How to calculate the integral $ int _ {- 10} ^ {10} frac {3 ^ x} {3 ^ { lfloor x rfloor}} dx $?

I have to calculate this integral:
$$ int _ {- 10} ^ {10} frac {3 ^ x} {3 ^ { lfloor x rfloor}} dx $$

I know this function to know. $$ 3 ^ {x- lfloor x rfloor} $$ is periodic with period $ T = 1 $ so I've rewritten the integral as $$ 20 int_ {0} ^ {1} frac {3 ^ x} {3 ^ { lfloor x rfloor}} dx $$

But the problem is that I can not understand how to calculate the final integral.

Any help is appreciated.

How to evaluate $ int_ {0} ^ { infty} ln ^ 2 (x) ln (1 + x) ln ^ 2 left (1 + frac {1} {x} right) frac { dx} {x} $

$$ I = int_ {0} ^ { infty} ln ^ 2 (x) ln (1 + x) ln ^ 2 left (1 + frac {1} {x} right) frac { mathrm dx} {x} $$

$ ln (1 + x) = x- frac {x ^ 2} {2} + frac {x ^ 3} {3} – cdots $

$$ int_ {0} ^ { infty} left (1- frac {x} {2} + frac {x ^ 2} {3} + cdots right) left[ln(x)lnleft(1+frac{1}{x}right)right]^ 2 mathrm dx $$

This integral takes the form of $$ J = int_ {0} ^ { infty} x ^ n left[ln(x)lnleft(1+frac{1}{x}right)right]^ 2 mathrm dx, n ge0 $$

$$ u = left[ln(x)lnleft(1+frac{1}{x}right)right]^ 2 $$

$$ u ^ {# 1} = frac {2 ln (x) ln (1 + 1 / x) left[(1+x)ln(1+1/x)-ln(x)right]} {x (1 + x)} $$

$$ v = frac {x ^ {n + 1}} {n + 1} $$

$$ J = frac {x ^ {n + 1}} {n + 1} left[ln(x)lnleft(1+frac{1}{x}right)right]^ 2- frac {2} {n + 1} int_ {0} ^ { infty} x ^ {n + 1} cdot frac { ln (x) ln (1 + 1 / x) left[(1+x)ln(1+1/x)-ln(x)right]} {x (1 + x)} mathrm dx $$

Wow … it gets too hard, I'm totally lost, no help.

inequality – Given two increasing continuous functions $ f, g $ prove that $ (ba) int ^ b_a f (x) g (x) dx> int ^ b_a f (x) dx int ^ b_a g (x) $ dx

This is the integral inequality of Chebyshev. The following proof is
http://imar.ro/journals/Mathematical_Reports/Pdfs/2010/2/Niculescu.pdf
(Theorem 3):

If $ f $ and $ g $ both increase (or decrease both) then
$$ tag {*}
0 le bigl (f (x) – f (y) bigr) cdot bigl (g (x) – g (y) bigr)
$$
for all $ x, y in [a, b]$. It follows that
$$
0 the int_a ^ b int_a ^ b bigl (f (x) – f (y) bigr) cdot bigl (g (x) – g (y) bigr) , dx dy \
= 2 (ba) int_a ^ bf (x) g (x) , dx – 2 left ( int_a ^ bf (x) , dx right) left ( int_a ^ bg (x) , dx right) ,.
$$

If $ f $ goes up and $ g $ goes down (or vice versa), then the
the inverse inequality is valid.

Yes equality then holds the equals worth in $ (*) $ for all $ x, y
in [a, b]$ (since $ f $ and $ g $ are supposed to be continuous).
In particular
$$
0 = bigl (f (a) – f (b) bigr) cdot bigl (g (a) – g (b) bigr)
$$
which means that (at least one of) $ f $ or $ g $ is constant.

calculation – $ left | frac {d ^ n} {dx ^ n} (x ^ 2-1) ^ n right | sqrt frac2 pi cdot frac {2 ^ nn!} { sqrt n} cdot frac1 { sqrt[4]{1-x ^ 2}} $, a binding for the Legendre polynomial

Question
CA watch
I)$$ left | frac {d ^ n} {dx ^ n} (x ^ 2-1) ^ n right | sqrt frac2 pi cdot frac {2 ^ nn!} { sqrt n} cdot frac1 { sqrt[4]{1-x ^ 2}}, $$ or equivalent $$ left | P_n (x) right | sqrt frac2 { pi n} cdot frac1 { sqrt[4]{1-x ^ 2}} text {(where $ P $ is the Legendre polynomial)}, $$ when $ x in (-1,1) $ and $ n in mathbb Z ^ + $.
ii) The constant term $ sqrt frac2 pi $ in (i) is the best.

I have successfully proved the statement (ii) using the asymptotic development of $ P_n $ given by Wikipedia:
$$ P_n ( cos theta) = sqrt frac2 { pi n sin theta} cos ((n + 1/2) theta- pi / 4) + O (n ^ {- 1} ). $$ Taking supremum limit,
$$ limsup_ {n to infty} | P_n ( cos theta) | sqrt n le sqrt frac2 pi sqrt[4]{1- cos ^ 2 theta}. $$
Replacing $ cos theta = x $we can see that (ii) is a weaker result of the above limit result. However, I do not know how to deduce the explicit limit of $ P_n (x) $.

calcul – Fresnel integral improper $ int_0 ^ infty sin (x ^ 2) dx $ question

I found an example of an inappropriate Fresnel Integral $ int_0 ^ infty sin (x ^ 2) dx $ calculation. He uses several substitutions. The next substitution is not clear to me.

I did not understand how to get the right side of the left side. What subtitling is done here?

$$ int_0 ^ infty frac {v ^ 2} {1 + v ^ 4} dv = frac {1} {2} int_0 ^ infty frac {1 + u ^ 2} {1 + u ^ 4} of. $$

How can I study the convergence of the incorrect integral $ int_ {2} ^ {+ infty} frac {dx} {x ln ^ 2x} $?

I have tried to apply comparison criteria and limit criteria (by reducing it to a form of $ p $-series) in vain.

Rather than discovering this particular solution, I would like to know how to approach even this kind of exercises.

to prove $ dx ^ 2 + dy ^ 2 + dz ^ 2 = dr ^ 2 + r ^ 2 d ( theta) ^ 2 + r ^ 2 (sin theta) ^ 2 d phi ^ 2 $

$ dx ^ 2 + dy ^ 2 + dz ^ 2 = dr ^ 2 + r ^ 2 d ( theta) ^ 2 + r ^ 2 (sin theta) ^ 2 d phi ^ 2 $

given $ x = rsin theta cos phi,
y = rsin theta sin phi
, z = r cos theta $

How can I do this?

computation – How do you integrate $ int_ {0} ^ { infty} frac {a cos {(cx)}} {a ^ 2 + x ^ 2} dx $?

This integral haunts me for some time, because it escapes all the methods of integration that I could find (substitution in u, integration by parts, trigonometric substitution and same method of Feynman). I realize that this is not elementary, but I can not find how to find the definite integral. I know you'll have to use Feynman's method, but I'm lost.

To be clear, I want to know How to integrate it, not what is the value.

dx

BlackHatKings: Proxies and VPN Section
Posted by: MervinROX
Post time: June 8, 2019 at 05:12.

Integration – Verification: $ left | int_0 ^ 1 f (x) dx right | leq frac {1} {2} int_0 ^ 1 | f (x) | dx. $

Let $ f (x) $ to be a function with $ f (x) $ continue on $[0,1]$. $ f (0) + f (1) = 0 $. Prove $$ left | int_0 ^ 1 f (x) dx right | leq frac {1} {2} int_0 ^ 1 | f (x) | dx. $$

Let $ x-1/2 = t $,then $ x = t + 1/2 $,$ dx = dt $. So
begin {align *}
left | int_0 ^ 1 f (x) dx right | & = left | int _ {- frac {1} {2}} ^ {~~ frac {1} {2}} f left t + frac {1} {2} right) dt right | \
& = left | left[tfleft(t+frac{1}{2}right)right]_ {- frac {1} {2}} ^ {~~ frac {1} {2}} – int _ {- frac {1} {2}} ^ {~~ frac {1} { 2}} tf & # 39; left (t + frac {1} {2} right) dt right | ~~~ & textit {integrating by parts} \
& = left | 0- int_ {0} ^ {1} left (x- frac {1} {2} right) f left (x right) dx right | ~~~ & textit {substituting $ t + 1/2 = x $} \
& leq int_0 ^ 1 left | left (x- frac {1} {2} right) f left (x right) right | dx \
& leq left | xi- frac {1} {2} right | int_0 ^ 1 | f (x) | dx ~~~ & textit {applying the first MVT to the integral} \
& leq frac {1} {2} int_0 ^ 1 | f (x) | dx
end {align *}

what is desired.