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Jun
17
comment The function $f(x) = \int_0^\infty \frac{x^t}{\Gamma(t+1)} \, dt$
I think it is Ramanujan's formula, i.e.: $$\int\limits_{0}^{\infty}\frac{z^{t} \, dt}{\Gamma(1+t)}=e^{z}-\int\limits_{0}^{\infty}\frac{e^{-z\tau}d\tau}{\tau(\ln^2 (\tau)+\pi^{2})},\,\,\, Re(z)\ge{0}$$
Jun
15
comment Evaluting: $\int\frac{1}{(1+\tan x)^2} dx$
@experimentX $$\frac{\sin x}{ \sin x + \cos x} - \frac{-\cos x}{\sin x + \cos x} = 1 \neq 2$$ ;)
Jun
15
answered Evaluating $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp(n x-\frac{x^2}{2}) \sin(2 \pi x)dx$
Jun
15
awarded  Enthusiast
Jun
13
comment Inequality between volume and its projections
@nikita2 even obvious facts need a proof in mathematics :)
Jun
13
comment Inequality between volume and its projections
@Potato sure it should be measurable.
Jun
13
awarded  Scholar
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 $\lim_{x \to \infty} \frac{1}{x} \int_x^{4x} \cos\left(\frac{1}{t}\right) \mbox {d}t$