Well?

dirichlets integral

[math] \int e^{-x^2} dx [/math]
is nonelementary as well, but
[math] \int_0^\infty e^{-x^2} dx [/math]
is known exactly.

What's the primitive of sin(x)/x ?
That's where I got stuck

are you from the other thread?

So what is it then, Mr. Smartass?

you fucktards you cant take the integral of sinx/x.
well you can but its theoretical

Fucking brainlets. ∫sin(t)/t dt = Si(t) in non-elementary functions.

so the exact answer is:
-10pi+20Si(30)

approximate:
-0.0808

Holy shit i guess i'm just as retarded at you guys.

I thought the lower boundary was burger+burger+bottle.

exact answer is:
-10pi

approximate:
-31.416

this is the only answer

other posts are brainlets

t. CERN scientist

^is correct if the beer mugs are treated as one variable

I think OP trolled us all into solving his homework.

nah i made the pic

>tfw planetoidbrain

I forget, fucking google it brainlet. it's that one trick where you square it and evaluate it as a double integral after transforming to polar coordinates.

>brainlet
>brainlet
>planetoidbrain
>brainlet

why is Veeky Forums so weird?

[math]\int_0^\infty \frac{\sin(u)}{u} du[/math]

Write [math]C_R: z = Re^{ia} \Rightarrow dz = iRe^{ia}da ,\;\; 0 \leq a \leq \pi[/math]
Then [math] C_r: z = \frac{1}{R} e^{ia} \Rightarrow dz = \frac{1}{R} i e^{ia}da ,\;\; -\pi \leq a \leq 0[/math]
Write [math] R_1: -R \leq x \leq -\frac{1}{R}[/math]
And [math] R_2: \frac{1}{R} \leq x \leq R[/math]

Consider the integral over the complex plane of the closed region [math]C = C_R \cup C_r \cup R_1 \cup R_2[/math]: [math]\oint_C \frac{e^{iz}}{z} dz[/math], which equals 0 by cauchy's theorem. Expanding the integral:

[math]\displaystyle 0 = \oint_C \frac{e^{iz}}{z} dz = \int_{C_R} \frac{e^{iRe^{ia}}}{Re^{ia}} iRe^{ia}da + \int_{C_r} \frac{e^{(1/R) ie^{ia}}}{(1/R) e^{ia}} i \frac{1}{R} e^{ia} da + \int_{R_1} \frac{e^{ix}}{x} dx + \int_{R_2} \frac{e^{ix}}{x} dx = i \int_{C_R} e^{iRe^{ia}} da + i \int_{C_r} e^{1/R i e^{ia}} da + \int_{R_1} \frac{e^{ix}}{x} dx + \int_{R_2} \frac{e^{ix}}{x} dx[/math]
[math]\displaystyle = i \int_0^\pi e^{iRe^{ia}} da + i \int_{-\pi}^0 e^{(1/R) i e^{ia}} da + \int_{-R}^{-1/R} \frac{e^{ix}}{x} dx + \int_{1/R}^R \frac{e^{ix}}{x} dx[/math]

Let's look at the first integral:
[math]i \int_0^\pi e^{iRe^{ia}} da = i \int_0^\pi e^{iR \cos(a)} e^{-R\sin(a)} da[/math], as [math]R \rightarrow \infty,\, e^{-R\sin(a)} \rightarrow 0[/math], so [math]i \int_0^\pi e^{iR \cos(a)} e^{-R\sin(a)} da \rightarrow 0[/math]

Now for the second:
[math]i \int_{-\pi}^0 e^{(1/R) i e^{ia}} da = -i \int_0^\pi e^{(1/R) i e^{ia}} da[/math], which becomes [math]- i \int_0^\pi 1 da = - i \pi[/math] as [math]R \rightarrow \infty[/math]

Taking the limit [math]R \rightarrow \infty[/math]
[math]0 = 0 -i\pi + \int_{-\infty}^0 \frac{e^{ix}}{x} dx + \int_0^\infty \frac{e^{ix}}{x} dx = \int_{-\infty}^\infty \frac{e^{ix}}{x} dx -i\pi \Rightarrow \int_{-\infty}^\infty \frac{e^{ix}}{x} dx = i\pi[/math]
[math]\Rightarrow \int_{-\infty}^\infty \frac{\cos(x)}{x}dx + i\int_{-\infty}^\infty \frac{\sin(x)}{x}dx = i\pi \Rightarrow \int_{-\infty}^\infty \frac{\sin(x)}{x}dx = \pi[/math]

By symmetry: [math]\int_0^\infty \frac{\sin(x)}{x}dx = \frac{\pi}{2}[/math]

We're all insecure about our intelligence

These both just look like autistic ramblings, I think you're full of shit

complex analysis is so beautiful
thanks friendo

It's just countour integration over this region.

en.wikipedia.org/wiki/Methods_of_contour_integration

nice psuedo rambling bullshit

using alot of symbols to explain intro level shit, hope you got a kick of of it

Come on guys, that’s too complicated. Given
[eqn]\int\ \frac{\sin x}{x}\ dx=\int\left (\int e^{-xy}\sin(x)\ dy\right )dx=\int \int e^{-xy}\sin(x)\ dx\ dy=\int \left (\frac{-ye^{-xy}\sin(x)\ -\ e^{-xy}\cos(x)}{1+y^{2}} \right )dy=\int \frac{1}{1+y^{2}}\ dy=\arctan y+C[/eqn]then [eqn]\int_{0}^{\infty}\frac{5\sin x}{x}\ dx=5\lim_{n\rightarrow \infty}\int_{0}^{n}\frac{\sin x}{x}\ dx=5\lim_{n\rightarrow \infty}\arctan y\ |_{0}^{n}=5\lim_{n\rightarrow \infty}\arctan n\ -\ \arctan 0=5\left [ \left ( \frac{\pi}{2} \right )\ -\ \left ( 0 \right ) \right ]=\frac{5\pi}{2} [/eqn] Now THAT, imho, is as simple as intro level shit gets. I got a picture for you too. I cleaned it up while typing.

wtf did you just spend your whole night working on an autism picture math problem?

hahh ahahahahah

Yes lolololol now do you think you could help me with this?

...

All these plebs falling for the elaborate homework thread

real, non memed, 100%, literal, in the flesh - Autism

AND STILL WRONG

It's almost like people browsing Veeky Forums would be interested in solving Science and Math problems. Go figure.

you are all clearly wrong
>absence of operator between two variables indicates addition
it's clearly
burger + 2(beer)^2=9
hence beer=sqrt(2)
brainlets

thank you for you informative post. it was well received and im a better person after reading it.

True.

However it doesn't actually change the final answer of the problem, since z^2 = 2 if you consider that and with the other assumption, 2z = 2, if you consider z = 1 from assuming it's 2z+2z.

dude no

you stupid

lrn2convergence fgt pls

The answer is nasty hangover and diarrhoea

>2017
>trying to heaven a thread
Ha!

Bottle * infinity / sgn(beer)

...

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>question is a parody of those Facebook math question images
>uses food and beverages as variables
An unnecessarily complex answer is the correct way to respond to this question. The only way to improve the answer or make it any more appropriate would be to apply even more difficult or excessive methods.

>beer=sqrt(2)
So beer is a completely artificial construct, devised by mathematicians who reject constructivism (another way of spelling "common sense") and whose only purpose is pacifying the masses who can't, won't and shan't learn rational trigonometry and the concept of quadrance? That seems about right.

>convergence
more like convirgin
that is you

Do we need to answer in fast food notation?

...

>trying to integrate sine of hotdog