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
2
votes
1 answer

Is the following "definition" for an improper integral of a function with two critical endpoints equivalent to the "standard" one?

The following are standard definitions. A function $f: [a,b) \rightarrow E$ (where $a < b \leq \infty$ and $E$ is a real Banach space) is called improperly Riemann integrable if it is Riemann integrable on $[a,b']$ for all $b' \in [a,b)$ and…
2
votes
4 answers

Study the convergence the improper integral $\int_1^\infty \frac1{\ln^7(x)}dx$

I know that $\int_1^\infty \frac1xdx$ diverges. I can probe that $\int_1^\infty \frac1{\ln(x)}dx$ diverges, as $\forall x>1:\frac1x<\frac1{\ln(x)}$ I also know that $\int_1^\infty \frac1{x^2}dx$ converges $ \therefore \int_1^\infty \frac1{x^7}dx$…
Yuta73
  • 123
2
votes
0 answers

An improper integral $\ln(x)/(1+x^2)$

The following improper integral converges or not: $$I=\int_0^\infty \frac{\ln(x)}{1+x^2}dx.$$ My attempt is as follows: $$I=I_1+I_2=\int_0^1 \frac{\ln(x)}{1+x^2}dx+\int_1^\infty \frac{\ln(x)}{1+x^2}dx.$$ For $I_1$, changing variable with $t=1/x$,…
Richkent
  • 1,151
2
votes
0 answers

Solve the special integral

I want to solve a integral which contains a shift version $$\int^{\infty}_{c}N [(1-e^{-1/t})]^{N-1} \frac{-1}{(t-c)^2}e^{-1/(t-c)}dt$$ This kind of integral has the form of normal integral $$ \int _{c} ^{\infty} {N [ f(x) ] ^{N-1}} f(x) ^{'}…
tan
  • 21
2
votes
3 answers

Improper Integral Question $\int_0^1 \ln\sqrt{x}dx $

I'm trying to compute this integral and check if it's an improper integral. What I did so far is to write the limit. $$\begin{align*} \int_0^1\ln\sqrt x\, dx &= x\ln\sqrt x - \int \frac12\, dx = x\ln\sqrt x - \frac12x + C…
Ofir Attia
  • 3,136
2
votes
3 answers

$\int^{\pi/2}_{0}\log|\sin x| \,dx = \int^{\pi/2}_{0}\log|\cos x| \,dx $

Prove that : $$\int^{\pi/2}_0 \log|\sin x| \,dx = \int^{\pi/2}_0 \log|\cos x| \,dx $$ I tried to cut the integral into a sum of parts and changing variable but it didn't work out right, i dont know how to solve this kind of problems in any other…
Lofaif
  • 361
2
votes
2 answers

Improper Integral $\int_0^1\frac{dx}{x^p}$

Is this integral convergent only for $p<1$? $$\int_0^1\frac{dx}{x^p}$$
2
votes
4 answers

Could you help me with this improper integral

How can I evaluate this improper integral? $$\displaystyle\int_0^{\infty}\frac{1}{x(1+x^2)}\,dx $$
2
votes
2 answers

Prove that: $\int_0^{\infty}\sin x\, dx=1$ and $\int_0^{\infty}\cos x\,dx=0.$

Recently I was solving some problems on Fourier transform and in one of the problems I encountered the following integral:$$\int_0^{\infty}\cos x\,dx.$$ Surely the integral does not converge and also they are not Riemann integrable, according to me.…
2
votes
1 answer

Result of $\int_{-\infty}^{\infty}\frac{\cos(ax)}{e^x+e^{-x}}$

I'm trying to validate if result of this integral is equal to: $$ \frac{\pi*ch(a\frac{\pi}{2})}{ch(\pi))+1}) $$ I'm trying to resolve it using the reside in $\frac{\pi}{2}$ but couldn't find a resolution to compare. Any help is most appreciated.
2
votes
3 answers

Improper integral $\int_0^\infty x\sin(x^3) dx$

I need to check if the following integral converges / diverges conditionally. $$\int_0^\infty x\sin(x^3) dx$$ I have tried integrating by parts and it didn't work. Will appreciate any help :).
2
votes
2 answers

find the real values of p and q such that following integral converges

$\displaystyle \int_0^1 x^p \left ( \ln \dfrac 1 x \right)^q \, dx $ given $f(x) = (-1)^qx^p(\log(x))^q$ if we are able to show that integral f(x) absolutely converges then integral converges using limit comparison test with $g(x) = 1/x^m$ where…
2
votes
1 answer

Convergence of $\int_0^\infty \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx$ without Gamma function

For what values of $a$ does the following integral converge: $$\int_0^\infty \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx$$ We have to study the integral at $0$ and $\infty$: $$\int_0^\infty \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx = \int_0^1…
VIVID
  • 11,604
2
votes
1 answer

The product of improper Riemann integrable function and a bounded continuous function

Let $f$ be an improper integrable function on an unbounded set $[a,+\infty)$ and $g$ is a bounded continuous function, then I hope to determine whether $fg$ is improper integrable or not? I think the answer is no, but I fail to find an example. I…
2
votes
3 answers

I think the solution to the improper integral $\int_{0}^{+\infty}\frac{1}{\left(1+x^{2}\right)^{m}}dx$ is wrong

Given $$\int_{0}^{+\infty}\frac{1}{\left(1+x^{2}\right)^{m}}dx$$ such that $m>0$. Study the convergence of the previous integral. Now, I've tried solving for $m$ using the comparison tests but have been unsuccessful. Luckily, I've found the solution…