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Replace $y \to \frac yT \implies dy \to \frac{1}{T}dy,$ Initially , $y = a$, now $y = Ta$ Similarly $y = Tb$ $e^{-Ty} \to e^{-Ty/T} = e^{-y}$ So the integral becomes, $$I = \lim_{T\to\infty}\int^{Tb}_{Ta}\frac{-e^{-y}\frac{1}{T}dy}{\frac{y}{T}}= \lim_{T\to\infty}\int^{Tb}_{Ta}\frac{-e^{-y}dy}{y}$$
I have no access to Luke's book "Integrals of Bessel Functions" that 萬雄彦 recommended in his/her answer. Using a CAS, I got some results. Naming $$I_{i,j}=\int_{0}^\infty e^{-ax}\, x^m\, J_i(bx)\, J_j(bx)\,dx$$ $$\color{blue}{I_{0,0}=a^{-(m+1)} \Gamma (m+1) \, _3F_2\left(\frac{1}{2},\frac{m+1}{2},\frac{m+2}{2};1,1;-\frac{4b^2}{a^2}\right)}$$ provided $b\... 0 Solving improper integral $$\int \frac{dx}{\sqrt{x(1-x)}}=\int \frac{dx}{\sqrt{(\frac{1}{2})^2-(x-\frac{1}{2})^2}}$$ (by completing square method) $$=\sin^{-1}(\frac{x-\frac{1}{2}}{\frac{1}{2}})+C$$ $$=\sin^{-1}(2x-1)+C$$ Hope it helps:) 0 Sorry this is not a clear answer. But I myself is in the process of following the derivations of some of the well known integrals involving Bessel functions, and you may find your answers in Luke's, "Integrals of Bessel Functions", p.314 and after. The pity is that you can't find in detail how they are derived. Maybe you should consult the original papers if ... 1 another solution from me : $$I=\int_{0}^{\infty }e^{-(x-\frac{a}{x})^2}dx\\ \\ =\frac{1}{2}\int_{0}^{\infty }(1+\frac{a}{x^2}+(1-\frac{a}{x^2})) .e^{-(x-\frac{a}{x})^2}dx\\$$ \ $$=\frac{1}{2}\int_{0}^{\infty }e^{-(x-\frac{a}{x})^2}d(x-\frac{a}{x})+\frac{1}{2}\int_{0}^{\infty }e^{-(x-\frac{a}{x})^2}d(x+\frac{a}{x})\\ \\$$ $$=\frac{1}{2}\int_{0}^{\infty }e^{-(... 2 Case 1 : p>0 In this case, the integral is divergent since for large enough x$$x^p(\ln x)^q>x^{p\over 2}$$Case 2: p<-1 In this case, the integral is convergent since for large enough x$$x^p(\ln x)^q<x^{p-1\over 2}$$Case 3: -1<p<0 In this case, the integral is divergent since for large enough x$$x^p(\ln x)^q>x^{p-1\... 2 Hint: put$y=\ln x$. You get$\int_{\ln 2} ^{\infty} e^{(1+p)y} y^{q} dy$. Can you handle this? The answer is either$p <-1$or$p=-1$and$ q<-1. 0 Here is an independent solution: \begin{align} I&=\int_0^1\frac{\ln y\ln^2(1+y)}{y}\ dy\overset{y=\frac{1-x}{x}}{=}\int_{1/2}^1\frac{\ln(1-x)\ln^2x-\ln^3x}{x(1-x)}\ dx\\ &=\underbrace{\int_{1/2}^1\frac{\ln(1-x)\ln^2x}{x}\ dx}_{IBP}-\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx-\underbrace{\int_{1/2}^1\frac{\ln^3x}{x}\ dx}_{-(\ln^42)/4}+\int_{1/2}^1\frac{\ln(1-x)... 3 The function doesn't have to be continuous. On the other hand, it must be assumed that the restriction off$to each interval$[c,b]$, with$c\in(a,b)$, is integrable. 3 Let $$y=sin^{-1}x\ \ \Rightarrow x= \sin y,\ \ dx= \cos y dy,$$ then $$I=\int_{0}^{1}\frac{sin^{-1}x}{x}dx=\int_{0}^{\frac{\pi }{2}}\frac{y}{siny}.cosy=\int_{0}^{\frac{\pi }{2}}\frac{y}{tany}dy.$$ Let $$I(a)=\int_{0}^{\frac{\pi }{2}}\frac{\arctan(a \tan y)}{\tan y}dy \Rightarrow I'(a)=\int_{0}^{\frac{\pi }{2}}\frac{dy}{1+(a \tan(y))^2}.$$ One can easily ... 8 There is no contradiction because$f$is not Riemann-integrable on$[0,1]$. The fact that the limit$\lim_{c \to 0}\int_c ^1f(x)dx$exists does not mean that$f$is Riemann-integrable on$[0,1]$. Note that Rudin says in the exercise that we can define the symbol$\int_0^1f(x)dx$to mean said limit in the case where we have a function on$(0,1]$. It does not ... 5 The implication$f$is integrable$\Rightarrow|f|is integrable is true for Lebesgue-integrable functions. But the function that is constructed in this example is not Lebesgue-integrable. Its improper Riemann integral exists. That is a different property that does not imply Lebesgue-integrability. 2 Your answer is correct. Let $$\forall x \in \mathbb{R}, \, g(x) = \int_a^x f(t) dt$$ As you noticed \begin{align} \displaystyle \int_a^\infty f(t) \, dt \text{ is convergent} &\iff (\exists L,\lim_{x \rightarrow + \infty} \int_a^x f(t) \, dt =L)\\ &\iff (\exists L,\lim_{x \rightarrow + \infty} g(x) = L) \end{align} Notice thatgis just function. ... 4 In fact, behind your question, there is a very interesting mathematical "character" which is $$PV \left( \frac{1}{x} \right).$$ We could avoid it, remaining with classical analysis tools as in this question whose interest is to introduce the concept of (Cauchy) Principal Value (abbreviated as "PV"). But the best way to attack rigorously this issue is to ... 0 Different approach \begin{align} I&=\int_0^1\left(\int_0^1\frac{xy\ln(xy)}{xy-1}\ dx\right)\ dy\overset{xy=z}{=}\int_0^1\left(\int_0^y\frac{z\ln(z)}{y(z-1)}\ dz\right)\ dy\\ &\int_0^1\left(\int_z^1\frac{z\ln(z)}{y(z-1)}\ dy\right)\ dz=\int_0^1\frac{z\ln^2(z)}{1-z}\ dz\\ &=\sum_{n=1}^\infty\int_0^1 z^n \ln^2(z)\ dx=2\sum_{n=1}^\infty\frac1{(n+1)^... 0 \begin{align} I&=\int_0^1 \int_0^1 \frac{xy\ln(xy)}{xy-1}\ dx \ dy\\ &=\int_0^1 \int_0^1 \frac{xy\ln(x)}{xy-1}\ dx \ dy+\int_0^1 \int_0^1 \frac{xy\ln(y)}{xy-1}\ dx \ dy\\ &=2\int_0^1 \int_0^1 \frac{xy\ln(x)}{xy-1}\ dx \ dy\\ &=2\int_0^1x\ln(x)\left(\int_0^1\frac{y}{xy-1}\ dy\right)\ dx\\ &=2\int_0^1x\ln(x)\left(\frac1x+\frac{\ln(1-x)}{x^2}... 2 part(1) $$first\ part\\ I=\int_{0}^{\infty }\frac{ln^2(tan(\frac{ax}{2}-\frac{\pi }{4}))}{x^2+1}dx=\int_{0}^{\infty }\frac{ln^2(cot(\frac{ax}{2}-\frac{\pi }{4}))}{x^2+1}dx\\ \\ we\ have\ ln^2(cot(\frac{ax}{2}-\frac{\pi }{4})=-2\sum_{n,m=1}^{\infty }(-1)^{m+n}\frac{sin(2m-1)ax.sin(2n-1)ax}{(2m-1)(2n-1))}\\ \\ =2\sum_{n,m=1}^{\infty }\frac{(-1)^{m+n}[cos(2m-... 1 No, it doesn't. Note that$$\frac{z-i}{1+iz}=-i$$and that therefore$$(z-i)^2\frac{e^{ixz}}{(1+iz)^2}=(-i)^2e^{ixz}=-e^{ixz}.$$3 The integral clearly diverges as x\to\infty. Notice that$$\left|\int_1^\infty\frac{e^{-x}\cos(bx)}x~\mathrm dx\right|\le\int_1^\infty e^{-x}~\mathrm dx=e^{-1}$$converges, while$$\int_1^\infty\frac1x~\mathrm dx=\ln(x)\bigg|_1^\infty=\infty$$diverges. 3 Another standard technique is the Laplace transform. \mathcal{L} is a self-adjoint operator, i.e.$$ \int_{0}^{+\infty} f(x)(\mathcal{L}g)(x)\,dx =\int_{0}^{+\infty} (\mathcal{L}f)(x)g(x)\,dx $$and since \mathcal{L}^{-1}\left(\frac{1}{x}\right)=1, \mathcal{L}\left(1-e^{-x}\cos(bx)\right)=\frac{1}{s}-\frac{s+1}{b^2+(s+1)^2}, the original integral ... 0 Noting$$ \int_{2\pi}^{2R}\frac{\sin x}{x}dx=\int_{\pi}^{R}\frac{\sin (2x)}{x}dx=-\frac{\cos(2x)}{2x}\bigg|_\pi^R-\frac12\int_{\pi}^{R}\frac{\cos (2x)}{x^2}dx $$one has \begin{eqnarray} &&\lim _{R\rightarrow \infty }\left(\int ^{2R}_{2\pi } \frac{\sin x}{x} dx-\int ^{R}_{\pi }\left(\frac{\sin x}{x}\right)^{2} dx\right)\\ &=&\lim _{R\... 0 Hints The equality you want to prove comes from integrating by parts. The term outside integrals comes from a double arc identity for the sine (\sin(2x) = 2\sin(x)\cos(x)). The term that was integrated was 1. Then, if you have the result, try replacing it on the limit you want to find. 0 From Ramanujan's Master Theorem we know.... \int_{0}^{\infty} x^{s-1} \sum_{0}^{\infty} \phi({k}){(-1)^k}\frac{x^k}{k!} dx=\Gamma({s})\phi({-s}) If we put \zeta(-k) in place of \phi(k) then we get... \int_{0}^{\infty} x^{s-1} \sum_{0}^{\infty} \zeta({-k}){(-1)^k}\frac{x^k}{k!} dx=\Gamma({s})\zeta({s}) or, \int_{0}^{\infty} x^{s-1} \sum_{0}^{\... 0 We want to solve the following integral:$$ I(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{1}{jt(1+jt)} e^{\frac{-jNat}{1+jt}} e^{jxt} dt $$To solve, we first recall a property of the Fourier transform. Namely:$$ F^{-1}\big\{ \frac{G(t)}{jt} \big\} = \int_{-\infty}^{x} g(\tau) d\tau $$where g(x) is the inverse Fourier transform of G(t), and ... 0 By additivity of the integral:$$\int_0^\infty f(x)\;dx = \int_0^1 f(x)\;dx + \int_1^\infty f(x)\;dx$$Note that f is symmetric: f(-x)=f(x). How can you then rewrite the integral? 1 The basic steps are all correct and well conceived. A few smallish comments: First split the integral at x=1, then consider absolute values. The integrand is negative on (0,1) and positive on (1, \infty). So the integrand needs to estimated a bit differently in step 3. The integral may be expressed in terms of the \Gamma function. It's value is ... 1 So I have found the solution. Our problem is:$$ I(x) = \frac{1}{2 \pi} \displaystyle\int\limits_{-\infty}^{\infty} \frac{1}{(1-jt)^{N}} e^{\frac{jaNt}{1-jt}} e^{-jxt} dt $$Where j = \sqrt{-1}, a \in \mathbb{R} and N \in \mathbb{Z_{++}}. Solution: The first thing to do is negate t. This gives us:$$ I(x) = \frac{1}{2 \pi} \displaystyle\int\... 2 b): The series converges iff\int_1^{\infty} \frac 1 {(\ln\,x )^{(\ln\,x )}} dx <\ \infty$. In this put$y=\ln\, x$. You will see that this transforms to$\int \frac {e^{y}} {y^{y}} dy$. 1 For (b) , consider the function$\frac{1}{(\log x)^{\log x}}$, and make the variable-change$y=\log x$, to see how a connection with the previous part emerges. 3 The theorem is proved by applying the second mean value theorem for integrals (given that$\Phi$is monotone and$f$is simply integrable). For$c_2 > c_1 > a$, there exists$\xi \in (c_1,c_2)such that \tag{*}\left|\int_{c_1}^{c_2}\Phi(x) f(x) \, dx\right| = \left|\Phi(c_1)\int_{c_1}^{\xi}f(x) \, dx + \Phi(c_2)\int_{\xi}^{c_2}f(x) \, dx \right| ... 0 As usual, trig identities are the answer: \begin{align} \int_0^\infty \sin(x^2)\sin(x)dx =& \frac{1}{2}\int_0^\infty\left[ \cos(x^2-x)-\cos(x^2+x)\right]dx \\ =& \frac{1}{2}\int_0^\infty\left( \cos\left[\left(x-\frac{1}{2}\right)^2 -\frac{1}{4}\right]-\cos\left[\left(x+\frac{1}{2}\right)^2 -\frac{1}{4}\right]\right)dx \\ = &\frac{1}{2}\left[\... 4 It is rather amusing that the substitution actually works. Here is an 'elementary' proof of its validity, which stays clear of any complex analysis, infinite expansion, or special function such as Basel. Decompose the integrand as below, I =\int_0^\infty \frac{\ln x}{(x+c)(x-1)}dx = \frac{1}{c+1} \int_0^\infty {\ln x}\left[ \frac{c-1}{(x+c)(x+1)} +\... 3 I'm going to word this quite loosely, since I think the general technique of bounding these bad boys is what's important here. Like the hint says, I'll treat the two cases separately. Case 1:b>0$Without loss of generality,$\int_1^\infty \frac{x^a}{1+x^b}dx = \int_1^\infty \frac{x^{a-b}}{x^{-b}+1}dx$since$x^{-b}$is never zero at any point on the ... 1 The Glasser's master theorem is a useful tool for the solution. First, use Euler's formula to decompose the sine term into the sum of exponentials. Then it boils down to computing the integral of the form $$J(p) = \int_{0}^{\infty} \exp\left( -a x^2 - \frac{p}{x^2} \right) \, \mathrm{d}x.$$ Assume for a moment that$a, p > 0$. Then by completing the ... 0 In order to do what you want to do, you will likely try to invoke the part of the Fundamental Theorem of Calculus which tells us that if$f(x)$is continuous over$[a,b]$then$\int_{a}^{b}f(x)dx$exists, and moreover, is equal to$F(b) - F(a),$where$F(x)$is any antiderivative of$f(x)$. However, even though you want to, you will not be able to ... 2 A more general way: Use the classic integral $$\int_0^\infty \frac{x^p}{a+x}\;dx=-a^p\frac{\pi}{\sin(\pi p)};-1<p<0$$ Then $$\int_0^\infty\frac{x^p}{(a+x)(c+x)}\;dx=\frac{\pi}{\sin(\pi p)}\frac{c^p-a^p}{c-a}$$ Now differentiate this with respect to$p$and compute the limit of the result of differentiation as$p$approaches$0to get \int_0^\... 13 Fixing a >0, we see that the integral \begin{align*} f(z) := \int_0^\infty \frac{\log x}{(x+a)(x+z)} dx. \end{align*} converges absolutely for z away from (-\infty,0]. So f defines an analytic function on \mathbb C \setminus (-\infty,0] (which can be checked by Morera's theorem and Fubini's theorem, for example.) This implies f(z) = \frac{\... 7 In fact, the equation $$\int_0^{\infty}\frac{\ln x\mathrm{d}x}{(x+a)(x+b)}=\frac{(\ln a)^2-(\ln b)^2}{2(a-b)}$$ holds for all distinct complexa$,$bnot lying on the nonpositive real axis, as follows immediately from the contour integral representation $$\int_0^{\infty}\frac{\ln x\mathrm{d}x}{(x+a)(x+b)}=-\int_H\frac{\mathrm{d}x}{2\pi\mathrm{i}}\frac{(\ln ... 2 Note that$$\lim_{x\to\infty}\frac{\frac{x^{1/3}}{\sqrt{1+x^2}}}{x^{-2/3}}=\lim_{x\to\infty}\frac x{\sqrt{1+x^2}}=1$$and that therefore your integral converges if and only if the integral \displaystyle\int_1^\infty x^{-2/3}\,\mathrm dx converges. But it does not: x^{-2/3}>x^{-1} and \displaystyle\int_1^\infty x^{-1}\,\mathrm dx diverges. 1 We can rewrite your limit to$$ \lim_{x\to \infty} \int_{0}^{x} e^{\lambda(t-x)} f(t)\,dt $$The e^{\lambda(t-x)} factor is always less than 1. And for an arbitrary N>0 we can even write$$ \begin{align} \int_{0}^{x} e^{\lambda(t-x)} f(t)\,dt &{}\le e^{-\lambda N}\int_{0}^{x-N} f(t)\,dt + \int_{x-N}^x f(t)\,dt \\&{}\le e^{-\lambda N} \int_{... 0 More generally lets consider $$I=\int_0^{\infty}\frac{\ln x}{a^2+x^2}\ dx$$ Letx=a\tan\theta\implies dx=a\sec^2\theta\ d\theta$$$I=\int_0^{\pi/2}\frac{\ln (a\tan\theta)}{a^2+a^2\tan^2\theta}\ a\sec^2\theta \ d\theta$$ $$=\int_0^{\pi/2}\frac{\ln (a) +\ln(\tan\theta)}{a}\ d\theta$$ $$=\frac{\ln (a)}{a}\int_0^{\pi/2}\ d\theta+\frac 1a\int_0^{\pi/2}\ln(\tan\... 15$$\bbox[10pt, border:2px, lightblue]{\int_0^\infty \frac{\ln x}{(x+c)(x-1)}dx=\frac{\pi^2+\ln^2 c}{2(1+c)},\ \ c>0}$$A nice solution can be found here due to Yaghoub Sharifi. Perhaps it might be into your interest to see a solution for the following integral:$$I(a,b)=\int_0^\infty \frac{\ln x}{(x+a)(x+b)}dx\overset{x\to \frac{ab}{x}}=\int_0^\infty \... 0 There is no need to do partial fractions on this one. Simply set$\frac{1}{t}=x$. Your new integral will be of the form$\frac{t}{t^2-1}$and that's simply a natural log. That's it! Can you finish from here? 2 Similar to eyeballfrog's answer, since for any$x \geq 1$,$\ln{(x)} \leq x^\frac 1e$$$\int_1^p \frac{\ln{(x)}}{x^2}\,dx \leq \int_1^p\frac {x^\frac 1e}{x^2}\,dx=\frac{e \left(p-p^{\frac{1}{e}}\right)}{(e-1) p} \lt \frac e {e-1}$$ 0 If the integral converges then$\int_{\epsilon}^{2\epsilon} \frac {\ln \, x } {x^{2}}\, dx$must tend to$0$as${\epsilon} \to 0$. However,$\int_{\epsilon}^{2\epsilon} \frac {\ln \, x } {x^{2}}\, dx \leq \ln\, (2 \epsilon) \int_{\epsilon}^{2\epsilon} \frac 1 {x^{2}}\, dx=\ln\, (2 \epsilon)[\frac 1 { \epsilon}-\frac 1 {2 \epsilon} ]\to -\infty$. 5 For the upper integral, note that$\ln(x) < \sqrt{x}$for all$x > 1$. For the lower integral, note that$\int_0^1 x^{-2}dx$already diverges, and$\ln(x)$just makes it diverge even harder. 0 This solution is thanks to @metamorphy's tips. We want to solve: $$I(t) = \int_{0}^{\infty} e^{-(x+a)} I_{0}(2 \sqrt{ax}) e^{jxt} dx$$ To start, we know that: $$I_{0}(x) = \sum_{k=0}^{\infty} \frac{\big(\frac{x}{2}\big)^{2k}}{k! \Gamma(k+1)}$$ Therefore: $$I_{0}(2 \sqrt{ax}) = \sum_{k=0}^{\infty} \frac{a^{k}x^{k}}{k! \Gamma(k+1)}$$ So then we ... 4 We can integrate by parts to solve it, however pay attention that we run into divergence issues if we take$\left(\frac{x^{n+1}}{n+1}\right)'=x^n$. To avoid that we will go with:$\left(\frac{x^{n+1}-1}{n+1}\right)'=x^n$. $$I(n)=\int_0^1 \left(\frac{x^{n+1}-1}{n+1}\right)'\text{li}(x) dx\overset{}=\underbrace{\left(\frac{x^{n+1}-1}{n+1}\right)\text{li}(x)\... 2 An alternate method using the Laplace transform: Let u = ax and s = (1-it)/a. Then this integral is$$ I= \frac{e^{-a}}{a}\int_0^\infty I_0(2\sqrt{u})e^{-su}du $$i.e. the Laplace transform of I_0(2\sqrt{u}). Since I_0(2\sqrt{u}) satisfies the differential equation z f''+f'-f = 0, its Laplace transform satisfies the differential equation s^2F'(s)... 1 An honest attempt that didn't lead to anything useful so far. Let's reduce the number of parameters:$$I=\int_0^{A} e^{-\frac{x^2}{K}}e^{-\frac{(A-x)^2}{K}}\frac{{\rm erf}(\lambda x)}{x}dxx=A y, \qquad \frac{A^2}{K} = \alpha^2 \ll 1, \qquad \lambda A = \beta \gg 1I(\alpha,\beta)=\int_0^1 e^{-2\alpha^2 (y^2+y+1/2)} \frac{{\rm erf}(\beta y)}{y}dy$...