2
votes
1answer
39 views

Asymptotic expansion of an integral with exponential decay and highly oscillating kernel [on hold]

I would appreciate if one can get the leading term of the following integral: $$I(x) = \large{\int}_0^\infty \frac{g(s)}{\sqrt{s^2 + \frac 1 4}}e^{- i x s- m x\sqrt{s^2 + \frac 1 4}}ds$$ as ...
0
votes
1answer
43 views

Convergence of $\int_0^{\infty}\left(\frac{x}{\ln x}\right)^c\frac{1}{1+x^{4c}}\ dx$

For which values of parameter $c\in\mathbb{R}$ is $$\int_0^{\infty}\left(\frac{x}{\ln x}\right)^c\frac{1}{1+x^{4c}}\ dx < \infty$$ Using the inequality $\ln x > \frac{x}{1+x}$ it can be shown ...
31
votes
5answers
732 views
+200

How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$

Please help me to find a closed form for this integral: $$I=\int_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx\tag1$$ I suspect it might exist because there are similar integrals having closed forms: ...
1
vote
1answer
41 views

How to prove that an integral converges

Let $(a_n)$, $(M_n)$ be sequences of positive real numbers such that ${a_n} \downarrow 0$, ${M_n} \uparrow \infty$ as $n\to\infty$. Let $\alpha>0$ and $\beta>1$. How to prove the following ...
1
vote
4answers
159 views

Convergent or Divergent Integral

Convergent or Divergent? $$\int_0^1 \frac {dx}{(x+x^{5})^{1/2}} $$ I have problem with the fact that if we have integration from 0 to a say and a to infinity. How does this change the way we do ...
0
votes
3answers
87 views

Convergence or divergence of the integral $\int_0^1 dx/\sin x $

Is this Convergent or Divergent $$\int_0^1 \frac{1}{\sin(x)}\mathrm dx $$ So little background to see if I am solid on this topic otherwise correct me please :) To check for convergence I can look ...
5
votes
1answer
89 views

Expressing $\int_{-\infty}^\infty dx/(x^2+1)^n$ in terms of Gamma function

How to prove this identity for $n>1/2$? $$\int_{-\infty}^{\infty}\frac{dx}{(x^2+1)^n}=\frac{\sqrt{\pi}\cdot \Gamma(n-\frac{1}{2}) }{\Gamma (n)}$$
1
vote
1answer
125 views

How to find this integral $\int_{0}^{\infty}\dfrac{f(x)}{g(x)}dx$ [duplicate]

show that: $$I=\int_{0}^{\infty}\dfrac{x^8-4x^6+9x^4-5x^2+1}{x^{12}-10x^{10}+37x^8-42x^6+26x^4-8x^2+1}dx=\dfrac{\pi}{2}$$ I found this : ...
0
votes
1answer
129 views

Finding the value of $\int_{0}^{1} \frac{\sin^2 x}{x^2}dx$

I would like to find the exact value of $$\int_{0}^{1} \frac{\sin^2 x}{x^2}dx.$$ First of all we know that it exists and must be $\hspace{0.1cm}$$\leq1$$\hspace{0.1cm}$ because$\hspace{0.1cm}$ ...
1
vote
2answers
40 views

How to recognize and evaluate improper integrals when the interval of integration is finite?

I do not understand improper integrals. Is $$ \int_1^e \frac{ \mathrm{dx}}{x(\ln x)^{1/2}}$$ an improper integral? Is $$ \int_0^2 \frac{\mathrm{dx}}{x^2+6x+8}$$ an improper integral? For both I ...
3
votes
0answers
78 views

Is this proof correct? Divergence of $\int_{1}^{\infty} \left| \frac{\sin x}{x} \right| \, \mathrm{d}x $

Problem: Show that $$ \int_{1}^{\infty} \left| \frac{\sin x}{x} \right| \,\mathrm{d}x $$ diverges. I know that there are many questions in which this problem is solved, but I want to know if my ...
3
votes
2answers
76 views

How to evaluate limits having infinity by infinity form

After taking an improper integral $\int_0^\infty \dots $ I arrived at $$\left({-x^2}e^{-\large\frac{x^2}{2a}}\,-2ae^{-\large\frac{x^2}{2a}}\right)\bigg|_{x=0}^{x=\infty}$$ Now I am trying to ...
2
votes
3answers
75 views

Evaluate $\frac{1}{a}\int_0^\infty{x^2}e^{-\frac{x^2}{2a}}\,dx$

Evaluate the following integral: $$\frac{1}{a}\int_0^\infty{x^2}e^{-\large\frac{x^2}{2a}}\,dx.$$
12
votes
5answers
296 views

The other ways to calculate $\int_0^1\frac{\ln(1-x^2)}{x}dx$

Prove that $$\int_0^1\frac{\ln(1-x^2)}{x}dx=-\frac{\pi^2}{12}$$ without using series expansion. An easy way to calculate the above integral is using series expansion. Here is an example ...
9
votes
3answers
126 views

Proof of $\int_0^\infty \frac{x^{\alpha}dx}{1+2x\cos\beta +x^{2}}=\frac{\pi\sin (\alpha\beta)}{\sin (\alpha\pi)\sin \beta }$

I found a nice formula of the following integral here $$\int_0^\infty \frac{x^{\alpha}dx}{1+2x\cos\beta +x^{2}}=\frac{\pi\sin (\alpha\beta)}{\sin (\alpha\pi)\sin \beta }$$ It states there that ...
9
votes
5answers
278 views

Prove that $\int_0^1\frac{1-x}{1-x^6}\ln^4x\,dx=\frac{16\sqrt{3}}{729}\pi^5+\frac{605}{54}\zeta(5)$

This integral comes from a well-known site (I am sorry, the site is classified due to regarding the OP.) $$\int_0^1\frac{1-x}{1-x^6}\ln^4x\,dx$$ I can calculate the integral using the help of ...
10
votes
3answers
202 views

Prove that $\int_0^1\frac{\ln(1-x)\ln^2x}{x-1}dx=\frac{\pi^4}{180}$

Prove that (please) $$\int_0^1\frac{\ln(1-x)\ln^2x}{x-1}dx=\frac{\pi^4}{180}$$ I've tried using Taylor series and I ended up with $$-\sum_{m=0}^\infty\sum_{n=1}^\infty\frac{2}{n(m+n+1)^3}$$ I am ...
1
vote
1answer
34 views

Transforming an improper integral to one with limits $0$ and $1$.

I´m working on transforming an improper integral to an integral with limit 0 and 1. I know I can use the following identities, but they just work for limits from 0 to infinity. Here are the ...
4
votes
3answers
148 views

Does the integral $\int_{a}^{b}\frac{dx}{\sqrt{(x-a)(x-b)}}$ exist?

What is the result of this integral $\displaystyle\int_{a}^{b}\dfrac{dx}{\sqrt{(x-a)(x-b)}}$ ? I have tried many possibilities like letting $\sqrt{(x-a)(x-b)}$=u or trying to make the denominator ...
1
vote
1answer
38 views

Convergence of the improper integral $\int_{0}^{\infty}\frac{x^{p-1}}{1+qx}dx$

I found that the following converges where ${0<p<1}$,and ${0<q}$, but I'm having some trouble where q is negative. Because it has some "blow up" point, it seems to diverge, but i'm not ...
0
votes
1answer
55 views

Existence of $\int_{0}^{1} \frac{\left\vert\,\log\left(x\right)\,\right\vert}{\sin\left(\sqrt{\,x\,}\,\right)}\,{\rm d}x$

$$ \mbox{Find out if the following improper integral exists:}\quad \int_{0}^{1}{\left\vert\,\log\left(x\right)\,\right\vert\over \sin\left(\sqrt{\,x\,}\,\right)}\,{\rm d}x $$ We have that ...
3
votes
0answers
101 views

Help on the Integration of $\int_0^{\infty} e^{-bx}\sin ax^2 \, \mathrm{d}x$.

I have had the misfortune of coming across the following integral, for real $b$ and $a > 0$: $$\int\limits_{0}^{\infty} e^{-bx} \sin\left(ax^{2}\right) \, \mathrm{d}x.\tag{1}$$ Naturally, I ...
2
votes
3answers
95 views

Evaluate $\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2} $

$$\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2} dx$$ My approach is to calc $$\int_{1}^{X} \frac{\ln{(2x-1)}}{x^2} dx$$ and then take the limit for the answer when $X \rightarrow \infty$ However, I must ...
-3
votes
1answer
40 views

How to find the integral of $\int \frac{GMm}{r^2}\,dr$ [closed]

I want to find the integral of: $$\int_R^\infty \frac{GMm}{r^2}\,dr$$
1
vote
4answers
70 views

Integration of $x/(x^2+1)$ from $-\infty$ to $\infty$

I am trying to find the area of this graph $\int_{-\infty}^\infty\frac{x}{x^2 + 1}$ The question first asks to use the u-substitution method to calculate the integral incorrectly by evaluating ...
-1
votes
1answer
83 views

Evaluate $\int_0^{+\infty } \frac{\log(t)}{1+t^2} \, dt$ [duplicate]

How can we compute $$I=\int_0^{+\infty } \frac{\log(t)}{1+t^2} \, dt$$ Mathematica gives $I=0$.
4
votes
3answers
274 views

Prove $\int_{\mathbb{R^{+}}} \frac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\frac{\pi}{32} (3\pi-4)^2$

How do you arrive at the result $$I=\displaystyle\int_{\mathbb{R^{+}}} \dfrac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\dfrac{\pi}{32} (3\pi-4)^2\ ?$$ Wolfram Alpha agrees numerically. I tried ...
0
votes
2answers
129 views

Improper integral of $\frac{\ln x}x$

Find $$\int_e^{\infty}\frac{\ln x}{x}\ dx$$ $A.\ \dfrac12$ $B.\ \dfrac{e^2}{2}$ $C.\ \dfrac{\ln(2e)}{2}$ $D.$ DNE (Does not exist) I tried doing this and this is where I've gone so far: $$\lim ...
1
vote
3answers
130 views

Evaluate $\int_0^\infty\frac{dl}{(r^2+l^2)^{\frac32}}$

How to evaluate the following integral $$\int_0^\infty\frac{dl}{(r^2+l^2)^{\large\frac32}}$$ The solution is supposed to look like this, unfortunately I can't derive it. $$ ...
1
vote
3answers
134 views

Value of the integral $\int_{\mathbb{R}} \frac{x\sin {(\pi x)}}{(1+x^2)^2}$

How do we evaluate the integral $$I=\displaystyle\int_{\mathbb{R}} \dfrac{x\sin {(\pi x)}}{(1+x^2)^2}$$ I have wasted so much time on this integral, tried many substitutions $(x^2=t, \ \pi x^2=t)$. ...
9
votes
2answers
177 views

An exercise from my brother: $\int_{-1}^1\frac{\ln (2x-1)}{\sqrt[\large 6]{x(1-x)(1-2x)^4}}\,dx$

My brother asked me to calculate the following integral before we had dinner and I have been working to calculate it since then ($\pm\, 4$ hours). He said, it has a beautiful closed form but I doubt ...
14
votes
1answer
261 views

Prove ${\large\int}_0^\infty\left({_2F_1}\left(\frac16,\frac12;\frac13;-x\right)\right)^{12}dx\stackrel{\color{#808080}?}=\frac{80663}{153090}$

I discovered the following conjectured identity numerically (it holds with at least $1000$ digits of precision). How can I prove it? ...
4
votes
3answers
153 views

Evaluate $\int_0^1\frac{x^a-x^{-a}}{x-1}dx$

I have heard that: $$\int_0^1\frac{x^a-x^{-a}}{x-1}dx=\frac1 a-\pi\cot(\pi a)$$ when $-1<a<1$. How would I prove this? That doesn't have an elementary indefinite integral, but the definite ...
2
votes
2answers
61 views

Duo Fresnel-like integrals $(??)$

I really wonder how I can prove the following integrals. $$\int_0^\infty \sin ax^2\cos 2bx\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}\left(\cos \frac{b^2}{a}-\sin\frac{b^2}{a}\right)$$ and ...
20
votes
1answer
341 views

Prove ${\large\int}_0^\infty\frac{\ln x}{\sqrt{x}\ \sqrt{x+1}\ \sqrt{2x+1}}dx\stackrel?=\frac{\pi^{3/2}\,\ln2}{2^{3/2}\Gamma^2\left(\tfrac34\right)}$

I discovered the following conjecture by evaluating the integral numerically and then using some inverse symbolic calculation methods to find a possible closed form: $$\int_0^\infty\frac{\ln ...
21
votes
3answers
520 views
+500

Integral ${\large\int}_0^\infty\frac{\ln x}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}\ \mathrm dx$

Please help me to evaluate this integral: $$ I={\large\int}_{0}^{\infty}{\ln\left(x\right) \over 1 + x}\, \,\sqrt{\,x + \sqrt{\,1 + x^{2}\,}\, \over 1 + x^{2}\,}\,\,{\rm d}x.\tag1 $$ Mathematica could ...
9
votes
2answers
230 views

Evaluate $\int_{0}^{\large\frac{\pi}{4}} \ln {(\sin x)}\cdot\ln {(\cos x)} \left(\frac{\ln{(\sin x)}}{\cot x}+\frac{\ln {(\cos x)}}{\tan x}\right)dx$

How do I find the value of this integral? $$I=\int_{0}^{\Large\frac{\pi}{4}} \ln {(\sin x)}\cdot\ln {(\cos x)} \left(\dfrac{\ln{(\sin x)}}{\cot x}+\dfrac{\ln {(\cos x)}}{\tan x}\right)dx$$ I tried ...
4
votes
1answer
50 views

Existence of improper integral

Prove that $$\int_{0}^{\infty} \frac{(\arctan x)^2}{x^2} dx$$ converges. This is my attempt: The above integral is equal to $$\int_{1}^{\infty} \frac{(\arctan x)^2}{x^2} dx + \int_{0}^{1} ...
1
vote
1answer
47 views

Existence of a function with certain integral properties

Does there exist a non-negative Borel-measurable function $g:\mathbb [1,\infty)\to[0,\infty)$ such that \begin{align*} \int_1^{\infty}g(y)^2\,\mathrm dy<&\,\infty,\\ ...
10
votes
2answers
302 views

A closed form of $\int_0^1\frac{\ln\ln\left({1}/{x}\right)}{x^2-x+1}\mathrm dx$

This integral has been bugging me since yesterday: $$\int_0^1\frac{\ln\ln\left({1}/{x}\right)}{x^2-x+1}\mathrm dx$$ I've tried substitution $y={1}/{x}$ and $e^y={1}/{x}$, but those didn't help ...
2
votes
1answer
91 views

Improper integral $\int_{0}^{\pi} \frac{x}{\sin x} dx$

Find out whether or not the following integral exists $$\int_{0}^{\pi} \frac{x}{\sin x} dx.$$ I'm pretty sure this integral doesn't exist but I can't seem to find a good way to prove this. It ...
6
votes
5answers
212 views

An improper integral : $\int_{0}^\infty {\ln(a^2+x^2)\over{b^2+x^2}}dx$

How to evaluate the following improper integral:$$\int_{0}^\infty {\ln(a^2+x^2)\over{b^2+x^2}}dx,$$ where $a,b>0$. I tried to suppose $$f(a)=\int_0^\infty {\ln(a^2+x^2)\over{b^2+x^2}}dx,$$ based ...
10
votes
3answers
192 views

Prove $\int_0^{\infty} \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}$

I have in trouble for evaluating following integral $$\int_0^{\infty} \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}$$ It seems really easy, but I ...
3
votes
2answers
128 views

Find $\lambda$ if $\int^{\infty}_0 \frac{\log(1+x^2)}{(1+x^2)}dx = \lambda \int^1_0 \frac{\log(1+x)}{(1+x^2)}dx$

Problem : If $\displaystyle\int^\infty_0 \frac{\log(1+x^2)}{(1+x^2)}\,dx = \lambda \int^1_0 \frac{\log(1+x)}{(1+x^2)}\,dx$ then find the value of $\lambda$. I am not getting any clue how to proceed ...
1
vote
3answers
32 views

Growth restriction for nonnegative, continuous functions whose integrals on $\mathbb{R}$ are bounded

When we have a nonnegative, continuous function $f(x)$ whose integral over all real numbers $\mathbb{R}$ is bounded, like: $$\int_{-\infty}^{\infty}f(x)dx = A< \infty $$ with $A \in \mathbb{R}$ ...
1
vote
0answers
154 views

$ \int_0^\infty (1+t^2)^{-s} (1+it)^{s'} 2t \; d t.$

The following integral bothers me since weeks: $$ \int_0^\infty (1+t^2)^{-s} (1+it)^{s'} 2t \; d t.$$ Has any body a suggestion for this integral. $Re\; s >0$ sufficiently large and $s'$ an ...
1
vote
1answer
63 views

Proving $\int^\infty_0 x^n e^{-x} \, dx = n!$

I was motivated by this question on the various applications of integration by parts to prove the following integral: $$\int^\infty_0 x^n e^{-x} \, dx = n!$$ Here's what I have done, I feel I am ...
2
votes
0answers
53 views

Fourier transform of a sinusoidal function

Let us consider following table which I want to calculate myself $$ x(t)=\frac{\sin(\omega_bt)}{\pi t}\quad\iff\quad X(j\omega)= \begin{cases} 1 & \text{if $|\omega|<\omega_b$}, ...
1
vote
2answers
75 views

Compute $\int_{0}^{\infty}e^{-tz}(z+d)^{n-1}dz$ as a function of $\Gamma(n)$

Is it possible to compute this integral $$\int_{0}^{\infty}e^{-tz}(z+d)^{n-1}dz$$ as a function of complete gamma $\Gamma(n)$. If possible, I'm looking for a closed form solution. Thanks!
0
votes
1answer
50 views

Evaluating integral involving Bessel function.

Evaluate $$\int_0^{\infty } \frac{2^{\frac{r}{\delta }} \left(2^{\frac{r}{\delta }}-1\right) r\ e^{-\frac{\alpha ^2+\left(2^{\frac{r}{\delta }}-1\right)^2}{2 \beta ^2}}\log 2 }{\beta ^2 \delta } ...