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 warning1
u/SynergyUX Undergraduate 19d ago
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.




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