2
votes
3answers
83 views

Cauchy principal value with log

How can I obtain the result of the following integration? $$\int_{-\infty}^{+\infty} \log \left(1+\frac{a^2}{x^2}\right)dx$$ Thank you!
1
vote
1answer
57 views

Assigning values to divergent integrals

I'm interested in the (obviously divergent) integral $$ \int_{-\infty}^\infty dx e^{-x f}\ ,$$ where $f$ is real. Is there any way to meaningfully assign a value to this integral? I was thinking of ...
0
votes
1answer
59 views

Calculus 2 Integral Question

I've been trying to resolve a calculus question and seem to be having troubles understanding exactly how to approach it. Some hints are supplied, but they don't exactly seem to help. Thanks to anyone ...
1
vote
2answers
41 views

Evaluating an integral over inifinty with polars leads to an integral of cosine over inifinity, how can this be resolved?

So I have the integral $$\int_0^\infty\int_0^\infty\frac{yx^2}{x^2 +y^2}e^{-(x^2 +y^2)} \,dx\,dy$$ And converting this into polars gives: $$\int_0^\infty r^2 e^{-r^2}\,dr ...
2
votes
1answer
51 views

Is this improper integral answer correct?

So I'm working on improper integrals and con/divergence and want some assurance that I've done the following correctly. $\int^∞_{-∞}cos(\pi t)$ As far as I'm aware this is convergent if and only if ...
2
votes
3answers
194 views

Why is the integral of sec^2(x) from 0 to pi infinity?

Why is it, if you take the integral of sec^2(x) from 0 to pi, my calculator returns "infinity" as the answer, but according to the second fundamental theorem of calculus, I got 0 with my own work. I ...
0
votes
2answers
34 views

Questions about hyperbolas and integration

I have a couple of questions regarding hyperbolas and their integrals. If it's too much, don't feel like you have to answer all 3 questions. My first question: The integral of a function like 1/x^2 ...
0
votes
3answers
36 views

Computing the limit of this function

So I have an improper integral: $$ \int_0^\infty \frac{13x}{x^2+1}-\frac{65}{5x+1} dx $$ I have solved the integral into this: $$ \lim_{t \to \infty} ...
0
votes
1answer
49 views

Integrals with infinite bounds sometimes written as limits, sometimes not?

When I saw Wikipedia's notation for the inverse Laplace transform, I became curious if there was a reason behind it. Is there a reason why Wikipedia writes the inverse Laplace transform as this ...
0
votes
1answer
53 views

Comparing Improper Integrals Involving Infinity

From my current understanding: $K>J$ and $L>K$ , therefore $L>K>J$. How can I compare the first integral $I$ ?
0
votes
1answer
574 views

Comparison Theorem for Integral Calculus

I have narrowed it down to C, E, and F, since we know that $1/x^{1/5}$ is always greater than the original function for all $x\geq 1$. However, the second set of conditions is more difficult to ...
1
vote
0answers
81 views

Hadamard regularization isn't working out

As part of an exercise in a grad course on "mathematical methods" (always such a helpful name), I've been asked to evaluate $I=\int_0^{1/2}{(x^2-x+c)^{-2}dx}$ as a Hadamard finite part integral for $0 ...
1
vote
4answers
839 views

Integration with infinity and exponential

How is $$\lim_{T\to\infty}\frac{1}T\int_{-T/2}^{T/2}e^{-2at}dt=\infty\;?$$ however my answer comes zero because putting limit in the expression, we get: $$\frac1\infty\left(-\frac1{2a}\right) ...
1
vote
3answers
205 views

Surface under $\frac{1}{x}$ is $\infty$, while surface under $\frac{1}{x^2}$ is $1$?

Since the antiderivative of $\frac{1}{x}$ is $\ln(|x|)$, the surface under the graph of $\frac{1}{x}$ with $x>1$ is $\infty$. However, the antiderivative of $\frac{1}{x^2}$ is $-\frac{1}{x}$, so ...