# Tag Info

### Asymptotic Expansion of Integrals via Integration by Parts - Leading Behavior

$$\int_1^\infty \frac{\cos(xt)}{t}\,dt=-\text{Ci}(x)$$ Expanded as series $$\int_1^\infty \frac{\cos(xt)}{t}\,dt=-\log (x)-\gamma+\sum _{n=1}^\infty (-1)^{n+1} \frac {x^{2n}}{ 2n\,(2n)! }$$
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### Evaluating $\lim_{t\to\infty}\left(\left(\log\left(t^2+\frac1{t^2}\right)\right)^{-1}\int_1^{\pi t}\frac{\sin^25x}{x}dx\right)$

Asssuming that $k$ and $t$ are both positive integers $$\int_1^{\pi t}\frac{\sin ^2(k x)}{x}\,dx=\frac{1}{2} (\text{Ci}(2 k)-\text{Ci}(2 k \pi t)+\log (\pi t))$$ Expanded as series, the numerator is ...
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### Cardinality of the set of arguments in integration

If one wants to take the union of all these finite sets and say that they are being evaluated "during integration", then the cardinality of such evaluations is indeed $\aleph_0$. This is ...

### Evaluating $\lim_{t\to\infty}\left(\left(\log\left(t^2+\frac1{t^2}\right)\right)^{-1}\int_1^{\pi t}\frac{\sin^25x}{x}dx\right)$

Consider $\int_1^ {\pi t} \frac{\cos^2(5x)}{x} dx$. Then this integral has the same growth behavior as $\int_1 ^{\pi t} \frac{\sin^2(5x)}{x} dx$ (that is, their quotient converges to 1; you can show ...
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1 vote
Accepted

### Asymptotic expansion of $\int_0^{2\pi} \frac{\cos(nt)}{1+t^2} dt$

Start with $$\frac 1{1+t^2}=\frac 1{(t+i)(t-i)}=\frac i 2 \left(\frac{1}{t+i}-\frac{1}{t-i}\right)$$ Use obvious changes of variable, expand the trigonometir functions to obtain the antiderivative in ...
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### evaluate $\int_{|z-e|=2} \frac{1}{(z-1)\log z}dz$

Here, $\mathrm{Log}z$ is the principal branch of the logarithm, defined in $\mathbb C\setminus\{x:x\le 0\}$. The function according to Cauchy integral formula: $$f(z)=\frac{z-1}{{\mathrm{Log}}\,z}$$ ...

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### Irrationality of $\sqrt{2}$ invoking some fact of definite integrals

Following is an alternative presentation of Jack D'Aurizio's answer that clarifies its relationship with Kostya_I's answer to a related question that I posed on MathOverflow. Kostya_I explained how ...
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### A common technique in number theory to evaluate integration

Calling this "dyadic summation" or "dyadic integration" or "dyadic decomposition" would all be understood. I'll note that generically it is not necessary to add a log ...
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### Error with Cauchy Integral Formula

Expanding in a series at $\theta = \frac{3\pi}{2}$ gives that the integrand is $$\frac{1}{2 (1 + \sin \theta)} = \frac{1}{\left(\theta - \frac{3 \pi}{2}\right)^2} + g(\theta),$$ for some integrable (...
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Accepted

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### Integral $\int_0^{\frac{\pi}{2}}\frac{\log\left(\sin x\right)}{\cos^2x+y^2\sin^2x}{d}x=-\frac{\pi}{2}\frac{\log\left(1+y\right)}{y}$

\begin{align} &\int_0^{\frac{\pi}{2}}\frac{\ln\left(\sin x\right)}{\cos^2x+y^2\sin^2x}{d}x\\ =& -\frac12\int_0^{\frac{\pi}{2}}\frac{\ln(1+\cot^2x)}{\cos^2x+y^2\sin^2x}{d}x\\ = & -\int_0^{\...
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Accepted

### Evaluating the sum $\sum_{n=1}^{\infty}{\frac{(-1)^n n}{n^2+1}}$

To evaluate the sum decompose the summand into partial fractions â€“ complexly: $$\frac i2\left(\sum_{n=1}^\infty\frac{(-1)^n}{1+in}-\sum_{n=1}^\infty\frac{(-1)^n}{1-in}\right)$$ Now use (6) on ...
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### How to integrate $\int_{0}^{1}\frac{x^{100}}{1+x}dx$?

As suggested in the comments, since the interval of integration is between $0$ and $1$, $$\frac{1}{1+x}=\sum_{n\geq 0}(-x)^n<\infty$$ and so, the integrand becomes \frac{x^{100}}{1+x}=\sum_{n\geq ...
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One way of doing it is; Let $u=x/a, v=y/b, w=z/c$. The Jacobian $\frac{d(x,y,z)}{d(u,v,w}$ for this transformation is : ${abc}$ So we are finding the volume bounded by : $u^2+v^2=w^2$ and multiplying ...