I'm trying to prove that:   $$F\,\{f'(x)\} = -i\omega F(\omega) \qquad (1) $$  where $\, F(\omega) = F\,\{f(x)\}$  This is my procedure so far:   $$F\,\{f'(x)\} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f'(t)e^{i\omega t} dt$$  Integrating by parts I obtained:   $$=\frac{1}{\sqrt{2\pi}} \big[ \space f(t) e^{i\omega t} \space \big|_{-\infty}^{\infty} -  \int_{-\infty}^{\infty} i\omega f(t)e^{i\omega t} dt \space \big]$$  Now, in order to (1) to be true I need to get:  $$ f(t)e^{i\omega t} \space \big|_{-\infty}^{\infty} =0 \qquad (2)$$  I developed it and got the following:  $$ f(\infty)e^{i\omega \infty} - f(-\infty)e^{-i\omega \infty } = f(\infty)e^{i\omega \infty}  - 0= f(t)e^{i\omega \infty} $$  I undarstand that the second term is $0$, since f(t) must have a value to accomplish Dirichlet conditions (and $e^{-\infty} = 0$),  but I don't see how $ f(\infty)e^{i\omega \infty}$ is $0$.
121726
Consider a function $f(t)$ with Fourier Transform $F(s)$. So $$F(s) = \int_{-\infty}^{\infty} e^{-2 \pi i s t} f(t) \ dt$$  What is the Fourier Transform of $f'(t)$? Call it $G(s)$.So $$G(s) = \int_{-\infty}^{\infty} e^{-2 \pi i s t} f'(t) \ dt$$  Would we consider $\frac{d}{ds} F(s)$ and try and write $G(s)$ in terms of $F(s)$?
107101
Given $X$ is integrable. Prove that $E(X) = \int_{0}^{\infty} [1-F(x)]dx - \int_{-\infty}^{0}F(x)dx$.  My 2ND solution Since $X$ is integrable, $E(X) = E(X^{+}) - E(X^{-}) = \int_{0}^{\infty} xdF_{X^{+}}(x) - \int_{0}^{\infty} xdF_{X^{-}}(x) = \int_{0}^{\infty} [F_{X^{-}}(x) - F_{X^{+}}(x)]dx$.   Now, since $F_{X^{-}}(x) = P(-min(X,0)\leq x) = 1 - P(\min(X,0)&lt;\ -x) = 1 - P(X&lt;-x)$  (since for $x\in (0,\infty), P(0&lt;-x) = 0$)   $= 1 - F_{X}(-x)$, $E[X] = \int_{0}^{\infty} [1 - F_{X}(-x) - F_{X^{+}}(x)] dx = \int_{0}^{\infty} 1 - F_{X}(x) - \int_{0}^{\infty} F_{X}(-x)dx$ (the last equality is because for $x\in (0,\infty), F_{X^{+}}(X) = F_{X}(x)$).   Finally, by using the change of variable $u=-x$ for the 2nd integral, with $u$ as a dummy variable, we easily get the RHS that we need (Q.E.D)  My question: Could someone help review the solution above to see if it's correct this time? Would really appreciate any input.
11285
Assume that $f$ be continuous on $\mathbb{R}$, $f'(x)$ exists for all $x\neq 0$, and $\lim_{x\rightarrow 0} f'(x)=1$. We need to show $f'(0)$ exist and is equal to $1$.  $f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}$, $\lim_{x\rightarrow 0}f'(x)=1\Rightarrow\lim_{x\rightarrow 0}\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=1$...am I going in the right direction? Please help.
106552
I'd really appreciate your help with the following problem:      Let $f$ be a integrable  function in $[a-1,b+1]$. I need to prove that:   $$\lim_{h\to 0} \int_{a}^{b} |f(x+h)-f(x)|dx=0 .$$   Basically I need to show that $\forall  \varepsilon  \exists  \delta .|h|&lt; \delta \to \left|\int_{a}^{b}|f(x+h)-f(x)|dx\right| &lt; \varepsilon $.  In addition, I know that because of $f$ being integrable I can write that $\sum \limits_{i=0}^{n-1}(\sup f-\inf f) \Lambda x_{i} &lt; \varepsilon$ for a division of $[a,b]$.  I tried to define $g=|f(x+h)-f(x)|$ and work with that, but it didn't lead me to any smart conclusion. Also I tried to put the limit inside of the integral and to use the derivative, but I'm not sure when I can do that.  Thank you for the help!
67960
I proved as follows:  Let $f(x) = \tan x -  x - \frac{x^{3}}{3}.$ Then $f '(x) = \sec^{2}x - 1 - x^{2}$ and $f''(x) = 2\sec^{2}x\tan x-x &gt; \tan x - x,$ since $\\tan x &gt; x,$ $f''(x) &gt; 0$ so $f'(x)$ is increasing. Since $f '(0) = 0,$ $f '(x) &gt; 0$ for all $x \in (0, \frac{\pi}{2}).$ So $f(x)$ is increasing. since $f(0) = 0,$ $f(x) &gt; 0$ for all $x \in (0, \frac{\pi}{2}).$  But my proof seems complex. Is there any simpler way to prove this?
57909
I'm trying to show that  $$\delta\big(f(x)\big) = \sum_{i}\frac{\delta(x-a_{i})}{\left|{\frac{df}{dx}(a_{i})}\right|}$$  Where $a_{i}$ are the roots of the function $f(x)$. I've tried to proceed by using a dummy function $g(x)$ and carrying out:  $$\int_{-\infty}^{\infty}dx\,\delta\big(f(x)\big)g(x)$$  Then making the coordinate substitution $u$ = $f(x)$ and integrating over $u$. This seems to be on the right track, but I'm unsure where the absolute value comes in in the denominator, and also why it becomes a sum.  $$\int_{-\infty}^{\infty}\frac{du}{\frac{df}{dx}}\delta(u)g\big(f^{-1}(u)\big) = \frac{g\big(f^{-1}(0)\big)}{\frac{df}{dx}\big(f^{-1}(0)\big)}$$  Can any one shed some light? Wikipedia just states the formula and doesn't actually show where it comes from.
114331
Let $X : \Omega \to \mathbb{N} $ be a r.v. Then $$ \mathbb{E} \{ X \} = \sum_{n=0}^{\infty} P(X &gt; n) $$ Attempt: $$ \mathbb{E} \{ X \} = \sum_j j P(X = j ) = \sum_j j(1 - P(X &gt; j) - P(X&lt;j ) )$$ But here it gets complicated. Am I on the right track?
81925
Question: Let $f:[0,1]\to\mathbb{R}$ be a continuous function such that $f(0)=0$ and $$\int_0^1f(x)dx=0.$$ Prove that $\exists a, b\in(0,1)$ such that $$\int_0^{a} xf(x)dx=0\text{ and }\int_0^bxf(x)dx=\frac{b^2f(b)}{2}.$$  My approach: Let $g:[0,1]\to\mathbb{R}$ be such that $$g(x)=x\int_0^xf(t)dt-\int_0^xtf(t)dt, \forall x\in[0,1].$$ By the first fundamental theorem of calculus we can conclude that $g$ is differentiable on $[0,1]$ and $$g'(x)=\int_0^xf(t)dt, \forall x\in[0,1].$$ Also, observe that $g(0)=0$ and $g(1)=-\int_0^1tf(t)dt$. Thus, by applying MVT to the function $g$ on the interval $[0,1]$, we can conclude that, $\exists c\in(0,1)$ such that $$g'(c)=\int_0^cf(t)dt=-\int_0^1tf(t)dt.$$ Observe that clearly three cases are possible, i.e, either$$\int_0^cf(t)dt&lt;0\text{ or }\int_0^cf(t)dt=0\text{ or }\int_0^cf(t)dt&gt;0.$$ Now let $h:[0,1]\to\mathbb{R}$ be such that $$h(x)=\int_0^xtf(t)dt, \forall x\in[0,1].$$ Please note that the part highlighted below is wrong, but still I have included it to just demonstrate my thinking process, as it might be of some help to others trying out this problem.  Observe that if $\int_0^cf(t)dt&lt;0$, then, $h(1)&gt;0$. This also implies that $\exists$ an open interval $(d,e)\in[0,c]$, such that $f(t)&lt;0, \forall t\in(d,e)$. Now select any point $c_1\in(d,e)$. Applying MVT to the function $h$ on the interval $[0,c_1]$, we can conclude that $\exists c_2\in(0,c_1)$ such that $$h'(c_2).c_1=f(c_2).c_2.c_1=h(c_1)-h(0)=h(c_1)&lt;0.$$ Now $h(c_1)&lt;0$ and $h(1)&gt;0$. Thus, by IVT we can conclude that $\exists a\in(c_1,1)\subseteq(0,1)$, such that $$h(a)=\int_0^af(t)dt=0.$$ A similar reasoning for the case when $\int_0^cf(t)dt&gt;0,$ shows that $\exists a\in(0,1)$, such that $$h(a)=\int_0^af(t)dt=0.$$ Now finally if $\int_0^cf(t)dt=0$, then we will have $h(1)=0$. Now if $f$ is identically equal to $0$ on $[0,c]$, then clearly $tf(t)=0, \forall t\in[0,c]\implies h(x)=0, \forall x\in[0,c].$ Thus choosing any point $x\in(0,c]$ and setting it as $a$, we will have $h(a)=0$ and we will be done in that case. Now if $f$ acquires both positive and negative values on $[0,c]$, then we can conclude that $\exists c_1,c_2\in(0,c)$, such that $f(c_1)&gt;0$ and $f(c_2)&lt;0$. Also, let us assume WLOG that $c_2&gt;c_1$.  I have not been able to make any significant approach other than this. Can someone help me out with this problem? Please note that a solution using integration by parts might not be possible, since $f$ is not a differentiable function.
59338
Let $f\in L^{2} (\mathbb R)$ with $\lim_{t\to \pm \infty} f(t)=0.$  Put, $F_{n} (x)= \frac{1}{2\pi} \int_{-n}^{n}e^{itx} f(t) dt, \ (n=1,2,...).$ Fix $\alpha \in (0, \infty)$ and we define $H_{n}(x)$ as follows:  $$\frac{1}{2\pi}\int_{-n}^{n} e^{itx} (f(t+\alpha) -f(t-\alpha))dt = (e^{-i\alpha x}- e^{i\alpha x})F_{n}(x) + H_{n}(x).$$     My Question: How to show $H_{n}(x) \to 0$ as $n\to \infty$ in $L^{2}(\mathbb R)$ ?   (I guess some where we need to use ; but I am bit confused, how to use it.)   Edit: @, pointed out below, and in MO, for the same,  I have tried little in that direction:  $$H_{n}(x)= \frac{1}{2\pi}\int_{-n}^{n} e^{itx} (f(t+\alpha) -f(t-\alpha))dt -(e^{-i\alpha x}- e^{i\alpha x}) \frac{1}{2\pi} \int_{-n}^{n}e^{itx} f(t) dt$$ we may write, $H_{n}(x)= \frac{1}{2\pi} e^{-i\alpha x} \int_{-n+\alpha}^{n+\alpha}e^{itx} f(t) dt- \frac{1}{2\pi} e^{i\alpha x} \int_{-n-\alpha}^{n-\alpha} e^{itx} f(t) dt-(e^{-i\alpha x}- e^{i\alpha x}) \frac{1}{2\pi} \int_{-n}^{n}e^{itx} f(t) dt;$ Or, we may write,  $H_{n}(x)= \frac{1}{2\pi} e^{-i\alpha x} \int_{-n+\alpha}^{n+\alpha}e^{itx} f(t) dt- \frac{1}{2\pi} e^{i\alpha x} \int_{-n-\alpha}^{n-\alpha} e^{itx} f(t) dt- \frac{1}{2\pi}\int_{-n-\alpha}^{n+\alpha} e^{itx} f(t+\alpha) dt + \int_{-n+\alpha}^{n+\alpha} e^{itx} f(t-\alpha) dt= I_{1}-I_{2}-I_{3}+I_{4};$  But then,(using one of the above form), we need to show:  $\lim_{n\to \infty}\int_{\mathbb R} |H_{n}(x)|^{2} dx=0.$  Now If I interchange the limit and integral (this may not be possible, here, I am doing without rigours), and then I can take limit inside the square, and then after taking limit, each integrals, $I_{i}, (i=1,2,3,4)$, will be becomes over $\mathbb R$; then I think, it will cancels each other;      Is this o.k or am I missing something, still, I don't see, where we need to use asymptotic behaviours of $f$ (and what does it mean ?, here is $f$ is arbitrary member of $L^{2}$) ? I am wondering, is it possible to get solution without, Plancherel or parsevel somewhere ?
69480
If $f(x)$ is a continuous function on $(-\infty, +\infty)$ and $\int_{-\infty}^{+\infty} f(x) \, dx$ exists. How can I prove that $$\int_{-\infty}^{+\infty} f(x) \, dx = \int_{-\infty}^{+\infty} f\left( x - \frac{1}{x} \right) \, dx\text{ ?}$$
208
I'm trying to prove that for $p,q&gt;0$, we have $$\int_0^1t^{p-1}(1-t)^{q-1}\,dt=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.$$    The hint given suggests that we express $\Gamma(p)\Gamma(q)$ as a double integral, then do a change of variables, but I've been unable thus far to express it as a double integral.  Can anyone get me started or suggest an alternate approach?  Note: This wasn't actually given to me as the $\Gamma$ function, just as a function $f$ satisfying $$f(p)=\int_0^\infty e^{-t}t^{p-1}\,dt$$ for all $p&gt;0$, but I recognized that. This is in the context of an advanced calculus practice exam.
88227
I've been trying to prove to myself that if $\Omega$ is an open connected set in $\mathbb{R}^n$, then if $f\colon\Omega\to\mathbb{R}^m$ is a differentiable function such that $f'(x)=0$ for all $x\in\Omega$, then $f$ is constant.   I've reduced the problem to just showing $f$ is locally constant on $\Omega$. Given $x_0\in\Omega$, I know that  $$ \lim_{x\to x_0}\frac{\|f(x)-f(x_0)-f'(x_0)(x-x_0)\|}{\|x-x_0\|}=\lim_{x\to x_0}\frac{\|f(x)-f(x_0)\|}{\|x-x_0\|}=0. $$  This implies $\lim_{x\to x_0}\|f(x)-f(x_0)\|=0$. So for any $\epsilon&gt;0$, there exists some open ball $B(x_0,\delta)$ around $x_0$ in $\Omega$ such that $\|f(x)-f(x_0)\|&lt;\epsilon$ for $x\in B(x_0,\delta)$. But since the choice of the open ball changes with $\epsilon$, I don't think I can conclude $\|f(x)-f(x_0)\|=0$ for $x\in B$. So this doesn't convince me that there actually exists a neighborhood of $x_0$ on which $f$ is constant.   How can you make the leap from zero derivative to locally constant?
116747
If this is a duplicate in any way, I'm very sorry.  I'm brushing up on some Complex Analysis with Special Functions in mind. Here's a problem I'm stuck on.     Evaluate the integral $$I=\frac{1}{2\pi i}\oint_{\lvert z\rvert=1}f(z)\, dz$$ over the counterclockwise oriented unit circle centred at $0$, where $$f(z)=\frac{2z}{z^2+\omega^2}$$ for $\omega\in\Bbb{C}$.   I'm pretty sure I know what happens when $\lvert \omega\rvert\neq 1$; it's simple: break it down into the two obvious cases. But what happens when $\lvert\omega\rvert =1$?    What I've done so far is let $z=\gamma (t)=e^{it}$ for $t\in [0, 2\pi ]$ and arrived at $$\frac{1}{\pi}\int_0^{2\pi}{\frac{dt}{1+\left(\frac{\omega}{e^{it}}\right)^2}},$$ hoping to use the derivative of $\operatorname{arctan}$, which isn't going to work, of course.  I'd like a few hints please :)  A hunch: I'm tempted to say that the integral does not exist for $\lvert\omega\rvert=1$ but I've no idea why.
40411
So far I've done this, but I don't know if it will help.  Let $\int_0^xf(t)dt=F(x)$.  Now, $1=\int xf(x)=F(x)x|_0^1-\int_0^1F(x)dx=1-\int_0^1 F(x)dx$. Then, $\int_0^1 F(x)dx=0$.  But I don't know how to keep going.
72067
Let ($\Omega$, $\cal{F}$, $\mu$) be a probability space and $f\in L^1(\Omega)$. Prove that  $$\displaystyle\lim_{p\to 0} \left[ \int_{\Omega}|f|^pd\mu \right]^{\frac{1}{p}}=\exp \left[ \int_{\Omega}\log|f| d\mu  \right],$$  where $\exp[-\infty]=0$. To simplify the problem, we may assume $\log|f|\in L^1(\Omega).$
29849
Let $x \in \mathbb R$  Find a closed form of $\int_0^\infty\dfrac{\cos(xt)}{1+t^2}dt$ .   Let me give some context: this is an exercise from an improper integrals course for undergraduates. My teacher was not able to give a valid solution without  resorting to Fourier series.   He actually wanted to prove first that the function is a solution of the ODE $$ y-y'' =0$$ without succeeding in proving it.   Anyway the result sould be $\dfrac{\pi}{2}e^{-x}$  Thanks for any help on this subject.
21812
I ask for the proof of the L'Hôpital rule for the indeterminate form $\frac{\infty}{\infty}$ utilizing the rule for the form $\frac{0}{0}$.   Theorem: Let $f,g:(a,b)\to \mathbb{R}$ be two differentiable functions such as that:  $\forall x\in(a,b)\ \ g(x)\neq 0\text{ and }g^{\prime}(x)\neq 0$ and $\lim_{x\to a^+}f(x)=\lim_{x\to a^+}g(x)=+\infty$ If the limit $$\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}$$ exists and is finite, then  $$\lim_{x\to a^+}\frac{f(x)}{g(x)}=\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}$$  My attempt:  Since $\lim_{x\to a^+}f(x)=+\infty$, $$\exists \delta&gt;0:a&lt;x&lt;a+\delta&lt;b\Rightarrow f(x)&gt;0\Rightarrow f(x)\neq 0$$  Let $F,G:(a,a+\delta)\to \mathbb{R}$, $F(x)=\frac{1}{f(x)}$, $G(x)=\frac{1}{g(x)}$. Then by the hypothesis $\lim_{x\to a^+}F(x)=\lim_{x\to a^+}G(x)=0$, $$\forall x\in(a,b)\ \ G(x)\neq 0\text{ and }G^{\prime}(x)=-\frac{1}{g^2(x)}g^{\prime}(x)\neq 0$$ The question is, does the limit $$\lim_{x\to a^+}\frac{F^{\prime}(x)}{G^{\prime}(x)}=\lim_{x\to a^+}\frac{-\frac{1}{f^2(x)}f^{\prime}(x)}{-\frac{1}{g^2(x)}g^{\prime}(X)}=\lim_{x\to a^+}\frac{g^2(x)f^{\prime}(x)}{f^2(x)g^{\prime}(x)}$$ exist?   The limit $$\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}$$ exists by the hypothesis but we don't know if the limit $\displaystyle\lim_{x\to a^+}\frac{g^2(x)}{f^2(x)}$ exists to deduce that the limit $$\lim_{x\to a^+}\frac{F^{\prime}(x)}{G^{\prime}(x)}$$ exists to use the L'Hôpital Rule for the form $\frac{0}{0}$.  EDIT: After discussing it with other users in the site, we came to the conclusion that this proof is only partial and can't logically be continued to yield the Theorem.  As a result, the rule for the $\frac{0}{0}$ form can't be used to proove the rule for the  $\frac{\infty}{\infty}$ form. Mr. Tavares and myself have already given two different proofs (with the pretty much the same main idea) of the Theorem in question using Cauchy's Mean Value Theorem. You can read them below. You can also read the proof Rudin gives for a stronger version of the Theorem (that does not suppose that $\lim_{x\to a^+}f(x)=+\infty$) in his book Principle of Meathematical Analysis. If you have any objections in either proofs please let me know. Thank you.