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Need help with $\int_{0}^{\infty} e^{- x} \ln \ln (e^{x} + \sqrt{e^{2 x} - 1}) d x$
I need help with this integral:
$\int_{0}^{\infty} e^{- x} \ln \ln (e^{x} + \sqrt{e^{2 x} - 1}) d x \approx 0.20597312051214...$
Is it possible to evaluated it in a closed form?

Feinsn 2022-06-22

How can I solve the equation...
$\log (\frac{π_{i}}{1 - π_{i}}) = \sum_{k = 0}^{K} x_{i k} β_{k} i = 1, 2, \dots, N$
How can I make the equation above the one below by taking " $e$ " to both sides.
Note that after taking $e$ to both sides of Eq. 1,
$\frac{π_{i}}{1 - π_{i}} = e^{\sum_{k = 0}^{K} x_{i k} β_{k}} .$
Thank you in advance.

Brenden Tran 2022-06-22

I have tried to solve a problem here and I've got an answer but I just don't know if it is the right one.
The problem is: For what conditions
$\int \frac{a x^{2} + b x + c}{x^{3} {(x - 1)}^{2}} d x$
represents a rational function.
First I have calculated the integral and the result is:
$(a + 2 b + 3 c) \ln | x | - \frac{2 c + b}{x} - (a + 2 b + 3 c) \ln | x - 1 | - \frac{a + b + c}{x + 1} - \frac{c}{2 x^{3}} + K$
where $K$ is a constant.
From here I have taken the conditions
$(a + 2 b + 3 c = 0) and ((b + 2 c! = o) or (a + b + c! = 0))$
which is the same as
$(a = - 2 b - 3 c) and (b! = - 2 c)$
So, from my result, the condition above must be satisfied in order for that function to be a rational function.
P.S != means not equal to
Can anyone tell whether the solution is correct or not and if it is not then help me with it ?

Emmy Dillon 2022-06-22

Calculating complex logarithm
I have to calculate the following log:
a) log(-4) b) log (3i)
I don't really know what to do..
a) $l o g (- 4) = l o g | - 4 | + i \cdot a r g (- 4) + 2 k i π = l o g 4 + ? ? + 2 k i π$
b) $l o g (3 i) = l o g | 3 i | + i \dot{a} r g (3 i) + 2 k i π = l o g (3) + ? ? + 2 k i π$

misurrosne 2022-06-22

Solve: $\frac{1}{2} \log_{\frac{1}{2}} (x - 1) > \log_{\frac{1}{2}} (1 - \sqrt[3]{2 - x})$
My try: Conditions identify: ${\begin{cases} x - 1 > 0 \\ 1 - \sqrt[3]{2 - x} > 0 \end{cases} \Leftrightarrow {\begin{cases} x > 1 \\ \sqrt[3]{2 - x} < 1 \end{cases} \Leftrightarrow x > 1$
$\frac{1}{2} \log_{\frac{1}{2}} (x - 1) > \log_{\frac{1}{2}} (1 - \sqrt[3]{2 - x}) \Leftrightarrow \log_{\frac{1}{2}} \sqrt{x - 1} > \log_{\frac{1}{2}} (1 - \sqrt[3]{2 - x}) \Leftrightarrow \sqrt{x - 1} - 1 + \sqrt[3]{2 - x} < 0$
I don't know how to solve this Any equation: $\sqrt{x - 1} - 1 + \sqrt[3]{2 - x} < 0,$ please help me solve that into the result, please guide me, thanks.

Layla Velazquez 2022-06-22

How do I solve this integral?
As stated the title, I get to a point which I can't do anything, and I'm sure I've made a mistake some where, here is my full working out:
$\int e^{i x} \cos (x) d x u = e^{i x} | u^{'} = i e^{i e} v = \sin (x) | v^{'} = \cos (x) e^{i x} \sin (x) - \int i e^{i x} \sin (x) d x + C e^{i x} \sin (x) - i \int e^{i x} \sin (x) d x + C u = e^{i x} | u^{'} = i e^{i e} v = - \cos (x) | v^{'} = \sin (x) e^{i x} \sin (x) - i (- e^{i x} \cos (x) + i \int e^{i x} c o s (x) d x) + C L e t \int e^{i x} \cos (x) d x = I I = e^{i x} (\sin (x) + i \cos (x)) - i^{2} I + C$
But ( $i^{2} = - 1$ ) so the equation should become:
$I = e^{i x} (\sin (x) + i \cos (x)) + I + C$
And this is where I'm stuck, I can't simply take $I$ away from both sides, that would make
$e^{i x} (\sin (x) + i \cos (x)) = 0$
What have a messed up in the process? And just to make it clear someone in a previous question didn't under understand what $v^{'}$ was, it's the same as $\frac{d v}{d x}$ , thank you in advance.

freakygirl838w 2022-06-22

How to find the domain of a complex rational function?
E.g.
$f (z) = \frac{3 z - i}{z}$
I understand that domain means $z \in C$ for which this is defined, but I don't know what to look for.
Perhaps:
Cannot be z=0 because of divizion of zero. But is that all?
Are the criteria for "defined" the same as for real rational functions?

Jeffery Clements 2022-06-22

Given the exponential equation $4^{x} = 64$ , what is the logarithmic form of the equation in base $10$ ?
Would it be $\frac{\log_{10} 64}{\log_{10} 4}$ ?

landdenaw 2022-06-22

I was going over some practice problems and got stuck with this one: I am supposed to find the maximum of the function:
$\frac{x}{x^{2} + 1}$
on the interval (0,4).
All I can think of is to try and plug in values from the given interval . Is there a more systematic approach to finding maxima of rational functions?

George Bray 2022-06-22

Why if a rational function has a horizontal asymptote, it excludes the possibility of this function to have an oblique azymptote, or viceversa?

Santino Bautista 2022-06-21

I cannot find a good explanation on how to represent a power series as a rational function. Thanks!
$2 - \frac{2}{3} x + \frac{2}{9} x^{2} - \frac{2}{27} x^{3} + . . .$

arridsd9 2022-06-21

Confusion regarding the Logarithmic function change of base formula
My textbook seems to be making a big leap when trying to prove the change of base formula for logarithms. If someone could help clear this up it would be very appreciated.
It starts with:
$b^{x \log_{b} (a)}$
and uses the power rule to get:
$b^{x \log_{b} (a)} = b^{\log_{b} (a^{x})}$
And it equates all this to:
$b^{x \log_{b} (a)} = b^{\log_{b} (a^{x})} = a^{x}$
Okay, I get it up to here, but then for me it leaps from that to this:
$\log_{a} (x) \cdot \log_{b} (a) = \log_{b} (a^{\log_{a} (x)}) = \log_{b} (x)$
And it says that divide through by $\log_{b} (a)$ to get the result.
What precisely has happened here? Could someone walk me through this step-by-step?
Thank you.

Summer Bradford 2022-06-21

Logarithmic functions?
$\log (x) + \log (x - 3) = \log (10 x)$
I have tried the following and not sure if I am doing it correctly...
1)
$\log (x) + \log (x - 3) = \log (10 x) ⟹ \log ((x) (x - 3)) = \log (10 x)$
$⟹ \log (x^{2} - 3 x) = \log (10 x) ⟹ x^{2} - 3 x = 10 x$
$x^{2} - 13 x = 0 ⟹ x = 0, x = 13$
2)
$\log (x) + \log (x - 3) = \log (10 x)$
$\log (x) + \log (x) - \log (3) = \log (10 x)$
$x + x - 3 = 10 x ⟹ x = - 3 / 8$
Are these approaches on the right path?

preityk7t 2022-06-21

Why can't I use product rule to derive x ln(3)?
The product rule is defined as
$(f \cdot g)^{'} = f^{'} \cdot g + g^{'} \cdot f .$
I have the following function $u (x) = x \cdot \ln (3)$ . I understand that you can derive it by implicit differentiation and have $\ln (3)$ as the result.
I, however, do not understand why I get the wrong result by applying the product rule:
$f (x) = x g (x) = \ln (3) f^{'} (x) = 1 g^{'} (x) = 1 / 3 D (f (x) * g (x)) = = 1 * \ln (3) + 1 / 3 * x = \ln (3) + 1 / 3 x \neq l n (3)$

George Bray 2022-06-21

how to find the integral of a rational logarithmic function
I can't seem to figure this one out,
it is:
$\int \frac{\ln (x)}{x} d x$
I substituted $u$ for $\ln (x)$ , so $u = \ln (x)$ and $d u = \frac{1}{x} d x$
then to find $x$ in terms of $u$ : $e^{u} = x$
so I get
$\int \frac{u}{e^{u^{2}}} d u$
from here I can't figure out where to go, I have tried playing around with the numbers but after a few hours I figured I'd ask someone here.
I sense that I must somehow get it to the form
$\int \frac{1}{x} d x,$
but i an not sure how to get a $1$ in the numerator.

Lucille Cummings 2022-06-21

How to get the derivative of $(\ln (x))^{\sec (x)}$ ?
? I know that the derivative of $\ln (x)$ is $\frac{1}{x}$ but what happens when you take it to an exponent of $\sec (x)$ ?

sedeln5w 2022-06-21

I am thinking if I could help with my current problem. Now I have a parameterized rational function $G (p, z)$ , where $p \in R^{n}$ denotes the coefficients (parameters) of the rational function, and $z$ denotes the indeterminate of the rational function which lies in complex domain.
I regard $G (p, z)$ as a mapping from $R^{n}$ to $G$ , where $G$ is a set of rational functions with indeterminate $z$ . Then I define that a property holds on a metric space $(G, d)$ if it holds on an open dense subset of $G$ .
However, I am wondering what conditions I should put on a parameter set $Θ \subseteq R^{n}$ , such that ${G (p, z) | p \in Θ}$ becomes an open subset of $G$ . Is making $Θ$ an open dense subset of $R^{n}$ sufficient?

Lydia Carey 2022-06-21

I am trying to decompose the following rational function: $\frac{1}{(x^{2} - 1)^{2}}$ in partial fractions (in order to untegrate it later).
I have notices that $(x^{2} - 1)^{2} = (x + 1)^{2} (x - 1)^{2}$
Therefore $\exists A, B, C, D$ s.t:
$\frac{1}{(x^{2} - 1)^{2}} = \frac{1}{(x - 1)^{2} (x + 1)^{2}} = \frac{A x + B}{(x + 1)^{2}} + \frac{C x + D}{(x - 1)^{2}}$
We then have
$A x + B = {\frac{1}{(x - 1)^{2}} |}_{x = - 1} ⟹ B - A = 1 / 4$
Same thing for $C x + D$ :
$C x + D = {\frac{1}{(x + 1)^{2}} |}_{x = 1} ⟹ C + D = 1 / 4$
How do I find A, B, C, D from here?

Gybrisysmemiau7 2022-06-21

When there are some conditions on the power of a rational function, how can we evaluate/prove the non-existence of the limit of the function? Here is a specific example:
Prove
$lim_{(x, y) \to (0, 0)} \frac{(x - y)^{p - 1} ((p - 2) x^{2} + 2 x y + p y^{2})}{(x^{2} + y^{2})^{2}} = 0$
if $p > 3$
Is this even solvable?

Jeffery Clements 2022-06-21

Suppose $C$ is an algebraic curve (which has singular points) over an algebraically closed field $k$ , and that $f$ is a rational function on $C$ . How does one defines the Weil divisor of $f$ ?
The problem is that the local rings of $C$ at singular points are not DVR's, so I do not have an obvious candidate for an order at a point.
Edit: Let me give an example, inspired by an answer from below. Suppose $C$ is curve $y^{2} = x^{3}$ . What would be the order of the rational function $x / y$ at the origin?

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