Complex Analysis – Prove the integral of $ f (z) = frac {1-e ^ {2iz}} {4z ^ 2} $ tends to $ 0 $ on the curve $ gamma = Re ^ {it} $ as $ R rightarrow { infty} $

I am asked to prove that the integral of $ f (z) = frac {1-e ^ {2iz}} {4z ^ 2} $ on the curve $ gamma = Re ^ {it} $ $ 0 leq t leq pi $, when $ R rightarrow { infty} $, go to zero. To do this I've tried $$ int_ gamma f (z) dz = int_0 ^ { pi} f ( gamma) gamma dt = int_0 ^ { pi} frac {1-e ^ {2iRe ^ {it }}} {4 (Re ^ {it}) ^ 2} Rie ^ {it} dt = int_0 ^ { pi} frac {1-e ^ {2iRe ^ {it}} {4R ^ 2e ^ { it}} idt $$
And now I need to see that $$ lim_ {R to infty} int_0 ^ { pi} frac {1-e ^ {2iRe ^ {}}} {4} ^ {it}} idt = 0 $$
However, I do not know how to solve this integral, so I can not continue.

calculation – Proof of the Dirac delta screening property with Integral Volume

The delta function of Dirac has the sieving property which says,

$ int _ {a} ^ {b} f (x) delta (x-x) dx = begin {case}
f (x) & a <x <b \
0 & otherwise
end {cases} $

I suspect by analogy that it also has a similar property for volume integrals,

$ int _ {V} f ( mathbf {r}) delta ( mathbf {r} – mathbf {r} & # 39;) d ^ {3} mathbf {r} & # 39; = begin {case}
f ( mathbf {r}) & mathbf {r } in V \
0 & otherwise
end {cases}. $

Can any one confirm this property – how could I prove it?

abstract algebra – Is a R module free from torsion and finite on a free base free on R if R is an integral domain?

Let $ R $ to be a Dedekind domain, which is not a prime ideal domain, for example
the ring of integers in a number field such that $ R = Bbb Z[sqrt{-6}]$.
then $ R $ will have a non-main ideal, $ I $ say. Then, as $ R $-module, $ I $ is finely generated,
without torsion, but not main. In the example above, we could take $ I = 2R + sqrt {-6} R $.

integration – The curve integral of the second type (can I change the order from $ dz $ to $ dx $ to write?)

Given an oriented curve $ Sigma $ which is smooth, suppose that the function $ f (x, y, z) $ is continuous on $ Sigma $can the curve integral of the second type be
$$
iint _ { Sigma} f (x, y, z) dzdx
$$

can be written as
$$
iint _ { Sigma} f (x, y, z) dxdz.
$$

I mean I can change the order of $ dz $ and $ dx $ in writing?

Thank you.

Integral sum of n radicals

For positive integers n, S (n) denotes the minimum value of the sum (sum) of k = 1 to n ((2k-1) ^ 2 + (a (k)) ^ 2)) ^ (1/2 where a (1), a (2), …. a (n) are positive real numbers whose sum is 17. if there exists a unique positive integer n for which S (n) is also an integer, then (n / 2) is

Disappearance of the Integral Sound Function and Its Derivative

assume $ f in C ^ 1 (R ^ +) $, $ int_0 ^ infty | f | ^ a < infty $ and $ | f & # 39; <$ M for some positive $ M $ and $ a $. Is it true that $ f (x) to0 $ as $ x to infty $?


I've tried to use the integration by parts $$ int_0 ^ x f (t) = xf (x) – int_0 ^ x tf (t) , dt $$
but in vain.

Why is the $ x ^ 2 $ integral of $ 0 to $ 5 not equal to the $ sqrt x $ integral of $ 0 to $ 25?

My calculator says the first is 41.67 and the last is 83.34. Why is that? Should not they be equal?

calculation and analysis – Integral limit exchange problems

I have an integrand that is a function of $ x in (t1, t2), y in (0, t1) $

P6[x_, y_, t1_, t2_] = (x * y * baby cot[(t1 - t2)/2]* Crib[(x - y)/2]^ 2 * Csc[(x - y)/2]^ 2) / 2

Now I integrate with x first, then with y and the plot,

I6x[x_, y_, t1_, t2_] = Integrate[P6[x, y, t1, t2], X]I6yx[x_, y_, t1_, t2_] = Integrate[I6x[x, y, t1, t2], y]Manipulate[
 Plot[{Re[I6yx[x, y, [Pi]/ 3, 2 [Pi]/ 3]],
I am[I6yx[Xy[I6yx[Xy[I6yx[xy[I6yx[xy[Pi]/ 3, 2 [Pi]/ 3]]}, {X, [Pi]/ 3, 2 [Pi]/ 3},
PlotRange -> {{-2, 2}, {-50, 50}}], {y, 0, [Pi]/ 3}]

Now, I exchange the integration command,

I6y[x_, y_, t1_, t2_] = Integrate[P6[x, y, t1, t2], y]I6xy[x_, y_, t1_, t2_] = Integrate[I6x[x, y, t1, t2], X]Manipulate[
Ground[{Re[I6xy[Xy[{Re[I6xy[Xy[{RĂ©[I6xy[xy[{Re[I6xy[xy[Pi]/ 3, 2 [Pi]/ 3]],
I am[I6xy[Xy[I6xy[Xy[I6xy[xy[I6xy[xy[Pi]/ 3, 2 [Pi]/ 3]]}, {X, [Pi]/ 3, 2 [Pi]/ 3},
PlotRange -> {{-2, 2}, {-50, 50}}], {y, 0, [Pi]/ 3}]

Do I get different behavior from plots? Why is it? More importantly, I incorporate real functions on real values, why do I have an imaginary value?

Quadratic equation with integral coefficient

Let $ a, b, c $ to be natural numbers, such as the roots of the equation $ ax ^ 2 + bx + c = 0 $ are distinct and both are in the meantime

  1. (0.1)
  2. (1,2)
  3. (2,3)

Find the minimum possible value of $ a, b, c.

For my part, I resolved for part 1, that is to say for separate roots between (0,1). But for the next two games, things get a bit too complicated.

Although there may be a similarity for a given part (0,1) in the stack exchange, there is no generalized method for solving other intervals of time. this type.

So please, help, I am new to stack the exchange.

Fourier series of exp (x) and its integral

I have to find the Fourier coefficients of $ f (x) = text {exp} (x) $ for $ -1 <x <1 $ and evaluate the value of this series to $ x = $ 2.

I've calculated the coefficients to
$$ a_n = frac {(e ^ 2-1)} {e pi ^ 2 n ^ 2 + e} (-1) ^ n $$ for the cosine-terms and
$$ b_n = frac {(1-e ^ 2) pi n} {e pi ^ 2 n ^ 2 + e} (- 1) ^ n $$
for the sinus-terms.

My first question would be, if these are correct?

Assuming they are correct, I get the next set

$$ f (x) = frac {a_0} {2} + sum_ {n geq1} frac {(- 1) ^ n} {e pi ^ 2 n ^ 2 + e}
left ((e ^ 2-1) cos ( pi n x) + (1-e ^ 2) pi n sin ( pi n x) right) $$

Now, how can I calculate its value for $ x = $ 2? In this case, the series becomes a little easier, but I do not know how to calculate its value.

And finally, when I integrate the term run by term, I do not really recover the old series anymore.

I am very grateful for any kind of help!

Thank you!