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bio website qoqosz.net
location Poland
age 25
visits member for 2 years, 2 months
seen Jul 20 at 8:05

I'm just a physics student.


Jun
13
accepted Inequality between volume and its projections
Jun
13
awarded  Student
Jun
13
asked Inequality between volume and its projections
Jun
12
comment Do improper integrals like $\int_{-\infty}^{+\infty} f$ converge if $xf(x)\rightarrow 0$?
In general the function you're integrating can have no limit at infinity. The same holds for $x f$. Take for example comb function d.pr/i/ElNd+ or if you want to deal with formulas, take sth like $f(x) = x^2 \exp (-x^8 \sin^2 (40x))$.
Jun
12
awarded  Critic
Jun
12
comment Limit finding of an indeterminate form
@Gigili just a typo :)
Jun
12
revised Limit finding of an indeterminate form
edited body
Jun
12
comment Limit finding of an indeterminate form
@AlexChamberlain take for example: $\lim_{x\to 0} \frac{\sin x}{x}$. When you're applying l'Hopital's rule, you're using the fact that $(\sin x)' = \cos x$ but it's the consequence of $\frac{\sin x}{x} \to 1$ when $x\to 0$.
Jun
12
answered Limit finding of an indeterminate form
Jun
10
awarded  Commentator
Jun
10
comment Evaluate $\lim_{x \to \infty} \frac{1}{x} \int_x^{4x} \cos\left(\frac{1}{t}\right) \mbox {d}t$
@stariz77 btw by using $3x \cos \frac{1}{c}$ you can also determine that symbol for $f$ is $\infty$ :)
Jun
10
comment Evaluate $\lim_{x \to \infty} \frac{1}{x} \int_x^{4x} \cos\left(\frac{1}{t}\right) \mbox {d}t$
In general we have: $$f'(c) = \frac{f(b) - f(a)}{b-a}$$ for some $c \in (a,b)$ if a<b. Now rewrite it as $(b-a) f'(c) = f(b) - f(a)$ and let $f(x) = \int^x \cos \frac{1}{t} \, dt$. In your case we have: $$(4x - x) \cos \frac{1}{c} = \int_x^{4x} \cos \frac{1}{t} \, dt$$
Jun
10
answered Evaluate $\lim_{x \to \infty} \frac{1}{x} \int_x^{4x} \cos\left(\frac{1}{t}\right) \mbox {d}t$
Jun
10
comment Evaluate $\lim_{x \to \infty} \frac{1}{x} \int_x^{4x} \cos\left(\frac{1}{t}\right) \mbox {d}t$
stariz77 right, sorry than :)
Jun
10
comment Evaluate $\lim_{x \to \infty} \frac{1}{x} \int_x^{4x} \cos\left(\frac{1}{t}\right) \mbox {d}t$
$f'(x) = \cos \left( \frac{1}{4x} \right) (4x)' - \cos \left( \frac{1}{x} \right) (x)'$ check this article on wikipedia - en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
Jun
10
comment Evaluate $A_r=\int_{{0}}^{\frac{\pi}{2}} \sin^{r}x \ \ dx$
First write: $A_r = \int_0^{\pi /2} (1 - \cos^2 x) \cdot \sin^{r-2} x \, dx = A_{r-2} - \int_0^{\pi/2} \cos^2 x \, \sin^{r-2} x \, dx$ and for integration by parts take: $u = \cos x, \; v' = \cos x \, \sin^{r-2} x$.
Jun
10
revised Equivalent of $ I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}\mathrm dx, n\rightarrow \infty$
deleted 3 characters in body
Jun
10
comment Equivalent of $ I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}\mathrm dx, n\rightarrow \infty$
@TenaliRaman you're right, I haven't noticed this divergence before. Thanks!
Jun
10
comment Equivalent of $ I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}\mathrm dx, n\rightarrow \infty$
@RagibZaman sure, but here $I_n$ behaves fine enough.
Jun
10
answered Equivalent of $ I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}\mathrm dx, n\rightarrow \infty$