troubles with finding where a inegral function is defined

[humble wasp]

Hello let F(x)= int from -1 to x of (-e^t)/(t-1)dt, how can i know the domain of the integral function?

I know the domain of the integrated function is R-{1}; from this how can i know the domain of F(x)?

[hushed gulch]

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[tall ruin]

Hi! Still need help with this?

[humble wasp]

yup thanks!

Also how can i briefly graph a integral function if its bounded like from -x to x?

whats the difference compared to when its between like "a and x"

[shell wing]

are you going to help

[tall ruin]

I finally got a free time. Would you also wanna discuss this @shell wing ?

[shell wing]

Sure where do you wanna start

[tall ruin]

Good morning. With the first question how can she know the domain of F(x), the key idea for domain of F is that since it’s the set of all x (integral from a to x), the path from -1 to x must not be in a point where the integrand becomes undefined. Here, the integrand is undefined at t=1

@shell wing @humble wasp

It’s defined on R{1}, right?

[humble wasp]

no, (-inifnity, 1)

thats why i was confused

[tall ruin]

The first part diverges so F(X) can’t be for any x>1. The function stops at 1. Can you pls do this step by step so I’d know where’s the confusing part is?

[shell wing]

@humble wasp

[humble wasp]

yea i dont get why x>1 doesnt work !

[shell wing]

Because the function gets unbounded in between

[humble wasp]

wdym

[shell wing]

like the area shoots to infinty

[tall ruin]

Ok think of it this way

[shell wing]

yes grace you tell her

[tall ruin]

You start at -1 and you move toward x. At t=1 there’s a vertical wall

[shell wing]

this wall

[tall ruin]

So if x>1, you must cross that wall to reach x

But you can’t cross it in a finite way so the “area” explodes there

Think of the vertical wall as the infinite spike

[shell wing]

@humble wasp keep up with her

[humble wasp]

yup yes

cause its not in the domain of f(t) right?

[humble wasp]

https://cdn.discordapp.com/attachments/1000234066324492368/1262981160150110219/cokebert.gif

[shell wing]

the limits of the integral

(-1,x)

[humble wasp]

ah yes

weell i think ill ask this exercise to my teacher just for clarity too

[tall ruin]

Yes but not just because it’s not in the domain. The discontinuity at t=1 is non integrable

In this case, you cross a point not in the domain and also the integral near that point diverges so x>1 is not allowed

Remember the rule that if the integrand has a discontinuity, if it’s removable maybe still ok to cross. But if the vertical asymptote like 1 / (t-c) it’s not ok to cross

Your case is non-integrable blow up. Your integrand -e^t / t-1 if it’s near t=1 it behaves like C / t-1 and the integral of 1 /(t-1) diverges logarithmically to infinite area so it’s not defined at 1 and area near 1 is infinite that’s why you can’t cross it.

[humble wasp]

ah alr i see

yea my teacher kinda explained it like that too

[tall ruin]

Lmk if it still confuses you because it can be

[shell wing]

Can some x after 1 not be in domain because the area might get canceled as there is a negative area there?

[humble wasp]

alr alr

[tall ruin]

Hi @shell wing

[shell wing]

hello

did i write it clearly?

[tall ruin]

Are you asking if the infinite positive and negative areas could cancel?

[shell wing]

yes

[tall ruin]

No. Cancellation can’t rescue this integral

The improper integrals don’t allow this cancellation because each side must converge separately.

[shell wing]

Is that a well known fact or is there a proof for this

[tall ruin]

The fact that for cancellation, u would need something like area → +∞ from one side, area → −∞ from the other side, and together they balance. But in this case for Riemann (improper integrals), each side must converge separately as what I stated earlier.

Try to split the integral at 1. The first one already diverges so we never even get to combine them.

[hushed gulch]

@humble wasp

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[humble wasp]

+close

[hushed gulch]

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[hushed gulch]

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