Is concealed difficulty an integral part of the cliffhanger scene form?

The “Cliffhangers” section in Masters of Umdaar says that the difficulties for rolls should be concealed until rolled against:

Of course, GMs, don’t reveal a difficulty for a specific approach until a player attempts it—let them stumble around to see which methods are more effective. (MoU 28, Cliffhangers: Running the Cliffhanger)

That stands out because it’s contrary to standard practices in Fate. On the one hand, that makes it feel like an optional playstyle preference note; on the other hand, it can be read as a deliberate and noteworthy departure from Fate norms to introduce a different sort of experience to the game.

Are concealed difficulties a crucial part of the cliffhanger concept or is this just a playstyle preference of the author? What difference does concealing difficulties make to the table experience when using cliffhangers?

Derivative of a Definite Integral

I have two equations as follows,

$ $ a = \int_0^1 dx \frac{c z_s^{d+1} x^d}{\sqrt{(1-(z_s/z_h)^{d+1} x^{d+1})(1-c^2 z_s^{2d} x^{2d})}} \tag{1}\label{1},$ $

\begin{align} S &= \frac{1}{4 z_s^{d-1}}\Bigg(-\frac{\sqrt{(1-c^2 z_s^{2d})(1-b^{d+1})}}{d-1} – \frac{1}{d-1} c^2 z_s^{2d} \int^1_0 dx x^d \sqrt{\frac{(1-(b x)^{d+1})}{(1-c^2(z_s x)^{2d})}}\ & -\frac{b^{d+1}(d+1)}{2(d-1)} \int^1_0 dx x \sqrt{\frac{(1-c^2(z_s x)^{2d})}{(1-(b x)^{d+1})}}\ & + b^{d+1}\int^1_0 dx \frac{x}{\sqrt{(1-(b x)^{d+1})(1-c^2(z_s x)^{2d})}}\Bigg) \tag{2}\label{2} \end{align}

where $ c=c(z_s)$ is a function of $ z_s$ , while $ a$ (I can fix a value for this), $ z_h$ , $ d$ (dimension) are constants, also $ b=z_s/z_h$ .

My goal is to obtain an expression for $ S$ independent of $ c$ , the conditions that can help with this requirement are,

$ $ \frac{dS}{dz_s} = 0 \tag{3}\label{3},$ $

and $ \eqref{1}$ . From $ \eqref{1}$ and $ \eqref{2}$ , we can take the derivative with respect to $ z_s$ and impose $ \eqref{3}$ on the derivative of $ \eqref{2}$ so that we can get expressions involving $ c$ and $ \frac{dc}{dz_s}$ in both the derivatives of $ \eqref{1}$ and $ \eqref{2}$ , then eliminate both $ c$ and $ \frac{dc}{dz_s}$ .

However, I tried to do the typical way as in the documentation but it does not produce the result I want.

d = 3; Integrate[(c[zs] zs^(d + 1) x^d)/((1 - x^(d + 1) (zs/zh)^(d + 1)) (1 - c[zs]^2 (zs x)^(2 d)))^(1/2), {x, 0, 1}] D[Integrate[(c[zs] zs^(d + 1) x^d)/((1 - x^(d + 1) (zs/zh)^(d + 1)) (1 - c[zs]^2 (zs x)^(2 d)))^(1/2), {x, 0, 1}] == 0, zs] 

Numerical contour integral

I am trying to compute the double integral for fixed $ m,z>0$ :

   Integrate[(Gamma[y/2] Sqrt[Gamma[3 - y]/Gamma[y]])/   Gamma[(3 - y)/2] z^(3 - y)    (Exp[-m x] - 1) x^(y - 3)/x, {x, 0, \[Infinity]},{y,    3/2 - I \[Infinity], 3/2 + I \[Infinity]}] 

The integral over $ x$ can be done analytically, and the result depends on the product $ mz$ , so there is effectively just one parameter. The $ x$ integral needs to be split into two regions I suspect, and in one region the contour of the $ y$ integral would need to be deformed so that it remains convergent. Since the $ y$ integral involves complicated branch cuts, I wanted to be able to do it numerically for a range of $ mz$ , to get a least a few digits of precision. I am having difficulty getting stable results numerically tho.

Error in nonlinearmodel fit for a function with Definite Integral and complex number

I am trying to fit a function fun and Here is the code which I am trying, Please find the data here dataset

Data = Import[    "E:\Shelender\codes\Mathematica\Aelastic \ relaxation\datat.asc"]; real = Data[[All, {1, 2}]]; imag = Data[[All, {1, 3}]]; w = 1.26*10^8; k = 1.38*10^-23;  f[H_, s_, d_] := ((1/(Sqrt[2*Pi]*s*H))*Exp[(-(Log[(H/d)])^2/(2*s^2))]) dynamic[x_?NumericQ, s_?NumericQ, d_?NumericQ, A_?NumericQ,    t_?NumericQ] :=   A*(1 - NIntegrate[      f[H, s, d]/((1 + (I*w*t*Exp[((H)/(k*x))]))), {H,        0, \[Infinity]}])  fit = ResourceFunction["MultiNonlinearModelFit"][    Rationalize[{real, imag}, 0],     ComplexExpand[ReIm@dynamic[x, s, d, A, t]],     Rationalize[{{A, 1.0*10^-4}, {t, 1.0*10^-12}, {d, 10}, {s, 0.25}},      0], {x}, PrecisionGoal -> 3, AccuracyGoal -> 3];  fit["ParameterTable"] Show[ListPlot[{real, imag}],   Plot[{fit[1, x], fit[2, x]}, {x, 0,     Max[real[[All, 1]], imag[[All, 1]]]}, PlotRange -> All],   PlotRange -> All] 

Although I am not getting any error but fit values are completely off

How to calculate this kind of double definite integral directly

Let $ D=\left\{(x, y) \mid x^{2}+y^{2} \leq \sqrt{2}, x \geq 0, y \geq 0\right\}$ , $ \left[1+x^{2}+y^{2}\right]$ represents the largest integer not greater than $ 1+x^{2}+y^{2}$ , now I want to calculate this double integral $ \iint_{D} x y\left[1+x^{2}+y^{2}\right] d x d y$ .

reg = ImplicitRegion[x^2 + y^2 <= Sqrt[2] && x >= 0 && y >= 0, {x, y}]; Integrate[x*y*Round[1 + x^2 + y^2], {x, y} ∈ reg] 

But the result I calculated using the above method is not correct, the answer is $ \frac{3}{8}$ , what should I do to directly calculate this double integral (without using the technique of turning double integral into iterated integral)?

Solving integral involving absolute value of a vector

I am trying to integrate the following in mathematica:
$ \int_0^r \frac{exp(-k_d(|\vec{r}-\vec{r_j}|+|\vec{r}-\vec{r_i}|)}{|\vec{r}-\vec{r_j}|\times|\vec{r}-\vec{r_i}|}r^2dr$ .
I have first defined, the following functions,
$ \vec p(x,y,z)= (x-x_j)\hat i + (y-y_j)\hat j+(z-z_j)\hat k$
$ \vec q(x,y,z)= (x-x_i)\hat i + (y-y_i)\hat j+(z-z_i)\hat k$ .
$ \vec r(x,y,z)=x\hat i + y\hat j+z\hat k $
Then I clicked the integration symbol in the classroom assistant panel and typed the integrand in the $ expr$ portion. While typing this, I have used $ Abs$ to take modulus of the functions $ \vec p(x,y,z)$ and $ \vec q(x,y,z)$ . I have included the limits as $ 0$ to $ Abs(r)$ and the $ var$ as $ r$ in the integration symbol. But when I press( Shift + Enter ) no output value is shown . Can anyone tell me where I have made mistake ?

Fitting an integral function given a set of data points

I have a set of measures of the resistivity of a given material at different thicknesses and I’m trying to fit them using the Fuchs-Sondheimer model. My code is:

data = {{8.1, 60.166323}, {8.5, 47.01784}, {14, 52.534961}, {15,     50.4681111501753}, {20, 39.0704975714401}, {30,     29.7737879177201}, {45, 22.4406}, {50, 15.2659673601299}, {54,     18.189933218482}, {73, 14.8377093467966}, {100,     15.249523361101}, {137, 15.249523361101}, {170,     10.7190970441753}, {202, 15.249523361101}, {230, 10.9744085456615}}  G[d_, l_, p_] := NIntegrate[(y^(-3) - y^(-5)) (1 - Exp[-yd/l])/(1 - pExp[-yd/l]), {y,0.01, 1000}];  nlm  = NonlinearModelFit[data, 1/(1 - (3 l/(2 d)) G [d, l, p]) , {{l, 200}, {p, 4}}, d, Method -> NMinimize] 

However it returns me these errors:

NIntegrate::inumr: The integrand ((1-E^(-(yd/l))) (-(1/y^5)+1/y^3))/(1-pExp[-(yd/l)]) has evaluated to non-numerical values for all sampling points in the region with boundaries {{0.01,1000}}. 
NonLinearModelFit: the function value is not a real number at {l,p} = {200.,4.} 

I think that the problem is in the way in which I defined the integral function G[d, l, p], because I had to fit a different set of data points with a different function of only one variable which I defined through the NIntegrate function and it gave me no error. Could anyone please help me?

Solve Improper Integrate using Residue theorem for this integral [migrated]

I am trying to solve this integral

$ \int_{-\infty}^\infty \frac{x\cdot sin(x)}{x^2+4} \cdot dx $

I have applied the residue theorem on a semicircle of radius $ R> 2$ , $ \gamma$ , so I have

$ \int_\gamma \frac{z\cdot sin(z)}{z^2+4} \cdot dz = \int_{-R}^R \frac{x\cdot sin(x)}{x^2+4} \cdot dx+ \int_{C_R} \frac{z\cdot sin(z)}{z^2+4} \cdot dz$

where $ C_R = \{ z : z=R e^{i \theta} , \theta \in (0,\pi)\}$ , but I cannot limit the second integral to eliminate this contribution when $ R \to \infty$ and I don’t know how I could do the integral in another way

Plot integral expression

Trying to plot in Mathematica with an integral as the iterator

Plot[x, {Integrate[1/Sqrt[0.31*x + 0.68*x^4 + 0.01*x^2], x], 0, 10^7}] 

But get the error that it the integral can’t be used as an iterator. On the other hand if I try

Plot[Integrate[1/Sqrt[0.31*x + 0.68*x^4 + 0.01*x^2], x], {x, 0, 10^7}] 

I get the ‘invalid integration variable or limit’ error.