I am trying to evalute this integral: $$ \int_{A} x^{m} y^{n} z^{p}(1-x-y-z)^{q}|d x \wedge d y \wedge d z| $$ A=R3 , x , y and z are positive and x+y+z&lt;=1 I used the change of values u=x+y+z , v=(z+y)/u and $\omega$=z/(y+z) which gave me x=u(1-v) y=vu(1-$\omega$) and z=uv$\omega$ The det of the matrix of the change of variables is: u^2*v However, I am struggling to find the boundries for u, v and $\omega$ and carrying the calculation. Any help would be appreciated.
123895
Let $A =$ {$(x,y,z) \in \mathbb{R}^3 |$  $x,y,z \geq 0$ and $x+y+z \leq 1$}.   Compute the integral $$\int_{A} x^m y^n z^p (1-x-y-z)^q dx dy dz$$  using new variables $u = x+y+z$, $v=\frac{y+z}{x+y+z}$, $w=\frac{z}{y+z}$.  Also, although it was not stated in the problem, I'm assuming that $m,n,p,q \in \mathbb{Z}$.   Since $f(x,y,z)=(x+y+z, \frac{y+z}{x+y+z}, \frac{z}{y+z}) = (u,v,w)$, $f^{-1}(u,v,w) = (u(1-v), uv(1-w), uvw)$  Thus J $= \begin{vmatrix} 1-v &amp; -u &amp; 0\\  v(1-w) &amp; u(1-w) &amp; -uv\\  vw &amp; uw &amp; uv\\ \end{vmatrix} = u^2v$  Skipping the part where I find the new limits of integration, and skipping some algebraic simplification we have:  $$\int_{A} x^m y^n z^p (1-x-y-z)^q dx dy dz$$   $$= \int_{0}^1 \int_{0}^1 \int_{0}^1 u^{m+n+p+2}(1-u)^q v^{n+p+1}(1-v)^mw^p(1-w)^n du dv dw$$  However, I have no  idea where to go from here. Any help would be much appreciated.
116570
This integral looks a lot like an elliptic integral, but with cubes instead of squares: $$I(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$$ Let's consider $a,b&gt;0$ for now. $$I(a,a)=\int_0^\infty \frac{dx}{x^3+a^3}=\frac{2 \pi}{3 \sqrt{3} a^2}$$ I obtained the general series solution the following way. Choose $a,b$ such that $a \geq b$, then: $$I(a,b)=\frac{1}{a^2} \int_0^\infty \frac{dt}{\sqrt{(t^3+1)(t^3+b^3/a^3)}}=\frac{1}{a^2} I \left(1, \frac{b}{a} \right)$$ $$\frac{b^3}{a^3}=p, \qquad I \left(1, \frac{b}{a} \right)=I_1(p)$$ $$I_1(p)=\int_0^\infty\frac{dt}{\sqrt{(t^3+1)(t^3+p)}}=2 \frac{d}{dp} J(p)$$ $$J(p)=\int_0^\infty\sqrt{\frac{t^3+p}{t^3+1}}dt=\int_0^\infty\sqrt{1+\frac{p-1}{t^3+1}}dt=$$ $$|p-1| \leq 1$$ $$=\sum_{k=0}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) (p-1)^k \int_0^\infty \frac{dt}{(t^3+1)^k}$$ Now this is the most problematic part. The first integral of this series diverges. However, it's a constant in $p$, so if we differentiate, it formally disappears: $$I_1(p)=2 \sum_{k=1}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) k (p-1)^{k-1} \int_0^\infty \frac{dt}{(t^3+1)^k}$$ Now, every integral in this series converges. The integtals can be computed using the Beta function, if we substitute: $$t^3=\frac{1}{u}-1$$ Finally, we rewrite: $$I_1(p)=\frac{\Gamma (1/3)}{3 \sqrt{\pi}} \sum_{k=1}^\infty \frac{k^2}{k!^2} \Gamma \left(k- \frac{1}{2}\right) \Gamma \left(k- \frac{1}{3}\right) (1-p)^{k-1}$$ Or, using the Pochhammer symbol:  $$I_1(p)=\frac{2 \pi}{3 \sqrt{3}} \sum_{k=0}^\infty \frac{(k+1)^2}{(k+1)!^2} \left(\frac{1}{2}\right)_k \left(\frac{2}{3}\right)_k (1-p)^k$$  My questions are:  Is the method I used valid (see the 'problematic part')? How to get this series into a Hypergeometric function form? Is there any 'arithmetic-geometric mean'-like transformation (Landen's transformation) for this integral? How to go about finding it?  If the method I used is correct, it can be used for any integral of the form ($m \geq 2$): $$I_m(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^m+a^m)(x^m+b^m)}}$$
63223
Given the integral $\int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx$  How can I evaluate it? Thanks!
73214
Does any one know the exact expression of the integral,  $$E_n(s)=\int_0^\infty \frac{r^{n-1}}{(1+r^2)^{\frac{s}{2}}}\,dr~~~~s&gt;n, n\in \mathbb{N}$$ or more generally, $$E_a(s)=\int_0^\infty \frac{r^{a-1}}{(1+r^2)^{\frac{s}{2}}}\,dr~~~~s&gt;a, a\in \mathbb{R}$$ For the special case $s=n, n+2$ I find out by induction that $$ E_{n-1}(n)=\frac{\omega_{n-1}}{2\omega_{n-2}}~~\text{and}~~E_{n-1}(n+2)=\frac{\omega_{n-1}}{2n\omega_{n-2}}. $$ where $\omega_{n-1} = \frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}$ is the  surface measure of the n-dimensional sphere of $\mathbb{R}^n$. Further result is welcome
118260
I am looking for some help spotting where I messed up some computations, it's been awhile... I'll show my work to my answer and then say what the correct answer is... Integrate: $$\int_0^{2\pi} \frac{\cos(3\theta)}{5-4\cos(\theta)}\mathrm{d}\theta$$  Let $z=e^{i\theta}$ for $0\leq \theta \leq 2\pi$. Then $|z|=1$ and  $$\cos(3\theta)=\frac{z^3+z^{-3}}{2}=\frac{z^6+1}{2z^3}$$ and  $$\cos(\theta)=\frac{z+z^{-1}}{2}$$ implies  $$\frac{\cos(3\theta)}{5-4\cos(\theta)}=\frac{z^6+1}{2z^3}\cdot \frac{z}{(-2z^2+5z-2)}$$ $$=\frac{z^6+1}{2z^2(-2z^2+5z-2)}=-\frac{z^6+1}{4z^2(z^2-\frac{5}{2}z+1)}=-\frac{z^6+1}{4z^2(z-1/2)(z-2)}.$$ Thus by the change of variable, $$\int_0^{2\pi} \frac{\cos(3\theta)}{5-4\cos(\theta)}\mathrm{d}\theta=\oint_{|z|=1} -\frac{z^6+1}{4z^2(z-1/2)(z-2)} \frac{\mathrm{d}z}{iz}$$ $$=-\frac{1}{4i}\oint_{|z|=1} \frac{z^6+1}{z^3(z-1/2)(z-2)} \mathrm{d}z.$$ Now obviously the poles of $f(z)=\frac{z^6+1}{z^3(z-1/2)(z-2)}$ are $z_0=0$, of order 3, and the simple poles $z_1=1/2$ and $z_2=2$. Only $z_0,z_1\in int \,\gamma$ where $\gamma=\{z\in \mathbb{C} : |z|=1\}$ (and $int \cdot$ is the interior of a set) thus by the Cauchy-residue theorem, we have $$\oint_{|z|=1} \frac{z^6+1}{z^3(z-1/2)(z-2)} \mathrm{d}z=2\pi i (Res(f,0)+Res(f,1/2)).$$  Now $$Res(f,0)=\frac{1}{2!} \lim_{z\to 0}\frac{\mathrm{d}^2}{\mathrm{d}z^2}z^3f(z)=\frac{1}{2}\lim_{z\to 0}\frac{\mathrm{d}^2}{\mathrm{d}z^2}  \frac{z^6+1}{(z-1/2)(z-2)}=\frac{21}{2}$$ by an elementary yet tedious calculus computation. I used wolfram alpha to be honest. Next, $$Res(f,1/2)=\lim_{z\to 1/2} (z-1/2)f(z)=\lim_{z\to 1/2}  \frac{z^6+1}{z^3(z-2)}=\frac{(1/2)^6+1}{(1/8)(-3/2)}=-\frac{65}{12}$$ Thus $$\oint_{|z|=1} \frac{z^6+1}{z^3(z-1/2)(z-2)} \mathrm{d}z=2\pi i\frac{61}{12}.$$ But, $$\int_0^{2\pi} \frac{\cos(3\theta)}{5-4\cos(\theta)}\mathrm{d}\theta=-\frac{1}{4i}\oint_{|z|=1} \frac{z^6+1}{z^3(z-1/2)(z-2)} \mathrm{d}z$$ $$=-\frac{1}{4i}2\pi i\frac{61}{12}=-\frac{61}{24}\pi.$$ However wolfram says the integral (in its original form) is equal to $\pi/12$. Where did I mess up?
81910
Let $\omega$ the (n-1) form on $\mathbb{R}^n$ $$\omega=\sum_{j=1}^{n}(-1)^{j-1}x_{j}dx_{1}\wedge\cdots\wedge \hat{dx_{j}}\wedge\cdots dx_{n}$$ Show that the restriction of $\omega$ to $S^{n-1}$ in precisely the volume for this sphere.  What I did was: $\omega$ never vanish on the sphere, because, defining $\sigma\in \Omega^{n-1}(S)$ for $$\sigma_{p}(v_{1},...,v_{n-1})=det(p,v_{1},...,v_{n-1})$$ and $i:S^{n-1}\rightarrow \mathbb{R}^{n}$ the inclusion function, then $\omega=i^{\ast}(\sigma)$ then $\omega\not=0$ and therfore is a volume form.   How proof that $\omega$ is the volume form?   The first thing that comes to mind is show that $\int_{S^{n-1}}\omega=Vol(S^{n-1})$ but I have serious problems with the definition, I think that is to much.   How see that $\omega$ is invariant on $\mathbb{R}^{n}$ under action of $O(n)$
110254
Let $f(z)$ be analytic in a closed set $A$ and bounded by contour $C$. Let $z_1,z_2,...,z_n$ be different arbitrary point in the interior of $C$ and let $w_n(z)=(z-z_1)(z-z_2)...(z-z_n) $. Prove the integral $$P(z)=\frac{1}{2\pi i} \int_C \frac{f(x)}{w_n(x)} \frac{w_n(x)-w_n(z)}{x-z} dx$$ is a polynomial of degree $(n-1)$ that is the same as $f(z)$ at $z_1,z_2,...,z_n$ points.  Note: P(z) is the Lagrange interpolating polynomial.  I think I have an idea for the proof, but I don't know how to write it formally.  Idea I think this formula $$f^n(z_0)=\frac{n!}{2\pi i} \int_J \frac{f(x)}{(x-z_0)^{n+1}} dx $$ will be useful because practically is telling us that P$(z)$ must be 'equal to' $\frac{f^n(z_0)}{n!} \left(\frac{w_n(x)-w_n(z)}{x-z} \right)$.  Now the rest is trying to proof that $\frac{f^n(z_0)}{n!} \left(\frac{w_n(x)-w_n(z)}{x-z} \right)$ must be something very similar to the Lagrange polynomial.  Is this a good idea? Could be demonstrated in another way? Thanks for any suggestions/hint/tip/help.
8145
I have been self-studying CA and find it very interesting. So, working through problems in a book I have, I ran across  $$\int_{0}^{\infty}\frac{\sin(ax)}{e^{2\pi x}-1}dx=\frac{1}{4}\coth(a/2)-\frac{1}{2a}$$  and $$\int_{0}^{\infty}\frac{\sin(ax)}{e^{x}+1}dx=\frac{1}{2a}-\frac{\pi}{2\sinh(\pi a)}$$  For the former, I wrote it as $\frac{e^{aiz}}{e^{2\pi z}-1}$ and used a rectangular contour with vertices $0, \;\ R. \;\ R+i, \;\ i$  $e^{2\pi z}-1$ has poles at $ni$. Of these, I think $i$ only lies within the contour.   Unless I am in error, I then calculated the residue at $i$ to be $\frac{e^{-a}}{2\pi}$  So, $2\pi i(\frac{e^{-a}}{2\pi})=ie^{-a}$  Now, where I get hung up is setting up the integrals around the contour. Two of which should tend to 0 as $R\to \infty$.  Here is what I done.  $$\int_{0}^{R}\frac{e^{iax}}{e^{2\pi x}-1}dx+\int_{0}^{\infty}\frac{e^{ai(R+iy)}}{e^{2\pi (R+iy)}-1}idy+\int_{R}^{0}\frac{e^{ai(x+i)}}{e^{2\pi (x+i)}-1}dx+\int_{\infty}^{0}\frac{e^{ai(iy)}}{e^{2\pi iy}-1}dy=-\sinh(a)$$  I am unsure of the limits on the second and fourth integrals.  I am not so sure this is correct. The second and fourth ones, which represent the vertical sides,  should tend to 0 as $R\to\infty$. I hope :).   This than gave me $(1-e^{-a})\int\frac{e^{iax}}{e^{2\pi x}-1}dx=-\sinh(a)$.  Which does not look correct. I did manage to solve this using series and $\pi csc(\pi z)$, but the contour I am unsure of.   Can someone lend a hand here?. Any advice on either would be appreciated.   For the other one, I only posted it because it looks similar, but I believe is actually more involved and 'tougher' if you will. If I can get one, perhaps I can manage to evaluate the other.  With that one, I think the same rectangle with the same vertices can be used, but $\pi i$ would have to be avoided. Perhaps a Principal Value in there somewhere. But, I was told to use vertices $0, \;\ R, \;\ R+2\pi i, \;\ 2\pi i$  Since I am relatively new to CA, setting up those integrals is the confusing part.    I can do easier types of contour integrals, but want to learn more about these more challenging ones.   Thanks for any assistance.
114331
Let $X : \Omega \to \mathbb{N} $ be a r.v. Then $$ \mathbb{E} \{ X \} = \sum_{n=0}^{\infty} P(X &gt; n) $$ Attempt: $$ \mathbb{E} \{ X \} = \sum_j j P(X = j ) = \sum_j j(1 - P(X &gt; j) - P(X&lt;j ) )$$ But here it gets complicated. Am I on the right track?
61584
I want to evaluate the integral $\int_{0}^{1}\int_{0}^{1}\sqrt{x^{2}+y^{2}}dxdy$ using the substitution $x=r\cos\theta$ and $y=r\sin\theta$ where the Jacobian is given as $|J(r, \theta)|=r$ and thus $dxdy=rdrd\theta$.  I need to set the range for $r$ and $\theta$. For $\theta$, it seems obvious that it ranges from $0$ to $\frac{\pi}{2}$ to cover all parts of the rectangle. But I am confused with the range for $r$. Would it be OK to set minimum $r$ as $0$ and maximum $r$ as $\sqrt{2}$, which is the diagonal of the square domain? I think I am confused with the fundamentals of integration.
84933
How can I change the variable in $$\int_{0}^{1}\int_{0}^{1} \frac{x-y}{(x+y)^3} dxdy$$ to computer the value  I tried to let $u=x+y$ and $v=x-y$  But I have no idea for next step   Can you give me some suggestion,thank you!!!
59194
Let us consider that $u\in C^2(\Omega)\cap C^0(\Omega)$ and satisfying the following equation .  $\Delta u=u^3-u , x\in\Omega$ and   $u=0 $   in  $\partial \Omega$  $\Omega \subset \mathbb R^n$ and bounded .   I need hints to find out what possible value $u$ can take ?  Thank you for ur hints in advance .
101754
For my complex analysis homework, I am trying to show that the integral of the real function $1/(1+x^n)$, for integer $n\ge2$, along the positive real line is $$\int_0^{\infty}\frac{dx}{1+x^n} = \frac{\pi}{n\sin(\pi/n)}.$$ The suggested approach is to consider the contour integral $$\int_{\gamma}f(z)dz = \int_{\gamma}\frac{dz}{1+z^n},$$ where $\gamma$ is a contour construced such that it traverses all of the positive real line, but only a single simple pole is included. So I started out by factoring the polynomial in the denominator, which has (distinct) roots $\omega_k=\exp(i[\pi+2k\pi]/n)$, so that $1+z^n=(z-\omega_0)(z-\omega_1)\cdots(z-\omega_{n-1})$. Using this, I attempt to find the residue at $z=\omega_0$ (which is the single pole included in $\gamma$ by construction) by $$\text{Res}_{\omega_0}f=\lim_{z\rightarrow\omega_0}\frac{1}{(z-\omega_0)\cdots(z-\omega_{n-1})}(z-\omega_0)= \frac{1}{(\omega_0-\omega_1)\cdots(\omega_0-\omega_{n-1})}.$$  I have found that for every $(\omega_0-\omega_k)=\exp(i\pi/n)-\exp(i\pi/n+2k\pi i/n)$, I can extract an $\omega_0$ from the $\omega_k$, yielding $(\omega_0-\omega_k)=\exp(i\pi/n)[1-\exp(2k\pi i/n)]$ such that we are left with $$\text{Res}_{\omega_0}f=\frac{1}{\prod_{k=1}^{n-1} e^{i\pi/n}[1-e^{2k\pi i/n}]}=\frac{1}{e^{i\pi(n-1)/n}\prod_{k=1}^{n-1} \left(1 - e^{2k\pi i /n}\right)}.$$  Now, since I know the final answer, I strongly believe that  $$\prod_{k=1}^{n-1}(1-e^{2k\pi i/n} )=n,$$ but I am having a very hard time trying to evaluate this product. Any pointers would be greatly appreciated.
121490
Peskin's QFT textbook   1.page 14  $$\int_0 ^\infty \mathrm{d}p\ p \sin px \ e^{-it\sqrt{p^2 +m^2}}$$  when $x^2\gg t^2$, how do I apply the method of stationary phase to get the book's answer.  2.page 27  $$\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}$$  where $r&gt;0$  3.page 27  $$\int_{m}^{\infty}\mathrm{d}E \sqrt{E^2-m^2}e^{-iEt}$$  where $m&gt;0$  I'm crazy about these integrals, but the textbook doesn't give the progress.
7522
I'm trying to understand a passage from the book: A Basic Course in Probability Theory, Rabi Bhattacharya Edward C. Waymire, in the page 21. The calculation is the following:  If $X$ is a random variable on $(\Omega,\mathcal{ F}, \mathbb{P} )$, then for any $p &gt; 0, x ≥ 0$, writing $x^p=p\int_0^x y^{p−1} dy$ in the formula $\mathbb{E}|X|^p = \int_Ω |X(\omega)|^p \mathbb{P} (d\omega)$ we  obtains:  $$ \mathbb{E}|X|^p=\int_{\Omega}\left(p\int_0^{|X(\omega)|}y^{p-1}dy   \right)d \mathbb{P} ~~ {\color{red}=^{??}}~ p\int_{0}^{\infty}y^{p-1}\mathbb{P}(|X|&gt;y) dy $$ I do not understand how the left side of the above equation became the right hand side. I know that the Fubini's theorem should be used but do not know the complete details.
56115
I have a matrix in companion form,    $$A=\begin{pmatrix}  0 &amp; \cdots &amp; 0&amp; -a_{0} \\  1 &amp; \cdots &amp; 0 &amp; -a_{1}\\ \vdots &amp;\ddots &amp; \vdots &amp;\vdots \\ 0 &amp;\cdots &amp; 1 &amp; -a_{n-1} \end{pmatrix}$$    where $A \in M_{n}$. I want to prove by induction that the characteristic polynomial is  $$p_{A}=t^{n}+a_{n-1}t^{n-1}+\cdots +a_{0}$$  The part that is confusing me is if we assume this hold for $A_{1}\in M_{n-1}$ how do we transform, or "add" to $A_{1}$ to get the n by n matrix $A\in M_{n}$? I hope this makes sense.
60077
For $n$ variables $x_1,\ldots,x_n$, the determinant $$  \det\left((x_i^{j-1})_{i,j=1}^n\right) =   \left|\begin{matrix}      1&amp;x_1&amp;x_1^2&amp;\cdots &amp; x_1^{n-1}\\      1&amp;x_2&amp;x_2^2&amp;\cdots &amp; x_2^{n-1} \\      \vdots&amp;\vdots&amp;\vdots&amp;\ddots&amp;\vdots\\      1&amp;x_{n-1}&amp;x_{n-1}^2&amp;\cdots&amp;x_{n-1}^{n-1}\\      1 &amp;x_n&amp;x_n^2&amp;\cdots&amp;x_n^{n-1}   \end{matrix}\right| $$ can be computed by induction; the image below says it shows that. I have done this before, if I submit this will I get marks?   MORE IMPORTANTLY how do I do it by induction? The "hint" is to get the first row to $(1,0,0,...,0)$.  I think there are the grounds of induction in there, but I'm not sure how (I'm not very confident with induction when the structure isn't shown for $n=k$, assume for $n=r$, show if $n=r$ then $n=r+1$ is true.  By the way the question is to show the determinant at the start equals the product $$\prod_{1\leq i&lt;j\leq n}(x_j-x_i)$$ (but again, explicitly by induction)