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To obtain the series in $a$, separate off the leading term proportional to $1/a^2 $ and expand the Gaussian instead of the hyperbolic cotangent: \begin{eqnarray} f(a)&=& 2\int_{0}^{\infty } dx\, x \ e …

I'll consider $$ \int_{-a}^{a} dy\int_{-a}^{a} dx\, \delta (x^2 +y^2 -s) $$ Focusing on the integration over $x$ first, we have $$ \delta (x^2 +y^2 -s) = \frac{1}{2\sqrt{s-y^2 } } \theta (s) \theta (s-y …

At any given, fixed $p=x+y$, the $q$-integration extends from the edge at $y=a$, from $x=p-a$, to the $x=y$ axis, at $x=y=\frac{p}{2} $. … The subsequent integration over $p$ extends from $p=0$ to $p=2a$. …

It seems your conjecture is true. Mathematica gives the result $$ (1 + 4^n (-1 + n) n \mbox{Beta} [1/2, -1 + n, 2 + n] - 4^n n (1 + n) \mbox{Beta} [1/2, 1 + n, n])/(2 (1 + n)) $$ in terms of the inc …

If you want to be mathematically precise, you'd have to say a bit more about what space you're doing this on and whether you're thinking of a Frechet derivative or a Gateaux derivative, etc., but for …

To begin with, in order to get a better sense of what the moving parts are here, separate the integration intervals into $y<x$ and $y>x$, and substitute $t=x-y$ or $t=y-x$ such that $h$ appears with argument …

Transforming to polar coordinates, $$ \int_{D}\dfrac{1}{\sqrt{(x-(1-\epsilon))^2 +y^2}} \frac{1}{\sqrt{1-\sqrt{x^2+y^2}}} \, dx \, dy = $$ $$ \int_{0}^{1} dr \frac{1}{\sqrt{1-r} } \int_{0}^{2\pi } d\p …

int_{0}^{1} dx \left( \sum_{k=0}^{n} (f(x))^k \right)^{\alpha } = \int_{0}^{1} dx \left( \frac{1-(f(x))^{n+1} }{1-f(x)} \right)^{\alpha } $$ Let's choose $f(x)=x^2 $ and $\alpha =1/2$, and substitute integration …

Since the specific treatment depends on the integration region, let's introduce lower and upper limits, $y_0 <y$, $$ I = \int_{y_0 }^{y} dx\, \frac{\exp \left[ -x -\frac{1}{2} \left( \frac{x-\mu }{\sigma … +\mu -y}{\sqrt{2} \sigma }\right) - \mbox{erf} \left( \frac{n\sigma^{2} +\mu -y_0 }{\sqrt{2} \sigma }\right) \right] \end{eqnarray*} Finally, for $|a|\in \ ]e^{y_0 } ,e^y [$, we can separate into two integration …

big( \frac{1}{2} \sum_{ij} z_i A_{ij} z_j \big)^{(N/2)} $$ from the exponential series on the left-hand side, since only the monomial containing each $z_i $ precisely once yields a non-zero result upon integration …

Multiplying by $\epsilon^{n} /n! $ and summing over $n$, we obtain the generating function \begin{eqnarray} M &=& \sum_{n=0}^{\infty } \frac{\epsilon^{n} }{n! } \int d^2 x\ (\mathbf x^T \mathbf F \mat …

It is not unreasonable that a massless field behaves in a way that is totally different from a massive one with arbitrarily small mass. Already at an elementary level, you can always perform a Lorentz …