## Asymptotic value of an integral using Mathematica

When I plug the integral
$$n \int_0^1 t^{r-1}(2t-t^2)(2-2t) dt$$, I get the following : $$2^{(-1 + 2 n) }nt^{(-1 + r) }(Beta[n, n] – 4 Beta[1/2, 1 + n, n])$$ where $$Beta(x,a,b)=\int_0^x u^{a-1}(1-u)^{b-1} du$$ is the incomplete beta integral.I want to know whether Mathematica can give me intermediate steps and,more importantly,how I can get the asyptotic limit of the value of the integral as n approches infinity.Lots of thanks for any help or hints in advance

## Solve an algebraic equation with an integral

I am trying to compute for the variable zm in terms of t which is written as an algebraic equation with an integral in it. The final answer should be zm = zm[t].

t - Integrate[(c[zm] z^(d - 1))/(f[z] Sqrt[f[z] + c[zm]^2 z^(2 d - 2)]), {z, 0, zm}] == 0

Just a note, c[zm] contains a negative sign inside the square root so that zm must be greater than zh in order for c[zm] to be real.

d = 3; zh = 2; c[zm_] := Sqrt[-(1 - zm^(d + 1)/zh^(d + 1))]/zm^(d - 1); f[z_] := (1 - z^(d + 1)/zh^(d + 1));  In[8]:= Integrate[(c[zm] z^(d - 1))/(f[z] Sqrt[f[z] + c[zm]^2 z^(2 d - 2)]), {z, 0, zm},   Assumptions -> zm > 2]  Out[8]= (1/64)*Pi*((-32 - 32*I) - (Sqrt[2*Pi]*zm*Sqrt[-16 + zm^4]*(-1 + Hypergeometric2F1[-(1/4), 1, 1/4, 16/zm^4]))/Gamma[5/4]^2)  Solve[t - Integrate[(c[zm] z^(d - 1))/(f[z] Sqrt[f[z] + c[zm]^2 z^(2 d - 2)]), {z, 0, zm}] == 0 , zm] 

I am not sure if using Solve can really find the expression, also the result of the integral contains an imaginary term, but it should not right since c[zm] is real from the Assumptions -> zm>2?

## An alternative command to compute a logarithmic integral

I am trying to see if Mathematica can calculate: $$\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+x)}{x}dx,$$ which has a well-known closed form. So I tried Integrate[Log[x]Log[1-x]Log[1+x]/x,{x,0,1}] but I waited for about 20 mins and it didn’t show any result, so I stopped the calculations. Is there a command that helps Mathematica calculate the meant integral within few mins?

## Help with NIntegrate settings to evaluate integral

I am trying to evaluate this integral: \begin{align*} \alpha_{2}=\int_{-\infty}^{1.645}\left[1-\Phi\left(\frac{\sqrt{25}}{\sqrt{15}} 1.645-\frac{\sqrt{10}}{\sqrt{15}} z_{1}\right)\right] \phi\left(z_{1}\right) d z_{1} \end{align*} where $$\Phi$$ is the Normal CDF and $$\phi$$ is the normal PDF. I know that the answer should be 0.03325.

I used the following code, but it doesn’t converge to an answer. Any suggestions?

pdf[x_] := PDF[NormalDistribution[0, 1], x]  cdf[x_] := CDF[NormalDistribution[0, 1], x]  NIntegrate[ 1 - cdf[Sqrt[25]/Sqrt[15] 1.645 - Sqrt[10]/Sqrt[15] x]* pdf[x], {x, -Infinity, 1.645}]  

which returns

NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.  NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in x near {x} = {-8.16907*10^224}. NIntegrate obtained 8.58263912306054315.954589770191005*^27949 and 8.58263912306054315.954589770191005*^27949 for the integral and error estimates. 

The following code in R gives me the correct answer:

inside <- function(z1, n1, n2, cv) {   nt <- n1 + n2   (1 - pnorm(sqrt(nt/n2) * cv - sqrt(n1/n2) * z1)) * dnorm(z1) }  additional.error <- function(n1, n2, cv) {   integrate(inside, lower = -Inf, upper = cv, n1 = n1, n2 = n2, cv = 1.645)\$  value }  additional.error(n1 = 10, n2 = 15, cv = qnorm(0.95)) $$$$ 

## 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 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,
$$\vec p(x,y,z)= (x-x_j)\hat i + (y-y_j)\hat j+(z-z_j)\hat k$$
Similarly,
$$\vec q(x,y,z)= (x-x_i)\hat i + (y-y_i)\hat j+(z-z_i)\hat k$$.
And,
$$\vec r(x,y,z)=x\hat i + y\hat j+z\hat k$$
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 ?