4 votes
Accepted

Integral containing floor function and derivative

$\int_1^{n} \lfloor u \rfloor f'(u)du=\sum\limits_{k=1}^{n-1}\int_k^{k+1} \lfloor u \rfloor f'(u)du=\sum\limits_{k=1}^{n-1}\int_k^{k+1} k f'(u)du=\sum\limits_{k=1}^{n-1}k[f(k+1)-f(k)]$. This can also ...
geetha290krm's user avatar
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2 votes

Show that, $\int_0^1\frac{\arctan x \operatorname{arctanh} x \ln x}{x}dx=\frac{\pi^2}{16}G-\frac{7\pi}{32}\zeta(3)$

Let $n$ be a natural number. Then, we have $$\int_0^1 \frac{\log^{2n-1}(x)\operatorname{arctanh}(x)\arctan(x)}{x}\textrm{d}x$$ $$=\frac{\pi}{4}\left(2^{-2 n-1}-1\right) \zeta (2 n+1) (2n-1)!$$ $$+\...
user97357329's user avatar
  • 5,376
2 votes

Integral containing floor function and derivative

Just note that $$\lfloor u \rfloor f'(u) = \sum_{k = 1}^{n-1} k\mathbb 1_{[k, k+1)}(u)f'(u).$$ Therefore, splitting your integral \begin{align} \int_{1}^n \lfloor u \rfloor f'(u) du &= \sum_{k = 1}...
Falcon's user avatar
  • 3,958
2 votes
Accepted

How to evaluate double integral: $\iint \frac{y}{x} \, dx \, dy$ if it is in the first quadrant and is bounded by: $y=0$, $y=x$, and $x^2 + 4y^2 = 4$

Let's take a look at the domain $$\Omega=\{(x,y)\in\mathbb{R}^2\colon 0\leq y\leq x,\, x^2+4y^2\leq4\}$$ and notice that $$x^2+4y^2\leq4\iff \frac{x^2}{2^2}+y^2\leq 1$$ In red we have $\Omega$, in ...
Davide's user avatar
  • 331
2 votes

How to integrate $\int_{0}^{1} \int_{0}^{1} \int_{0}^{1}\frac{x^{4a - 1} \ln(x)}{\sqrt{yz} \cdot (1 + x^{2a}z + yzx^{4a} + yx^{2a})} \,dx \,dy\,dz$

$$I=\dfrac1{4a^2}\int\limits_0^1\int\limits_0^1\int\limits_0^1\dfrac{x\ln(x)}{\sqrt{yz}(1+yx)(1+zx)}\text dx\text dy\text dz$$ $$=\dfrac1{4a^2}\int\limits_0^1\int\limits_0^1\int\limits_0^1\dfrac{x\ln(...
Yuri Negometyanov's user avatar
2 votes

Fundamental calculus theorem and derivative

Yes, almost everywhere (by Lebesgue's Fundamental Theorem of Calculus, thanks @peek-a-boo) when $f(t) \cdot g(t)$ is integrable, and in general for continuous functions. Define $h(t) = f(t) \cdot g(t)$...
Mel's user avatar
  • 156
1 vote

What U substitution is being used here? (direct integral calculation of charge)

This integral is best handled by trigonometric substitution of the form $x_s=y\tan(\theta)$, $dx_s=y\sec^2(\theta) d\theta$ or $x_s=y\sinh(\phi)$, $dx_s=y\cosh(\phi)d\phi$. The resulting integrals ...
DarkLordOfPhysics's user avatar
1 vote
Accepted

How do we evaluate the hypergeometric type series where $\delta_{n}=\large \int_{_{0}}^{^{N}} {{u} \choose {n}}{{N-u} \choose {N-n}} \textit{ du }$

On the surface, since $\sum_{n=0}^N{{u} \choose {n}}{{N-u} \choose {N-n}}=1$ (there is only one nonezero element with a value of $1$), the answer is $B$, i.e. $\sum_{n=0}^N\delta_n=N$. See this post ...
Math-fun's user avatar
  • 9,487
1 vote

Fundamental calculus theorem and derivative

The requirement is that $f(x)\cdot g(x)$ be continuous function over $[a,b]$ where $a < b$ are constants and $x \in [a,b]$. Counter example is $f(x) = 1, g(x) = \dfrac{1}{x}$ over $[a,b] = [-1,1]$.
Wang YeFei's user avatar
  • 6,242
1 vote

How to evaluate $\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}dx$ in an elegant way?

Let $I$ denote the integral in the question. Then we can write \begin{align*} I &= \int_0^1\frac{\log^2 (x) \log(1-x^2)-\log^2(x)\log(1+x)}{1+x}dx \\ &= \int_0^1\frac{\log^2 (x) \log(1-x^2)}{1+...
I hate over moderation's user avatar
1 vote

Calculating asymptotics of integral $B(r) = \int_0^\infty \frac{2}{\pi} \frac{k^2 \sin(k r )}{k^3 + a} dk$

This is probably an unsatisfying answer, but at least it should produce the correct answer. Computational software (I used Mathematica) can evaluate the integral: $$ B(r) = \frac1{\pi^{3/2}\sqrt3} G_{...
Greg Martin's user avatar
  • 77.4k
1 vote

Show that $\int_{-\infty}^\infty \frac{e^x}{e^{2x}+e^{2a}}\frac{1}{x^2+\pi^2}dx = \frac{2\pi e^{-a}}{4a^2+\pi^2}-\frac{1}{1+e^{2a}}$

Suppose $a\ge 0$. Let $\displaystyle f(z) = \frac{e^z}{(e^{2z}+e^{2a})(z^2+\pi^2)}$. Then $f(z)$ will have poles $$ z=\pi i, z_k=(k+\frac12)\pi i+a,k=0,1,2,\cdots.$$ in the upper half plane. Let $R_n=\...
xpaul's user avatar
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1 vote

About an integral from MIT Integration Bee 2024

For the question (2), we can use power series after splitting the interval into two. $$ \begin{aligned} (PV)\int_0^{\infty} \frac{x^{s-1}}{1-x^k} d x&=\int_0^1 \frac{x^{s-1}}{1-x^k} d x+\int_1^{\...
Lai's user avatar
  • 19.6k

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