Evaluating ∫∞−∞e−x2dx\int_{-\infty}^{\infty}e^{-x^2}dx using polar coordinates. [duplicate]

This question already has answers here: Proving ∫∞0e−x2dx=√π2 (19 answers) Closed 6 years ago. I need to express the following improper integral as a double integral of x and y and then, using polar coordinates, evaluate it. I=∫∞−∞e−x2dx Plotting it, we find a Gaussian centered at x=0 which tends to infinity to both sides. We … Read more

Finer asymptotic estimate of an integral

I’m studying the asymptotic behaviour for large $n\in \mathbb N$ of $\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}$ Using the substitution $u=(n+1)\ln(t)$, $$\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}=\frac{1}{n+1}\int_0^\infty\exp(\frac{u}{n+1})\frac{1}{1+\exp(u)}du$$ And, using dominated convergence, one has $\displaystyle \int_0^\infty\exp(\frac{u}{n+1})\frac{1}{1+\exp(u)}du=\ln(2)+o(1)$, so that $$\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}=\frac{\ln(2)}{n}+o(\frac{1}{n})$$ I’m looking to refine that estimate, that is to say get a sharper estimate of $\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}-\frac{\ln(2)}{n}$. It amounts to … Read more

Convergence of improper integral ∫π60x√1−2sinxdx\int_{0}^{\frac{\pi}{6}}\dfrac{x}{\sqrt{1-2\sin x}}dx

I’m trying to determine whether the following improper integral is convergent or divergent. ∫π/60x√1−2sinxdx At first, I substituted t=π2−x and then I used 1−12×2≤cosx. But I couldn’t determine. 🙁 Second attempt, I used sinx≤x on [0,π6]. But I couldn’t determine. :-[ Could you give me some advice? Thanks in advance. Answer Since (√1−2sinx)′=−cosx√1−2sinx, using integration … Read more

Improper integration problem

$$ \int \limits^{\infty }_{0}\frac{\tan^{-1}\left( x\right) + \tan^{-1}\left( \frac{\alpha x +\beta }{\beta x -\alpha } \right) }{x^{2}+1} dx$$. For $ \alpha, \beta >0$… My question is how we can evaluate this improper integration above?. Actually I tried partial fraction and many other ways but I couldn’t complete. How I can find the value of this integration?. … Read more

Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$

Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$ $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx=\lim\limits_{t\to\infty}\int_1^{t} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$ Partial integration can’t solve the integral $\int \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$. What substitution (or other methods) would you suggest? Answer For $x \in [1,\infty)$, you have $$0 \le \frac{1}{x\sqrt[3]{x^2+1}} \le \frac{1}{x^\frac{5}{3}}$$ and $\int_1^\infty \frac{dx}{x^\frac{5}{3}}$ is convergent. AttributionSource : Link , Question Author : user300045 , … Read more

Improper integral involving trigonometric function

I was wondering what happens when evaluating an improper integral involving a trigonometric function where the denominator is a rational function with a zero at x=0. The example I have in mind is \int_{-\infty}^{\infty}\frac{sin(ax)}{x(x-i)(x+i)} dx If I rewrite the sine in terms of the exponential and then evaluate two integrals, one for the upper half-plane … Read more

Integral of Lorentzian type with trigonometric function

Consider the following Riemann integral ∫∞0dxα2(x−x0)2+α2sin[(x−x1)t]x−x1 with the displacements x0,x1∈R and broadening α∈R. The (claimed) solution should be π[α2δ2+α2+e−αtδ2sin(δt)−α2cos(δt)δ2+α2] with δ=x0−x1. Is an approximation involved, and how do I evaluate this integral? Answer Start from a simplified form involving fewer constants, to grasp the idea; we need ∫+∞−∞(11+t2)sinttdt; notice that the integrand is the product … Read more

For which a>0a>0 does ∫∞adx(x2−a)4a\int_a^\infty \frac{\mathrm{d}x}{(x^2-a)^{4a}} converge?

As the title suggests, I need help finding a>0 for which the following improper integral converges: ∫∞adx(x2−a)4a So, at first I thought I would just do this: f(x)=1(x2−a)4a Then I wanted to find a function g(x) such that f(x)≤g(x) for any x∈[a,∞). I thought I could use something like g(x)=1(a2−a)4a but I am not sure … Read more

Evaluation of an integral

How would one prove that: ∫π/20ln(1+cosθ)cosθdθ=π28 This is what I did. ∫π/20ln(1+cosθ)cosθdθ=∫π/201cosθ∞∑n=1(−1)n−1cosnθndθ=∞∑n=1(−1)n−1n∫π/20cosn−1θdθ=∞∑n=1(−1)n−12nB(12,n2) No clue how to proceed. I am also interest in seeing another approach that does not use Taylor expansion. Answer I give some steps here, and I leave it to you to fill in the details. Start with f(a)=∫π/20ln(1+acosθ)cosθdθ. Then f′(a)=∫π/2011+acosθdθ=⋯=2√1−a2arctan(√1−a√1+a) Integrating, we … Read more

Improper integral ∫1−00+0dx(4−3x)√x−x2dx\int _{0+0}^{1-0}\frac{dx}{\left(4-3x\right)\sqrt{x-x^2}}\:dx

How do I solve this? ∫1−00+0dx(4−3x)√x−x2dx I know it’s a type 3 improper integral, and I’m having issues with these. I think that I need to write it as a sum of limits and then try to compute the values of those limits and the value of the sum would be my improper integral value. … Read more