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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.

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How to compute $\int_{2/\pi}^{+\infty }\ln\cos(1/x)\,dx$?

What it says in the title. If $$I=\int_{2/\pi}^{+\infty }\ln\left({\cos\left({\frac{1}{x} }\right) }\right) \, \mathrm dx,$$ how could I proceed in order to find the value of $I$?
Healg
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Help finishing this exercise!

Given the following functions: $$ F(t)= \int_0^\infty e^{-tx}\dfrac{\sin{x}}{x}\,dx, \quad t>0$$ $$ F_s(t)= \int_0^s e^{-tx}\dfrac{\sin{x}}{x}\,dx, \quad t \geq 0, s>0$$ Show that $F$ is well-defined in $(0,+\infty)$ and it is differentiable in…
José D.
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"Not converging" vs. diverging improper integral

Take the improper integral $$\int^1 _{-\infty} \cos \pi x \; dx $$ From which it is clear that: $$\lim_{b \to -\infty} \int^1 _{b} \cos \pi x \; dx = -\frac{1}{\pi}\sin b \pi$$ The integral oscillates between $\frac{1}{\pi}\text{ and…
ptrcao
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Prove $\frac{\sqrt\pi z^v}{2^v~\Gamma\left(v+\frac{1}{2}\right)}\int_0^\infty e^{-z\cosh t}\sinh^{2v}t~dt=\int_0^\infty e^{-z\cosh t}\cosh vt~dt$?

How to prove $\dfrac{\sqrt\pi z^v}{2^v~\Gamma\left(v+\dfrac{1}{2}\right)}\int_0^\infty e^{-z\cosh t}\sinh^{2v}t~dt=\int_0^\infty e^{-z\cosh t}\cosh vt~dt$ ? Does some formulae in http://dlmf.nist.gov/5.12 helpful?
Harry Peter
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Test for convergence of improper integrals $\int_{0}^{1}\frac{\sqrt{x}}{(1+x)\ln^3(1+x)}dx$ and $\int_{1}^{\infty}\frac{\sqrt{x}}{(1+x)\ln^3(1+x)}dx$

I need to test if, integrals below, either converge or diverge: 1) $\displaystyle\int_{0}^{1}\frac{\sqrt{x}}{(1+x)\ln^3(1+x)}dx$ 2) $\displaystyle\int_{1}^{\infty}\frac{\sqrt{x}}{(1+x)\ln^3(1+x)}dx$ I tried comparing with…
atomoutside
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Evaluate $\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$

Evaluate this definite integral: $$\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$$
Soham Chowdhury
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Improper integral convergence problem comparison

Does the following improper integral converge ? $$\int _1^{\infty}\:\cfrac{e^{1/x}-1}{x}dx$$ I have tried to compare it to some known improper integrals but with no luck. Thanks for helping.
Richy65
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Convergence of $\int_{-\infty}^\infty \frac{1}{1+x^6}dx$

Okay, so I am asked to verify the convergence or divergence of the following improper integrals: $$\int_{-\infty}^\infty \frac{1}{1+x^6}dx$$ and $$\int_1^\infty \frac{x}{1-e^x}dx$$ Now, my first attempt was to use comparison criterion with $$\int…
user191919
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improper integral question - $\int_{1}^{\infty}\!e^{-x}\ln x\,dx$

I ran into this integral question: does this integral converge: $$\int_{1}^{\infty}\!e^{-x}\ln x\,dx$$ ? Thank you very much in advance, Yaron
user76508
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$\int_0^1 \frac{{f}(x)}{x^p} $ exists and finite $\implies f(0) = 0 $

Need some help with this question please. Let $f$ be a continuous function and let the improper inegral $$\int_0^1 \frac{{f}(x)}{x^p} $$ exist and be finite for any $ p \geq 1 $. I need to prove that $$f(0) = 0 $$ In this question, I really…
Simba
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Help with closed form of $\int_0^\infty\frac{\tanh(ax)}{e^x-1}dx$

I have been trying to find the value of:$$\int_0^\infty\frac{\tanh(ax)}{e^x-1}dx=\int_0^\infty\frac{e^{2ax}-1}{(e^{2ax}+1)(e^x-1)}dx$$ Under u-substitution: Let $u=e^{ax}$, $x=\frac{\log(u)}{a}$, $dx=\frac{du}{ua}$ $$\int_1^\infty…
aleden
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Why does the integral of 1/x diverge?

Why does the integration below diverge? \begin{equation} \int_{-\infty}^\infty\frac{1}{x}dx \end{equation} I know this integral diverge from $-\infty$ to $0$ (or $0$ to $\infty$). But I don't understand why these two integrals are not the same.…
Orient
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Convergence of $\int_1^\infty e^{-\ln^2(x)}dx$.

I`m interested in the convergence of the integral : $$\int_1^\infty e^{-\ln^2(x)}dx$$ I've tried using algebraic identities and some substitutions which lead me no where. Some examples to what I tried : $$\int_1^\infty e^{-\ln^2(x)}dx=\int_0^\infty…
Sar
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Does the integral $\int\limits_0^{\infty}\frac{\sin^{2}x}{x^{2}\ln(1+\sqrt x)} dx$ converge?

Does the integral $$\int\limits_0^\infty \frac{\sin^{2}x}{x^{2}\ln(1+\sqrt x)} dx$$ converge? It's easy to check that $\int\limits_1^{\infty}\frac{\sin^{2}x}{x^{2}\ln(1+\sqrt x)} dx$ does converge, but I couldn't find the right method for either…
gbi1977
  • 389
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Show an improper integral $\int_0^1 \frac{dx}{\sqrt{x^3-x}}$ converges

I would like to show that the following improper integral converges, but it's been a while since I've done this sort of calculus and I'm drawing a blank: $$ \int_0^1 \frac{dx}{\sqrt{x^3-x}} $$ My first thought was to try and simplify the expression…
chris
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