The hard integral thread.

Yes, that's essentially the method that I used. (and PRSOM)

However, I'm beginning to get a little suspicious. You seem to have memorised a whole load of useful results; you can exchange the order of integrals in the blink of an eye; you handle Laplace and Fourier methods with ease ... so ... can the the rumour really be true? That the so-called KUMMER* project is not a myth? That in 1957, the Americans cryogenically preserved the brain of John von Neumann, and have recently managed to "light it up" again? Perhaps we'll never know, but when Kummer posts, what I'm hearing is:

"Hee-ee-e-re's Johnny!!"

(*Kold 'Ungarian Man Must Evaluate Residues - yeah, I know - acronym technology wasn't too impressive in the 1950s)

can you show your workings?

Yes, that's essentially the method that I used. (and PRSOM)

However, I'm beginning to get a little suspicious. You seem to have memorised a whole load of useful results; you can exchange the order of integrals in the blink of an eye; you handle Laplace and Fourier methods with ease ... so ... can the the rumour really be true? That the so-called KUMMER* project is not a myth? That in 1957, the Americans cryogenically preserved the brain of John von Neumann, and have recently managed to "light it up" again? Perhaps we'll never know, but when Kummer posts, what I'm hearing is:

"Hee-ee-e-re's Johnny!!"

(*Kold 'Ungarian Man Must Evaluate Residues - yeah, I know - acronym technology wasn't too impressive in the 1950s)

Evaluate $\displaystyle \int_0^\infty \frac{e^{-x^{n+1}}-e^{-(2^{n+1}x^{n+1})}}{x} \ dx$ for $n \in \mathbb{N}$

Any hints? I don't have a clue. Presumably some series expansion to start?

Not you! I've seen you on TPIT a few years ago and your skills were beastly enough back then. I shudder to think what they are now...

Evaluate $\displaystyle \int_0^{\frac{\pi}{2}} \frac { e^{-\frac{\tan x}{\sqrt{3}}}-e^{-\sqrt{3}\tan x} } {\sin 2x} \ dx$

It's easy to show that $\displaystyle \frac{1}{2}\int_a^b \sec^2{x} e^{-y \tan{x}} \, dy =\frac { e^{-a \tan x}-e^{-b\tan x} } {\sin 2x}$. Now,

So the answer to your integral is $\frac{1}{2} \log(3).$

...

Evaluate $\displaystyle \int_0^{\frac{\pi}{2}} \frac { e^{-\frac{\tan x}{\sqrt{3}}}-e^{-\sqrt{3}\tan x} } {\sin 2x} \ dx$

Hey! Hope all is well!

One of mine (undergrad level)

This time we'll use $\displaystyle \frac{1}{x^2} = \int_0^{\infty} te^{-xt}\;{dt}$ - we have:

$\begin{aligned} \displaystyle I & = \int_0^{\infty} \frac{e^{-x}}{x^2}\left(3x+e^{-3x}-1\right)\;{dt} = \int_0^{\infty} \int_0^{\infty} te^{-xt-x}\left(3x+e^{-3x}-1\right)\;{dt} \;{dx} \\& =\int_0^{\infty} \int_0^{\infty} te^{-xt-x}\left(3x+e^{-3x}-1\right)\;{dx} \;{dt} = \int_0^{\infty} \frac{9t}{(1+t)^2(4+t)} \\& = \log(256)-3.\end{aligned}$

I have a notificiation everytime this thread is used, since integration is nicez
But integration should be banned 12-6am.


Posted from TSR Mobile

This time we'll use $\displaystyle \frac{1}{x^2} = \int_0^{\infty} te^{-xt}\;{dt}$ - we have:

$\begin{aligned} \displaystyle I & = \int_0^{\infty} \frac{e^{-x}}{x^2}\left(3x+e^{-3x}-1\right)\;{dt} = \int_0^{\infty} \int_0^{\infty} te^{-xt-x}\left(3x+e^{-3x}-1\right)\;{dt} \;{dx} \\& =\int_0^{\infty} \int_0^{\infty} te^{-xt-x}\left(3x+e^{-3x}-1\right)\;{dx} \;{dt} = \int_0^{\infty} \frac{9t}{(1+t)^2(4+t)} \\& = \log(256)-3.\end{aligned}$

That first result, did you just know it? Or did you have to find it/figure it out before employing it. Unless it is a known result in the integral world?
Also your successive lines are a page of work to me lol. I hope you working is not how it is presented on here, it is too good and hard to be that short of a human mind.


Posted from TSR Mobile

I constructed the first one; it's not something that I recalled from memory. Its proof is simple. But obviously don't take it the way it's presented For example, I could have added more details when integrating with respect to x.