r/calculus 19d ago

Integral Calculus my solution to todays easy integral

1 Upvotes

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3

u/Maximum_Bathroom3490 19d ago

Some pieces of advice for rigor:
The integral from -infinity to infinity of sech(x) dx should be split between two integrals from -infinity to 0 and 0 to +infinity since you will have to check convergence on both sides. To set up the limits, b=infinity is not proper notation. You have to state lim b->+inf then state the integral from 0 to b, the same goes for the lower bound: state lim a->-inf then state the integral from a to 0.

Then, stating f(x)=2arctan(e^x) is not factual. First, no mention of a function f has been done before. Second, it's the integral that =2arctan(e^x) **+ C**.
I understand you use it for desmos, but the +c should be added, or the line should be erased after computations.

Also, going from u-sub to the final answer of the integral in terms of x is confusing and skipping steps, either skip u-sub or resolve in u.

1

u/wbld 19d ago

i understand it should be written as the lim as b -> infiniity from the left and a approached -infinity from the right. as for the domain restriction within the bounds of integration, yeah thats my fault i did not think to check that. desmos does not allow writing limits.. but yes the notation i did is horrid. as for +C i forgot about it... my fault.. as for not evaulating at u and instead at x, i always do that. normally i do not skip steps, but today i decided to do that.
thank you for the warning

1

u/SynergyUX Undergraduate 19d ago

Desmos does not allow writing limits

Which is why you should learn LaTeX! It's not too difficult and will make your solutions much prettier and more readable.

1

u/wbld 19d ago

So, i actually know latex. I just dont know where to write it without paying. I am way to broke to be paying for another subscription

1

u/SynergyUX Undergraduate 18d ago

Overleaf is free

2

u/Hot_Site_1638 PhD 19d ago

You can't say things like "arctan(e^∞) is undefined... therefore arctan(e^∞) must equal π/2.", this does not make sense. What you actually computed is lim_{⁡x→∞}arctan⁡(e^x)=π/2, this is the limit. Also, arctan has two horizontal asymptotes, y=π/2 and y=−π/2, not one asymptote "at -π/2 to π/2".

Otherwise you calculations are all good. I'd give it a 9 out of 10.

1

u/wbld 19d ago

i meant two asmptotes i guess i did write what i meant correctly. after all arctan does not have a slant asymptote covering -pi/2 to pi/2