My Math Forum Integrals

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 April 28th, 2013, 07:59 AM #1 Newbie   Joined: Apr 2013 Posts: 11 Thanks: 0 Integrals Judge if those integrals converge: $a)\ \int_0^{+\infty}x^{17} e^{-\sqrt{x}}dx$ $b)\ \int_1^{+\infty}\frac{\cos x}{\sqrt[3]x}dx$ $c)\ \int_0^1\frac{sin x}{x^{1.5}}dx$ $d)\ \int_0^1 \frac{1}{\sqrt[3]{1-x^3}}$ Thank you.
 April 28th, 2013, 12:48 PM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Integrals [color=#000000]I think it is easy to prove that the last two converge, c. It is obvious that we have a problem at point 0, so knowing that close to zero $\sin(x)\approx x$ for a small ?, $\int_{0}^{\epsilon}\frac{\sin(x)}{x^{\frac{3}{2}}} \;\mathbb{d}x\approx \int_{0}^{\epsilon}\frac{x}{x^{\frac{3}{2}}}\;\mat hbb{d}x=\int_{0}^{\epsilon}\frac{1}{\sqrt{x}}\;\ma thbb{d}x<\infty=$, so the integral $\int_{0}^{1}\frac{\sin(x)}{x^{\frac{3}{2}}}\;\math bb{d}x$ converges. d. $\int_{0}^{1}\frac{1}{\sqrt[3]{1-x^3}}\;\mathbb{d}x\leq \int_{0}^{1}\frac{1}{\sqrt{1-x}}\;\mathbb{d}x=\int_{0}^{1}\frac{1}{\sqrt{x}}\;\ mathbb{d}x<\infty=$, so again this integral converges.[/color]
 April 28th, 2013, 01:44 PM #3 Global Moderator   Joined: May 2007 Posts: 6,641 Thanks: 625 Re: Integrals 1) converges. Transform by y^2 = x and get an integrand 2(y^35)exp(-y) 2) diverges. Concatonate pairs of intervals of length ? (starting at ?/2) and you have an integrand ~ 1/x^(2/3) as x -> ?.
April 28th, 2013, 02:03 PM   #4
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Re: Integrals

Quote:
 2) diverges. Concatenate pairs of intervals of length ? (starting at ?/2)
[color=#000000]Since $\int_{0}^{\infty}\frac{\cos(x)}{\sqrt[3]{x}}\;\mathbb{d}x=\frac{\Gamma\left(\frac{2}{3}\ri ght)}{2}$ I don't see why the same one starting from 1, should diverge. [/color]

 April 29th, 2013, 06:29 AM #5 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Integrals [color=#000000]It was easier than I expected. b. $\int_{1}^{\infty}\frac{\cos(x)}{\sqrt[3]{x}}\;\mathbb{d}x=\frac{\sin(x)}{\sqrt[3]{x}}{\Huge|}_{1}^{\infty}+\frac{1}{3}\underbrace{\ int_{1}^{\infty}\frac{\sin(x)}{x^{\frac{4}{3}}}\;\ mathbb{d}x}_{\rm I}=-\sin(1)+\frac{1}{3}\textrm{I}$ $|\textrm{I}|\leq \int_{1}^{\infty}\frac{|\sin(x)|}{x^{\frac{4}{3}}} \;\mathbb{d}x\leq \int_{1}^{\infty}\frac{1}{x^{\fbox{\frac{4}{3}}}}\ ;\mathbb{d}x<\infty$ look at the boxed number, this is the case $\int_{1}^{\infty}\frac{1}{x^{\fbox{p}}}\;\mathbb{d }x$ which converges for p>1. For the first integral you could say that it converges since the expotential in the denominator is always greater that the every power of x and $\lim_{x\to \infty}\frac{x^{17}}{\mathbb{e}^{\sqrt{x}}}=0$ (note that here we have a problem at infinity since that limit at 0 is 0). So you could say that for a very big number M, for all $x\geq M\Rightarrow \mathbb{e}^{\sqrt{x}}>x^{10^{6}}\Rightarrow \frac{1}{\mathbb{e}^{\sqrt{x}}}<\frac{1}{x^{10^6}} \Rightarrow \int_{M}^{\infty}\frac{x^17}{\mathbb{e}^{\sqrt{x}} }\;\mathbb{d}x<\int_{M}^{\infty}\frac{1}{x^{10^6-17}}\;\mathbb{d}x<\infty$ so all the given integrals a,b,c,d converge. [/color]
April 29th, 2013, 02:11 PM   #6
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Re: Integrals

Quote:
Originally Posted by ZardoZ
Quote:
 2) diverges. Concatenate pairs of intervals of length ? (starting at ?/2)
[color=#000000]Since $\int_{0}^{\infty}\frac{\cos(x)}{\sqrt[3]{x}}\;\mathbb{d}x=\frac{\Gamma\left(\frac{2}{3}\ri ght)}{2}$ I don't see why the same one starting from 1, should diverge. [/color]
The formula for ?(2/3) puzzles me.
The usual definition I am familiar with has an integrand of exp(-x)/³?x.

April 29th, 2013, 02:29 PM   #7
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Re: Integrals

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 Originally Posted by mathman The formula for ?(2/3) puzzles me. The usual definition I am familiar with has an integrand of exp(-x)/³?x.
$\int_{0}^{\infty}\frac{\cos(x)}{\sqrt[3]{x}}\;\mathbb{d}x=\mathfrak{Re}\left(\int_{0}^{\in fty}\frac{\mathbb{e}^{-\mathbb{i}x}}{\sqrt[3]{x}}\;\mathbb{d}x\right)=\mathfrak{Re}\left(\int_{ 0}^{\infty}x^{\frac{2}{3}-1}\mathbb{e}^{-\mathbb{i}x}\;\mathbb{d}x\right)=\mathfrak{Re}\lef t(\frac{\Gamma \left(\frac{2}{3}\right)}{2}-\frac{\mathbb{i} \pi }{\Gamma \left(\frac{1}{3}\right)}\right)=\frac{\Gamma\left (\frac{2}{3}\right)}{2}$

April 30th, 2013, 09:37 AM   #8
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Re: Integrals

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 Originally Posted by Z $\mathfrak{Re}\left(\int_{0}^{\infty}x^{\frac{2}{3}-1}\mathbb{e}^{-\mathbb{i}x}\;\mathbb{d}x\right)=\mathfrak{Re}\lef t(\frac{\Gamma \left(\frac{2}{3}\right)}{2}-\frac{\mathbb{i} \pi }{\Gamma \left(\frac{1}{3}\right)}\right)$
Hi Z , could you illustrate what you have done here?

May 4th, 2013, 01:05 PM   #9
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Re: Integrals

Quote:
Originally Posted by zaidalyafey
Quote:
 Originally Posted by Z $\mathfrak{Re}\left(\int_{0}^{\infty}x^{\frac{2}{3}-1}\mathbb{e}^{-\mathbb{i}x}\;\mathbb{d}x\right)=\mathfrak{Re}\lef t(\frac{\Gamma \left(\frac{2}{3}\right)}{2}-\frac{\mathbb{i} \pi }{\Gamma \left(\frac{1}{3}\right)}\right)$
Hi Z , could you illustrate what you have done here?

May 4th, 2013, 02:14 PM   #10
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Re: Integrals

Quote:
Originally Posted by zaidalyafey
Quote:
 Originally Posted by zaidalyafey What about a hint ?
[color=#000000]
See here. I didn't respond earlier because I needed time to do the computations. I used mathematica to get the result I posted earlier, so I didn't have an answer by that time.[/color]

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