# Tagged Questions

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### A closed-form of $\frac{1}{2}\int_0^\infty\left[\frac{x^2\cos x}{\cosh 2x-\cos x}-\frac{2x^2}{e^{4x}-2e^{2x}\cos x+1}\right]\,dx$

I am looking for a closed-form of this integral $$\frac{1}{2}\int_0^\infty\left[\frac{x^2\cos x}{\cosh 2x-\cos x}-\frac{2x^2}{e^{4x}-2e^{2x}\cos x+1}\right]\,dx$$ I ...
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### Integral of $\ln(x)\operatorname{sech}(x)$

How can I prove that: ...
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### 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: ...
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### cosine integral

Show that $$\int_0^x \frac{1-\cos(t)}{t}=\gamma+\ln(x)-\operatorname{Ci}(x)$$ where $$\operatorname{Ci}(x)=-\int_x^\infty \frac{\cos(t)}{t} \, dt$$ and gamma is an euler-mascheroni constant. I did as ...
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### improper integral containing $\sqrt{\cos x-\dfrac{1}{\sqrt 2}}$ in the denominator

How do i find the value of this integral-- $$I=\displaystyle\int_{0}^{\pi/4} \frac{\sec^2 x \ dx}{\sqrt {\cos x-\dfrac{1}{\sqrt 2}}}$$ I came across this integral too in physics.
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### 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 : ...
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### Solving integral that contain exponential function and lower incomplete gamma function

I have the following integral; $$y=\int_0^\infty\frac{e^{-xf}}{m+x}\gamma(a,hx)~dx$$ where $f,m,h\in\mathbb{R}^+$ , $a\in\mathbb{N}$ , $\gamma\left(a,h x\right)$ is the lower incomplete gamma ...
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### Dealing with an integral: can we go any farther?

I meet an integral, but it is beyond my ability. $${\rm I}\left(a\right) = \int_{a}^{1}{\arcsin\left(\,\sqrt{\,{1 - x^{2} \over 1 - a^{2}}\,}\,\right) \over x + 1}\,{\rm d}x, 0\le a <1.$$ I can ...
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### Definite integral $\int_{0}^{\infty}e^{-u}\frac{1}{\left(\sqrt{1+(h+u)^{2}}\right)^{5}}du$

Hi guys I have the following definite integral to solve: $$\int_{0}^{\infty}e^{-u}\frac{1}{\left(\sqrt{1+(h+u)^{2}}\right)^{5}}du$$ is it possible to obtain an analytic expression? And if not why? ...
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### An Improper Integral

I need help with this integral: $\Large {\int_0^\infty \frac{dx}{x\sqrt{1+x}}}$ What I did: Substitute $\sqrt {1+x} = t$. Then the integral turns into $\int_1^\infty 2dt/(t^2-1)$. Now I replaced ...
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### 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$\hspace{0.1cm}$ ...
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### Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$

Here is a challenging one maybe some would like a go at. Show that: ...
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### Closed form for integral of integer powers of Sinc function

(Edit: Thank you Vladimir for providing the references for the closed form value of the integrals. My revised question is then to how to derive this closed form.) For all $n\in\mathbb{N}^+$, ...
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### Evaluate $\frac{1}{a}\int_0^\infty{x^2}e^{-\frac{x^2}{2a}}\,dx$

Evaluate the following integral: $$\frac{1}{a}\int_0^\infty{x^2}e^{-\large\frac{x^2}{2a}}\,dx.$$
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### 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 ...
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### 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 ...
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### 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 ...
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### Show that $\int^{\infty}_0 \frac{\sin^4 x}{x^4}=\frac{\pi} 3$. [duplicate]

Show that $$\int^{\infty}_0 \left( \frac{\sin x}{x}\right)^4=\frac{\pi} 3$$ Although I know the integral with the index is $1$ and $2$, I have no idea on this one. Please help.
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### A probabilistic integral $\int_{-\infty}^{\infty}e^{-x^2/2\sigma^2}\arcsin\left(1-2\left|\lfloor x\rceil-x\right|\right)\,dx$

In my probabilistic studies, a tough integral appeared. Note that $\lfloor x\rceil$ is not the floor function; it is the nearest integer function. Up to some constants, it appears in a Buffon-like ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### Convergence of improper integral of $\ln f(x)$

Is there something know about the convergence of $\int_0^1 \ln f(x)dx$ for $f(x)$ continous on $\left(0,1\right)$ and both limits exists, i.e. $\lim_{x\to 0} f(x)$ and $\lim_{x\to 1} f(x)$ ? I ...
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### 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 ...
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### 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 ...
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### 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 ...
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### How to identify continuity or discontinuity of an [Definite] integral?

How can I figure out whether an improper integral converges based on the discontinuities in the integrand? For instance, these two both have discontinuities within the intervals of integration, and ...
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### $\int_0^\infty (1+t^2)^{-s} (1+it)^{s'} 2t \; d t.$

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 ...
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### Choose appropriate contour for a complex integral

I have a problem to solve integral $$I = \int^{\infty}_0 \frac{\mathrm{d}x}{(x-z)(1+x^2)^{\kappa+2}}$$ I can solve the same integral with borders $-\infty$ to $\infty$ using residue theorem but ...
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### Evaluation of $\int_0^1 \frac{\log(1+x)}{1+x}\log\left(\log\left(\frac{1}{x}\right)\right) \ dx$

I need some hints, clues for getting the closed form of $$\int_0^1 \frac{\log(1+x)}{1+x}\log\left(\log\left(\frac{1}{x}\right)\right) \ dx$$
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### fourier transform of scaled function

let us consider following example one thing which i did not understand is where absolute value of $a$ came from?ok if we have $\int^{\infty}_{-\infty} x(a*t)*e^{-j\omega*t}dt$ then we may have ...
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### Laplace transform of : $t^{\gamma-1} F(\alpha,\beta,\delta,\frac{t}{d})$, where $F$ is the Gauss' hypergeometric function

What is the Laplace transform of : $t^{\gamma-1} F(\alpha,\beta,\delta,\frac{t}{d})$, where $\gamma >0$ and $F$ is the Gauss' hypergeometric function. Note that I have the Laplace transform of : ...
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### Closed form for $f(x)=\int_{0}^{+\infty}e^{it^{x}}dt$?

Let $x>1$ and $f(x)=\int_{0}^{+\infty}e^{it^{x}}dt$. Does this integral have a closed form ? Fist point, the integral converges. Indeed let $u=e^{it^{x}}$ and $v=\frac{-i}{x}t^{1-x}$ we have ...
### How to calculate the integral $\int_{-1}^{1}\frac{dz}{\sqrt[3]{(1-z)(1+z)^2}}$?
The integral is $I=\displaystyle\int_{-1}^{1}\dfrac{dz}{\sqrt[3]{(1-z)(1+z)^2}}$. I used Mathematica to calculate, the result was $\dfrac{2\pi}{\sqrt{3}}$, I think it may help.