Let $α∈(0,∞)$. Show that $Γ(α)≔\int_0^∞ x^{α-1}\mathrm e^{-x}\,dx$ exists as a Lebesgue integral, and that $Γ(α+1)=αΓ(α)$.
For $x∈[0,1]$, since $α-1>-1$, $x^{α-1}$ is integrable on $[0,1]⇒x^{α-1}\mathrm e^{-x}≤x^{α-1}$ is integrable on $[0,1]$.Since $\lim_{x→∞}\mathrm e^{-x}x^n=0$, it is bounded by some constant $C⇒\mathrm e^{-x}≤Cx^{-n}⇒x^{α-1}\mathrm e^{-x}≤Cx^{α-1-n}$.
For $n>α$, $x^{α-1-n}$ is integrable on $[1,∞)⇒x^{α-1}\mathrm e^{-x}≤x^{α-1-n}$ is integrable on $[1,∞)$. $$Γ(α+1)=\lim_{n→∞}\int_0^nx^αe^{-x}dx=-\lim_{n→∞}(n^αe^{-n}-αΓ(α))=αΓ(α)$$
We know that $\frac{\sin x}x$ is not integrable over $(1, ∞)$. Deduce, or prove otherwise, that neither of the following functions is integrable over the given intervals:
Apply Thm 5.8. (Substitution) with $f(x)=\frac{\sin x}x$ and $g(x)=x^2,\frac1{x^2}$
- $\frac{\sin\left(x^2\right)}x$ over $(1, ∞)$
- $\frac1x\sin\frac1{x^2}$ over $(0,1)$
- $\frac{\sin\left(x^2\right)}x=\frac12f(g(x))g'(x)$ is not integrable over $g^{-1}(1,∞)=(1,∞)$.
- $\frac1x\sin\frac1{x^2}=-\frac12f(g(x))g'(x)$ is not integrable over $g^{-1}(1,∞)=(0,1)$.
By comparing terms in binomial expansions, or otherwise, show that
\[
\left(1+\frac{x}{n}\right)^n≤\left(1+\frac{x}{n+1}\right)^{n+1},
\]
for $n≥2, x≥0$. Use the MCT to deduce that
\[
\lim_{n→∞}\int_1^2\left(1+\frac{x}{n}\right)^{-n} \,dx=\mathrm e^{-1}-\mathrm e^{-2}
\]
For $0≤x<n$,\begin{align*}\left(1+\frac{x}{n}\right)^n&=
1+x+\left(1-\frac1n\right)\frac{x^2}{2!}+\left(1-\frac1n\right)\left(1-\frac2n\right)\frac{x^3}{3!}+⋯\\&≤1+x+\left(1-\frac1{n+1}\right)\frac{x^2}{2!}+\left(1-\frac1{n+1}\right)\left(1-\frac2{n+1}\right)\frac{x^3}{3!}+⋯=\left(1+\frac{x}{n+1}\right)^{n+1}\end{align*}$∴\left(1+\frac{x}{n}\right)^{-n}≥\left(1+\frac{x}{n+1}\right)^{-(n+1)}$ and $\lim_{n→∞}\left(1+\frac{x}{n}\right)^{-n}=\mathrm e^{-x}$. By MCT and FTC\[\lim_{n→∞}\int_1^2\left(1+\frac{x}{n}\right)^{-n}\,dx=\int_1^2\mathrm e^{-x}\,dx=-\mathrm e^{-x}|_1^2=\mathrm e^{-1}-\mathrm e^{-2}\]
Use the binomial expansion of $(1-x)^{-k}$ to show that $\left(1-\frac{u}{n}\right)^{-n}≥\left(1-\frac{u}{n+1}\right)^{-(n+1)}$ for $n≥2$ and $0≤u<n$. Hence, or otherwise, prove that
\[
\lim_{n→∞}n^α \int_0^1 x^{α-1}\mathrm e^{-n β x}(1-x)^n\,dx=(β+1)^{-α}Γ(α)
\]
where $α>0, β>-1$, and $Γ(α)=\int_0^∞\mathrm e^{-u} u^{α-1}\,du$.
\begin{align*}\left(1-\frac{u}{n}\right)^{-n}&=1+u+\left(1+\frac1n\right)\frac{u^2}{2!}+\left(1+\frac1n\right)\left(1+\frac2n\right)\frac{u^3}{3!}+⋯\\&≥1+u+\left(1+\frac1{n+1}\right)\frac{u^2}{2!}+\left(1+\frac1{n+1}\right)\left(1+\frac2{n+1}\right)\frac{u^3}{3!}+⋯=\left(1-\frac{u}{n+1}\right)^{-(n+1)}\end{align*}
$∴\left(1-\frac{u}{n}\right)^n≤\left(1-\frac{u}{n+1}\right)^{n+1}$ and $\lim_{n→∞}\left(1-\frac{u}{n}\right)^n=\mathrm e^{-u}$
\begin{align*}
\lim_{n→∞}n^α \int_0^1 x^{α-1}\mathrm e^{-nβx}(1-x)^n\,dx&=\lim_{n→∞}\int_0^1 u^{α-1}\mathrm e^{-βu}\left(1-\frac un\right)^n\,du&\text{Let }u=nx\\
&=\int_0^1 u^{α-1}\mathrm e^{-(β+1)u}\,du&\text{By MCT}\\
&={1\overβ+1}\int_0^1 \left(t\overβ+1\right)^{α-1}\mathrm e^{-t}\,dt&\text{Let }t=(β+1)u\\
&=(β+1)^{-α}Γ(α)&\text{By definition of Γ(α)}
\end{align*}
Show that $0≤\frac{x}{1+x^α}≤1$ for $α>1, x≥0$, and deduce that
\[
\lim_{n→∞}\int_0^{2π} \frac{n x \sin x}{1+n^α x^α}\,dx=0
\]
Clearly $\frac{x}{1+x^α}≥0$. For $\frac{x}{1+x^α}≤1$, multiply by $1+x^α>0$, we get $x-1≤x^α$.If $x≤1$, we have $x-1≤0≤x^α$; if $x>1$, since $α>1$, we have $x-1<x<x^α$.
Replace $x$ with $nx$, we get $0≤\frac{nx}{1+n^αx^α}≤1⇒\left|\frac{nx\sin x}{1+n^αx^α}\right|≤1$. By Bounded Convergence Theorem, \[ \lim_{n→∞}\int_0^{2π}\frac{nx\sin x}{1+n^α x^α}\,dx=\int_0^{2π}\lim_{n→∞}\frac{nx\sin x}{1+n^α x^α}\,dx=\int_0^{2π}0=0 \]
Prove that
\[
\lim_{n→∞}\int_0^{n^2}n\left(\sin\frac xn\right)\mathrm e^{-x^2}\,dx=\frac12
\]
Rewrite LHS using characteristic functions
\[
I=\lim_{n→∞}\int_0^∞χ_{(0,n^2)}n\left(\sin\frac xn\right)\mathrm e^{-x^2}\,dx
\]
Since $\left|\sin\frac xn\right|≤\frac xn\,∀x≥0$
\[0≤n\left|\sin\frac xn\right|\mathrm e^{-x^2}≤x\mathrm e^{-x^2}\]
By DCT and $\lim_{n→∞}n\left(\sin\frac xn\right)=x$
\[
I=\int_0^∞x\mathrm e^{-x^2}\,dx=-\left.\frac{\mathrm e^{-x^2}}2\right|_0^∞=\frac12
\]
Let $α>-1$. Show that $x^α\log x$ is integrable over $(0,1)$, and
\[
\int_0^1 x^α \log x\,dx=-(1+α)^{-2}
\]
Deduce that for $β>-1$, $x^β(1-x)^{-1}\log x$ is integrable over $(0,1)$, and
\[
\int_0^1 x^β(1-x)^{-1}\log x\,dx=-\sum_{n=1}^∞(n+β)^{-2}
\]
For $0<ε<1$, integrating by parts
\begin{align*}
\int_ε^1x^α\log x\,dx&=\left.\frac{x^{α+1}}{α+1}\log x\right|^1_ε-\frac1{α+1}\int^1_εx^α\,dx
\\ &=-\frac{ε^{α+1}}{α+1}\logε-\left[\frac1{(α+1)^2}-\frac{ε^{α+1}}{(α+1)^2}\right]
\\&=-\frac{ε^{α+1}}{α+1}\logε+\frac{ε^{α+1}}{(α+1)^2}-\frac1{(α+1)^2}
\end{align*}
The intervals $[ε,1]$ increases to $[0,1]$. Apply Corollary 4.3.[Baby MCT] to $-x^α\log x>0$ we have$$\int_0^1 x^α\log x\,dx=\lim_{ε→0}\left[-\frac{ε^{α+1}}{α+1}\logε+\frac{ε^{α+1}}{(α+1)^2}-\frac1{(α+1)^2}\right]=-\frac1{(α+1)^2}$$Applying MCT for series to $-x^β(1-x)^{-1}\log x=\sum_{n=1}^∞-x^{n+β-1}\log x$\[ -\int_0^1 x^β(1-x)^{-1}\log x\,dx=-\sum_{n=1}^∞\int_0^1x^{n+β-1}\log x\,dx=\sum_{n=1}^∞(n+β)^{-2}<∞ \]So $x^β(1-x)^{-1}\log x$ is integrable over $(0,1)$.
Show that for $n>0$, $\mathrm e^{-nx}\sin x$ is integrable over $[0, ∞)$, and
\[
\int_0^∞\mathrm e^{-nx}\sin x\,dx=\frac1{1+n^2}
\]
Deduce that for $0≤a≤1,\left(\mathrm e^x-a\right)^{-1}\sin x$ is integrable over $[0, ∞)$, and
\[
\int_0^∞\left(\mathrm e^x-a\right)^{-1}\sin x\,dx=\sum_{n=1}^∞\frac{a^{n-1}}{1+n^2}
\]
$\left|\mathrm e^{-nx}\sin x\right|≤\mathrm e^{-nx}$ and $\int_0^∞\mathrm e^{-nx}\,dx=1/n$, by comparison, $\mathrm e^{-nx}\sin x$ is integrable on $[0,∞]$. Integrating by parts,
\begin{align*}
I=\int_0^∞\mathrm e^{-nx}\sin x\,dx&=-\frac1n\mathrm e^{-nx}\sin x|_0^∞+\frac1n\int_0^∞\mathrm e^{-nx}\cos x\,dx\\
&=\frac1n\int_0^∞\mathrm e^{-nx}\cos x\,dx\\
&=-\frac1{n^2}\mathrm e^{-nx}\cos x|_0^∞-\frac1{n^2}\int_0^∞\mathrm e^{-nx}\sin x\,dx\\&=\frac1{n^2}-\frac1{n^2}I\\⇒I&=\frac1{1+n^2}
\end{align*}$0≤a≤1,x>0⇒0≤a\mathrm e^{-x}<1⇒\left(\mathrm e^x-a\right)^{-1}=\left(1-a\mathrm e^{-x}\right)^{-1}\mathrm e^{-x}=\sum_{n=1}^∞a^{n-1}\mathrm e^{-nx}$By ratio test, $\sum_{n=1}^∞\frac{a^{n-1}}{1+n^2}$ converges. By Lebesgue’s Series Theorem \[ \int_0^∞\left(\mathrm e^x-a\right)^{-1}\sin x\,dx=\sum_{n=1}^∞a^{n-1}\int_0^∞\mathrm e^{-nx}\sin x\,dx=\sum_{n=1}^∞\frac{a^{n-1}}{1+n^2} \]
Let $a∈(0,1)$. Prove that $\frac{\mathrm e^{-ax}-\mathrm e^{(a-1)x}}{1-\mathrm e^{-x}}$ is an integrable function on ℝ, and that
\[
\frac12\int_{-∞}^∞\frac{\mathrm e^{-ax}-\mathrm e^{(a-1)x}}{1-\mathrm e^{-x}}\,dx=\frac1a+\frac1{a-1}+\frac1{a+1}+\frac1{a-2}+\frac1{a+2}+\frac1{a-3}+…
\]
$f(-x)=\frac{\mathrm e^{-ax}-\mathrm e^{(a-1)x}}{1-\mathrm e^{-x}}$ Dividing by $-\mathrm e^x$ we get $f(-x)=f(x)$. So we only need to consider $[0,∞)$.Since $\mathrm e^{-kx}>0$, for all $k$, $(\mathrm e^{-ax}-\mathrm e^{(a-1)x})\mathrm e^{-kx}$ have same sign. By MCT for series\begin{align*}\int_0^∞\frac{\mathrm e^{-ax}-\mathrm e^{(a-1)x}}{1-\mathrm e^{-x}}dx&=\int_0^∞\sum_{k=0}^∞(\mathrm e^{-ax}-\mathrm e^{(a-1)x})\mathrm e^{-kx}dx\\&=\sum_{k=0}^∞\int_0^∞(\mathrm e^{-ax}-\mathrm e^{(a-1)x})\mathrm e^{-kx}dx\\&=\sum_{k=0}^∞\frac1{a+k}+\frac1{a-k-1}\end{align*} $\left|\frac1{a+k}+\frac1{a-k-1}\right|=\frac{|2a-1|}{(a+k)(1-a+k)}≤\frac{|2a-1|}{k^2}⇒$RHS converges$⇒\frac{\mathrm e^{-ax}-\mathrm e^{(a-1)x}}{1-\mathrm e^{-x}}$ is integrable on $[0,∞)$.
Prove that $\int_0^∞ \cos x \arctan(λx)\mathrm e^{-x}\,dx→\fracπ4$ as $λ→∞$.
For $x>0$, the non-negative sequence $\left(\arctan(λx)\right)_{λ≥1}$ increases to $\fracπ2$, by MCT and $ℒ\{\cos x\}(s)=\frac s{s^2+1}$\[\lim_{λ→∞}\int_0^∞\cos x\arctan(λx)\mathrm e^{-x}\,dx=\fracπ2\int_0^∞(\cos x)\mathrm e^{-x}\,dx=\fracπ2ℒ\{\cos x\}(1)=\fracπ4\]
Let $J_0(x)=\frac2π\int_0^{π/2}\cos(x\cosθ)\,dθ$.
- Show that $J_0$ is differentiable on ℝ.
- Show that the function Γ, as defined in Q1, is differentiable on $(0,∞)$.
- (1) Since $\cos(x\cosθ)$ is bounded, it is integrable on $(0,\fracπ2)$ for all $x∈ℝ$.
(2) $\frac∂{∂x}\cos(x\cosθ)=-\cosθ\sin(x\cosθ)$ exists for all $x∈ℝ,θ∈(0,\fracπ2)$.
(3) $\left|\frac∂{∂x}\cos(x\cosθ)\right|≤\cosθ$ for all $x∈ℝ,θ∈(0,\fracπ2)$ and $\int_0^{π/2}\cosθ\,dθ=1$.
By Thm 7.5 $J_0$ is differentiable on ℝ. - $Γ(x)≔\int_0^∞ t^{x-1}\mathrm e^{-t}\,dt$ exists for all $x∈(0,∞)$ as shown in Q1.
$\frac∂{∂x}t^{x-1}\mathrm e^{-t}=t^{x-1}\mathrm e^{-t}\log t$ exists for all $t,x∈(0,∞)$. Consider $(0,1)$ and $(1,∞)$ separately:
$\left|t^{x-1}\mathrm e^{-t}\log t\right|≤t^{x-1}\log t$ and $t^{x-1}\log t$ is integrable over $(0,1)$ by Q7, so $t^{x-1}\mathrm e^{-t}\log t$ is integrable over $(0,1)$.
Since $\lim_{t→∞}\mathrm e^{-t/2}\log t=0$, it is bounded by some constant $C$.\begin{align*}∫_1^∞(t^{x-1}\mathrm e^{-t/2})\mathrm e^{-t/2}\log t\,dt&<C∫_1^∞t^{x-1}\mathrm e^{-t/2}\,dt\\&=C2^x∫_1^∞u^{x-1}\mathrm e^{-u}\,du \text{substitute }u=t/2\\&<C2^xΓ(x)\end{align*}By Thm 7.5 Γ is differentiable on $(0,∞)$.
Let $f(x,y)=y^3\mathrm e^{-y^2x},F(y)=\int_0^∞f(x,y)\,dx$. Calculate $F'(0)$ and $\int_0^∞\frac{∂f}{∂y}(x,0)\,dx$.
How do your answers relate to the theorem about differentiating through integrals?
\begin{gather*}F(y)=y^3\int_0^∞\mathrm e^{-y^2x}\,dx=\left.y^3⋅\frac{\mathrm e^{-y^2x}}{-y^2}\right|_{x=0}^∞=y^3⋅\frac1{y^2}=y⇒F'(0)=1\\\frac{∂f}{∂y}(x,0)=\lim_{y→0}\frac{f(x,y)}y=\lim_{y→0}y^2\mathrm e^{-y^2x}=0⇒\int_0^∞\frac{∂f}{∂y}(x,0)\,dx=0\end{gather*}So $F'(0)>\int_0^∞\frac{∂f}{∂y}(x,0)\,dx$. This agrees with the corollary Fatou's lemma: $F'(0)≥\int_0^∞\frac{∂f}{∂y}(x,0)\,dx$.How do your answers relate to the theorem about differentiating through integrals?
We will show $f$ has no bounding function $g$ that satisfies the conditions of DCT:
Suppose $∃g:g(x)≥\left|\frac{∂f}{∂y}(x,y)\right|∀{|y|}<δ$, let $y=x^{-1/2}$ then ${|y|}<δ⇔\left|x^{-1/2}\right|<δ⇔{|x|}>δ^{-2}$ \[g(x)≥\left|\frac{∂f}{∂y}(x,x^{-1/2})\right|=e^{-1}x^{-1} ∀{|x|}>δ^{-2}\]so $g(x)$ is not integrable on $[0,∞)$.
By differentiating through the integral sign, evaluate the following integrals:
- $\int_0^∞\frac{\mathrm e^{-x} \sin t x}x\,dx$
- $\int_0^{π/2}\log\left(a^2\cos^2x+b^2\sin^2x\right)\,dx$, where $a,b>0$.
- $\frac{d}{dt}\int_0^∞\frac{\mathrm e^{-x} \sin t x}x\,dx=\int_0^∞\mathrm e^{-x} \cos t x\,dx=ℒ\{\cos tx\}(1)=\frac1{t^2+1}$
The integral is 0 when $t=0$. Therefore $\int_0^∞\frac{\mathrm e^{-x} \sin t x}x\,dx=\int_0^t\frac1{τ^2+1}\,dτ=\arctan t$ - Let $I(a,b)=…$ Then\begin{align*}
\frac{∂I}{∂a}&=\int_0^{π/2}\frac{2a\cos^2x}{a^2\cos^2x+b^2\sin^2x}\mathrm dx,\\
\frac{∂I}{∂b}&=\int_0^{π/2}\frac{2b\sin^2x}{a^2\cos^2x+b^2\sin^2x}\mathrm dx,
\end{align*}By Complex Analysis sheet 5 Q10,\[\int_0^{2π}\frac{dx}{a^2\cos^2x+b^2\sin^2x}=\frac{2π}{ab}\]By substitution $x↦π+x$ we find the integral on $[π,2π]$ and $[0,π]$ are equal, so\[\int_0^π\frac{dx}{a^2\cos^2x+b^2\sin^2x}=\fracπ{ab}\]By substitution $x↦π-x$ we find the integral on $[\fracπ2,π]$ and $[0,\fracπ2]$ are equal, so\[\int_0^{π/2}\frac{dx}{a^2\cos^2x+b^2\sin^2x}=\fracπ{2ab}\]So we get $b\frac{∂I}{∂a}+a\frac{∂I}{∂b}=π$. Also it is easy to see $a\frac{∂I}{∂a}+b\frac{∂I}{∂b}=π$.
For $a≠b$, this linear system has unique solution $\frac{∂I}{∂a}=\frac{∂I}{∂b}=\fracπ{a+b}⇒I=π\log(a+b)+C$.
For $a=b$, $I(a,b)=π\log a$, in particular $I(1,1)=0$. By continuity $I=π\log\left(a+b\over2\right)$.
Let $K(t)=\int_1^∞\frac{\cos(tx)}{x^2}dx$. Show carefully that $K'(t)=\frac{K(t)-\cos t}t$ for $t>0$.
A formal differentiation under the integral suggests that
\begin{align*}
-\int_1^∞{\sin(tx)\over x}dx&={1\over t}\left(\left.\cos(tx)\over x\right|_1^∞+\int_1^∞{\cos(tx)\over x^2}dx\right)\\
&=\frac{K(t)-\cos t}t
\end{align*}But this integral doesn't exist. Taking a step back, rewrite $K$ using Integration by parts:
\begin{equation}\label1
K(t) = -{\sin t\over t}+{2\over t}\int_1^∞ {\sin(tx)\over x^3}dx
\end{equation}
Since $\left|\sin(tx)\over x^3\right|≤\frac1{x^3}$, the integral $G(t)≔\int_1^∞ {\sin(tx)\over x^3}dx
$ exists for all $t$.$\sin(tx)\over x^3$ is differentiable wrt $t$ for all $x$. Since $\left|\cos(tx)\over x^2\right|≤\frac1{x^3}$ and $1\over x^3$ is integrable over $[1,∞)$, by Thm 7.5 $$ G'(t)=\int_1^∞{\cos(tx)\over x^2}dx=K(t)$$ Differentiating \eqref{1} we have \begin{align*} K'(t)&= {\sin t-t\cos t\over t^2} + {2\over t}K(t) -{2\over t^2}G(t)\cr &={\sin t-t\cos t\over t^2}+ {2\over t}K(t) -{2\over t^2}\left({t\over 2}K(t)+{\sin t\over 2} \right)\cr &= -{\cos t\over t}+{1\over t}K(t). \end{align*}
Alternatively, substitute $x↦x/t$ $$K(t)=tf(t) f(t)=\int_t^∞\frac{\cos x}{x^2}dx$$ By FTC $f'(t)=-\frac{\cos t}{t^2}⇒K'(t)=f(t)+tf'(t)=\frac{K(t)-\cos t}t$