝐧𝐜𝐨𝐫𝐩𝐨𝐫𝐚𝐭𝐞𝐬, 𝐞𝐱𝐭𝐞𝐧𝐝𝐬, 𝐣𝐮𝐬𝐭𝐢𝐟𝐢𝐞𝐬, and 𝐬𝐮𝐫𝐩𝐚𝐬𝐬𝐞𝐬 Euler's substitutions. In Euler's substitutions, the choice of signs based on the domain must be made manually, whereas in the USM, the supporting theorems prescribe which sign to use according to the domain. Moreover, the USM shows that Weierstrass substitutions and the use of complex exponentials for integration are merely two sides of the same coin. The USM not only 𝐮𝐧𝐢𝐟𝐢𝐞𝐬 these two techniques into one, but also 𝐠𝐞𝐧𝐞𝐫𝐚𝐥𝐢𝐳𝐞𝐬 them.

USM: geometriadominicana.blogspot.c

#math #calculus #integrals #technique #new #halfangleapproach #symmetry

2024-11-06

Y el último video de esta tanda de 5 videos sobre integración. Aquí resuelvo un ejemplo con substitución trigonométrica, ¡harta diversión! youtu.be/7TGZ4gzeTQM #calculo #matemáticas #maths #calculus #integrales #integrals

Pustam | पुस्तम | পুস্তম🇳🇵pustam_egr@mathstodon.xyz
2024-07-20

An excellent general result.

If \(\Re(s)>1\),
\[\displaystyle\int_0^\infty\frac{\ln x}{x^s+1}~\mathrm dx=\frac{\pi^2}{4s^2}\left[\sec^2\left(\frac{\pi}{2s}\right)-\csc^2\left(\frac{\pi}{2s}\right)\right]\]

Special cases:
\[\displaystyle\int_0^\infty\frac{\ln x}{x^2+1}~\mathrm dx=0\]
\[\displaystyle\int_0^\infty\frac{\ln x}{x^3+1}~\mathrm dx=-\frac{2\pi^2}{27}\]
\[\displaystyle\int_0^\infty\frac{\ln x}{x^4+1}~\mathrm dx=-\frac{\pi^2}{8\sqrt2}\]
\[\displaystyle\int_0^\infty\frac{\ln x}{x^5+1}~\mathrm dx=-\frac{4\pi^2}{25}\left(\frac{2+\sqrt5}{5+\sqrt5}\right)=-\frac{(5+3\sqrt5)\pi^2}{125}\]

#Integral #Integrals #GeneralResult #GeneralResults #Result #Results #Logarithms #Logarithm #Integration #DefiniteIntegral #Calculus #IntegralCalculus

Pustam | पुस्तम | পুস্তম🇳🇵pustam_egr@mathstodon.xyz
2024-05-13

Take a look at this interesting integral. #MathsChallenge [Hint: use the properties of the Jacobi theta function of the third type \(\vartheta_3(z,q)\)]

\[\boxed{\displaystyle\int_0^{\frac{\pi}{4}}\dfrac{1+2\displaystyle\sum_{n\geq1}e^{-n^2\pi x}}{1+2\displaystyle\sum_{n\geq1}e^{-n^2\pi/x}}\ \mathrm{d}x=\sqrt\pi}\]

#MathChallenge #IntegralChallenge #InterestingIntegral #WeirdIntegral #Integral #Integrals #DefiniteIntegral

Pustam | पुस्तम | পুস্তম🇳🇵pustam_egr@mathstodon.xyz
2024-05-12

Integral challenge!

\[\displaystyle\int_0^\infty\ln\left(1+\dfrac{\cosh\alpha}{\cosh x}\right)\ dx=\dfrac{\pi^2}{8}-\dfrac{\arccos^2(\cosh\alpha)}{2}\]

#Integral #Integrals #IntegralChallenge #HyperbolicFunction #HyperbolicCosine #Logarithm #DefiniteIntegral

Pustam | पुस्तम | পুস্তম🇳🇵pustam_egr@mathstodon.xyz
2024-05-12

Interesting integral! #Challenge
\[\displaystyle\int_0^1\dfrac{\ln (x)\ln(1-x)}{x(1-x)}\operatorname{Li}_2(x)\ dx=5\zeta(2)\zeta(3)-8\zeta(5)\]
Where \(\operatorname{Li}_2(x)\) denotes the dilogarithm (or Spence's function), and \(\zeta(x)\) denotes the Riemann zeta function.

#ZetaFunction #Zeta #Dilogarithm #SpenceFunction #Polylogarithm #Integral #DefiniteIntegral #Integration #Integrals #RiemannZetaFunction #Logarithm #Function #LogarithmicFunction

Coach Pāṇini ®paninid@mastodon.world
2024-03-21

@ThinkingSapien @whybird

I also studied the same in the course of electrical engineering education.

In terms of #calculus, I think that derivatives *are* intuitive.

But #integrals?

Integrals are #CounterintuitiveMagic.

Markus Redekermrdk@mathstodon.xyz
2024-03-16

The integral

\[ \int_0^\infty x^k e^{-x} dx = k! \]

is used as motivation for the gamma function and in the irrationality proof of \( e \). But it can also be used for the transformation \( T \) defined by

\[ T f = \int_0^\infty f(x) e^{-x} dx. \]

If \( f = \sum_i f_i x^i \) is a power series and everything converges, it is transformed to \( T f = \sum_i f_i i! \). One can therefore say that \( T \) evaluates \( f \) at the factorial and write

\[ T f = f(!). \]

Are there other unusual places at which one can evaluate a function?

#Mathematics #PowerSeries #Integrals

2024-02-27

youtube.com/watch?v=MwVBzE7Z5g
Huh maybe if I saw this back in 1st year Uni I wouldn't've failed Calc 1 and would've stayed in the CS program? Could've been a dev. But nah, swapped over to IS&T degree (already met its math req). Kinda was more what I wanted to do anyway though. I still sometimes do wonder.
#Integrals #Calculus #Integration

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