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8

A Laplace transform is easier than integrating by parts. Consider the function $$F_n(s) = \int_{0}^\infty t^{4n+3}e^{-st}dt = \frac{(4n+3)!}{s^{4n+4}}.$$ Then $I(n) = \mathrm{Im}[F_n(1-i)]$. Since $$F_n(1-i) = \frac{(4n+3)!}{(1-i)^{4n+4}} = \frac{(4n+3)!}{(-4)^{n+1}} \in \mathbb R,$$ $I(n) = 0$.

6

The function is Riemann integrable. Even though the maximum and minimum in one/two of the intervals in any partition are different,this difference vanishes when you let the lengths of the intervals in the partition tend to $0$.

4

First define for any integer $n$ $${{A}_{n}}=\int_{0}^{\infty }{{{x}^{n}}{{e}^{-x}}\sin \left( x \right)dx}\quad and\quad {{B}_{n}}=\int_{0}^{\infty }{{{x}^{n}}{{e}^{-x}}\cos \left( x \right)dx}$$ Now the substitution $u={{x}^{1/4}}$ reduces the integral in question to $4{{A}_{4n+3}}$. Using integration by parts any one can verify the recurrence ...

3

Yes, $f$ is bounded. That is part of the definition of Riemann-integrable function: if $f$ is Riemann-integrable on $[a,b]$ then $f|_{[a,b]}$ is bounded. And if you want to prove that $F$ is continuous at $x$, then you can assume without loss of generality that you are working on an interval $[a,b]$ with $b>x$. Unless $x$ the largest element of $I$. in ...

2

The change of variables $y=x^n$ removes the external factor of $n$, putting all the dependence inside the integral: $$n\int_{0}^{1}\frac{x^{n}}{x^{n}+x+1} \,dx = n \int_{0}^{1}\frac{y}{y+y^{1/n}+1} \tfrac1ny^{1/n-1} dy = \int_{0}^{1}\frac{y}{y^{2-1/n}+y+y^{1-1/n}} \,dy.$$ The integrand is an increasing function of $n$, so by the monotone convergence ...

2

This is false. If $f$ is any constant function then the hypothesis is satisfied. If the constant value of $f$ is not in the range of $g$ then we cannot have $f(c)=g(c)$. For a specific example take $f \equiv 1$ and $g \equiv 2$.

2

Your example doesn't work because $f$ is continuous. Take $f(x)=-1$ for $-1 \leq x \leq 0$ and $f(x)=1$ for $0 < x \leq 1$. If you compute $F$ you will see that the right hand derivative of $F$ at $0$ is $+1$ and the left hand derivative is $-1$.

2

$f$ is bounded on $[0,2]$ and continuous on $[0,2] \setminus \{1\}.$ Hence, $f$ is Riemann- integrable.

2

Using contour integration, with the contour being a 'pie slice', composed of lines $$z_1(t) = t,\qquad t\in[0,R]$$ $$z_2(t) = t(1-i),\qquad t\in[0,\frac{R}{\sqrt{2}}]$$ $$z_3(t) = Re^{it},\qquad t\in[-\frac\pi 4,0]$$ and consideringing $$\oint_C z^{4n+3} e^{-z} dz = 0$$ you can prove that $$(1-i)^{4n+4}\int_0^\infty u^{4n+3} e^{-u(1-i)} du = \int_0^\... 2 Assuming only that f is Riemann integrable and g is differentiable and monotone, this is straightforward to prove with the additional condition that g' is continuous on [c,d]. We assume without loss of generality that g is non-decreasing. Consider a partition P: c = x_0 < x_1 < \ldots < x_n = d and the Riemann sum$$\tag{1}S(P,fg')= ...

2

Let $$I_n = \int_0^\infty x^n \sin (\sqrt{x}) \exp (\sqrt{x}) \, dx, \qquad n \in \mathbb{N}.$$ After enforcing a substitution of $x \mapsto \sqrt{x}$ one has $$I_n = 4 \int_0^\infty x^{4n + 3} e^{-x} \sin x \, dx.$$ The following useful property for the Laplace transform will now be employed to evalaute the integral: $$\int_0^\infty f(x) g(x) \, ... 2 The following will also evaluate the integral for real values of n: With the substitution you mentioned, we have$$I\left( n \right)=4\int_{0}^{\infty }{{{u}^{4n+3}}\sin \left( u \right)\exp \left( -u \right)du} = 4 \Im \left\{ \int_{0}^{\infty }u^{4n+3}\exp \left( u(i-1) \right)du \right\}=4 \Im \{J(n)\}$$With s=4(n+1) and the series representation of ... 1 The finitenss of the improper Riemann integral \int_{-\infty}^{\infty} f(x)dx implies that given \epsilon >0 we can find \Delta such that \int_{-\infty}^{-\Delta} f(x)dx <\epsilon and \int_{\Delta}^{\infty} f(x)dx<\epsilon. For |h| <1 the function f(x+h)-f(x) vanishes for |x| >\Delta +1. Hence \int_{\mathbb R} |f(x+h)-f(x)|dx=\... 1 M_j is the maximum value of f on [x_{j-1},x_j] and m_j is the minimum value on the interval. To find these values we can find the points, if any, where the derivative is 0. M_j is then the maximum of the values of f at these points together with the values at x_{j-1} and x_j. In this example f'(x)=0 only when x=0. M_1, for example, is ... 1 The function g is bounded, say |g(t)|\leq M for all t\in [0,1/2]. This means that$$ |g_{2}(t)|= \left|\int_0^t g_1(s)\,ds\right|\leq \int_0^t|g_1(s)|ds\leq \int_0^t M\,ds= tM $$This gives us:$$ |g_3(t)| = \left|\int_0^tg_2(s)\,ds\right|\leq \int_0^t|g_2(s)|ds\leq\int_0^t sM\,ds = \frac12t^2M  Now generalize the pattern and finish with a proof, say ...

1

Your counterexample won't work because the derivative of $F$ is just $f$ on $(a,b)$. Take $f\colon[0,2]\to \Bbb R$ as the step function with $f(x)=0$ for $x<1$ and $f(x)=1$ for $1\le x\le 2$. Then $f$ is Riemann integrable, and $F$ is a piecewise linear function which is $0$ from $0$ to $1$, and then $F(x)=x-1$ for $1\le x\le 2$. In particular $F$ is not ...

1

Proceed by induction on $n$. For the inductive step, applying the inductive hypothesis to your result from integration by parts, we get $f_{n+1}^{(n)}(x) = F(x)$, so $f_{n+1}^{(n+1)}(x) = F'(x) = f(x)$.

1

We can show the following: Let $a,b\in\mathbb R,a<b$ and $f:[a,b]\rightarrow \mathbb R$ Riemann-integrable. Let $g:[a,b]\rightarrow \mathbb R$ such that $M=\{x\in[a,b];f(x)\neq g(x)\}<\infty$ is finite. Then $g$ is Riemann-integrable with $\int\limits_{a}^{b}\!f(x)\,\mathrm{d}x=\int\limits_{a}^{b}\!g(x)\,\mathrm{d}x$. (Notice that we do not need $f$ or ...

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