Given theorem on existence and uniquenes of solutions to ordinary differential equations (Picard's theorem) there is only one local solution if the Lipschitz condition is satisfied, in fact it is when ...

This integral can be expressed with help of the residue theorem: $$I=\oint \frac{\cos(z)}{z(e^{z}-1)}dz=2\pi i \sum_{k}\; \mathrm{Res}(f,a_k)$$ There is only one removable singularity at $z=0$ and you ...

Let's write $$\int_0^\infty \frac{x \ln(x)}{x^4+4x^2+1}dx=\int_0^1\frac{x \ln(x)}{x^4+4x^2+1}dx+\int_1^\infty \frac{x \ln(x)}{x^4+4x^2+1}dx$$ For the first integral we change the variable $x\to z=1/x$ ...

It is more convenient to calculate the integral with respect to $z$ first and then with respect to $x$ and $y$. We have $$\int_{0}^{\infty}\frac{1}{(1+x^2 z^2)(1+y^2 z^2)} dz=\frac{x \operatorname{... View answer 1 votes$$\lim_{n\to \infty}(\sqrt{n^2+6b}-\sqrt{n^2-n})=\lim_{n\to \infty}\frac{n^2+6b-(n^2-n)}{\sqrt{n^2+6b}+\sqrt{n^2-n}}=\lim_{n\to \infty}\frac{6b+n}{\sqrt{n^2+6b}+\sqrt{n^2-n}}=\\ \lim_{n\to \infty}\...

Since $34 max(p,q,r)$ is even, it means that $p q + q r + r p$ must be even since $2018$ is even. Now, all three numbers of $p q, q r, r p$ cannot be odd, thus at least one is even, i.e. say $p=2$, ...

Since there is no straightforward approach to plotting "curves" depending on three parameters one can use Manipulate with three controls. We have three complex numbers a0, a1, a2 but since they lie on ...

Method I Since we have a fraction going to a non-vanishing value given its denominator is going to 0 we have to assume that its numerator also tends to 0, therefore we should solve : Reduce[Limit[...

First of all I strongly recommend a great two-volume book by R.Courant and D.Hilbert Methods of Mathematical Physics, vol.1 (linear algebra, series expansions, integral equations, calculus of ...

There are a few books written in Mathematica and dealing with mathematical research (to some extent), they are really interesting, e.g. : "Mathematica in Action" by Stan Wagon (a ...