Questions tagged [improper-integrals]

Questions involving improper integrals, defined as the limit of a definite integral as an endpoint of the interval of integration approaches either a specified real number or $\infty$ or $-\infty$, or as both endpoints approach limits.

An improper integral is defined as the limit of a definite integral as an endpoint of the interval of integration approaches either a specified real number or $\infty$ or $-\infty$, or as both endpoints approach limits.

Specifically, an improper integral is a limit of the form:

$$\lim_{b\to \infty} \int_{a}^{b} f(x) \ dx \,,\ \lim_{a \to -\infty} \int_{a}^{b} f(x) \ dx$$ or of the form $$\lim_{c \to b^{-}} \int_{a}^{c} f(x) \ dx \,,\ \lim_{c \to a^{+}} \int_{c}^{b} f(x) \ dx$$

in which one takes a limit in one or the other (or sometimes both) endpoints.

Often, we can compute values for improper integrals, even when the function cannot be integrated in the conventional sense (as a Riemann integral, for instance), because of a singularity in the function or because one of the bounds of integration is infinite.

7820 questions
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Test for divergence of $\int_{0}^{\infty} \frac{\sin^2(x)}{x}dx$ without evaluating the integral

I would like to prove that $$\int_{0}^{\infty} \frac{\sin^2(x)}{x}dx$$ diverges without actually evaluating the integral. Is there a convergence test from calculus or real analysis that can show that this integral diverges? Thanks. Edit: Someone…
user484604
5
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4 answers

Why the integral $\int_0^3 \frac{1}{x^2-6x+5}dx$ doesn't exist?

Why the integral $\int_0^3 \frac{1}{x^2-6x+5}dx$ doesn't exist? $$ I=\frac{1}{2} \lim_{t\to 1^{-}} [\arctan(2-3/2) - \arctan(-3/2)] + \lim_{t\to 1^+}[\arctan(0/2) - \arctan(1-3/2)] $$ I can find $\arctan(0)= \pi/2$ So why the answer in my book is…
user32104
  • 527
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Any idea how to attack this integral $\int_0^{\infty} p^2 e^{-\beta\sqrt{p^2+1}} \, dp$

I need to do this integral: $\int_0^{\infty} p^2 e^{-\beta\sqrt{p^2+1}} \, dp$ ($\beta > 0$) Any technique how to do this integral? (apparently not possible in closed form) or at least produce the result as a rapidly converging series. A closely…
alfC
  • 501
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How to compute $\int_0^{\frac{\pi}{2}}\frac{\arctan(\sqrt{\tan(x)})}{\tan(x)}dx$

I have been asked to compute the integral $$\int_0^{\frac{\pi}{2}}\frac{\arctan(\sqrt{\tan(x)})}{\tan(x)}dx$$ I have been told that it converges, and we only need its value. I tried the substitutions $ u=\tan(x)$, $\;\; v^2=\tan(x) $ . I thought…
5
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show $\int_{1}^{\infty} \frac{1}{\sqrt{x^4-x}} dx$ converges

Show $\int_{1}^{\infty} \frac{1}{\sqrt{x^4-x}} dx$ converges. I was able to show $\int_{2}^{\infty} \frac{1}{\sqrt{x^4-x}} dx$ converges, comparing it with the function $\frac{1}{x^{3/2}}$. I have trouble showing that $\int_{1}^{2}…
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Proof that improper integral $\int_{0}^{\infty} e^{-x}(1+\sin(x^2)) dx$ exists

I want to prove that the following integral exists: $$ \int_0^\infty e^{-x}\left(1+\sin\left(x^2\right) \right) dx $$ First I tried to calculate it by splitting it up (multiplied it out and then separated it at the sum symbol) but then I got…
jogico
  • 51
5
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1 answer

Conditionally convergent integral

It is known that in a conditionally convergent series, the terms can be rearranged so as to output any desired value. Thus, a conditionally convergent series is said to be undefined. My question is about the following integral: $\int\limits_{ -…
zokomoko
  • 338
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Evaluate the integral, $\int_0^{\infty} \frac{x^2}{(x+c)^{3/2}}\,e^{-x^2} dx$, where c>0

This integral came up in a research problem I'm working on, but I haven't had much luck calculating it. I suspect that the integral doesn't have a very clean form, but if anyone knows of an easy substitution or some elementary way to evaluate this…
Sam
  • 61
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how do i check the convergence of $\int^{b}_{a}\frac{dx}{(x-a)^p}$

Given question is to check the convergence of $$\int^{b}_{a}\frac{dx}{(x-a)^p}$$ I have managed to solve it till $$\lim_{\epsilon\to0}\frac{(b-a)^{1-p}}{1-p}-\frac{\epsilon^{1-p}}{1-p}$$ According to value of $p$ how to check convergence and…
5
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1 answer

Evaluating $\int_0^{\infty} \frac{\sin xt \sin yt \cos zt}{t^2} \, dt$

The problem is to evaluate the improper integral $I = \int_0^{\infty} \frac{\sin xt \sin yt \cos zt}{t^2} dt$. This can be written as $\int_0^{\infty} dt \int_0^y \frac{\sin xt \cos st \cos zt}{t} ds$, noting that $\int_0^y \cos(st) ds = \frac{\sin…
larryh
  • 207
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Intuition of $\int_{-\infty}^{\infty}$?

What's the intuition of the improper integral $$\int_{-\infty}^{\infty}$$ Is it really integral over the entire domain $\mathbb{R}$?
mavavilj
  • 7,270
5
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$\int_{-\infty}^{\infty} e^{tx} \frac{1}{\pi(1+x^2)} \mathrm dx$

which ways to compute this integral: $$ \int_{-\infty}^{\infty} e^{tx} \frac{1}{\pi(1+x^2)} \mathrm dx $$ for different cases of t. t can be any value in $\mathbb{R}$. I was stuck at how to get the antiderivative of the integrand. Thank you!
Jessie
  • 87
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Convergence of doubly infinite improper integral for odd functions.

I was working on this integral: $$\int_{-\infty}^{+\infty} \frac{x \, dx}{1+x^2}$$ Calculations shows that the limits DNE, and therefore the integral diverge. I used Mathematica and found the same result. But, the integrand is an odd functions,…
5
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1 answer

How does one show that $\int_0^\infty \left|\frac{\sin x} x\right| \, dx=\infty$?

In many place one finds accounts of how to evaluate $$ \int_0^\infty \frac{\sin x} x\,dx = \underbrace{\lim_{a\to\infty}\int_0^a}_{\text{Why view it this way?}} \frac{\sin x} x\, dx. $$ And it gets asserted that the reason why the integral must…
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