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Difficult Improper Integrals in Real Analysis.

bw0young0math

New member
Jun 14, 2013
27
Hello. I'm studying improper integrals in real analysis. However, two problems are very difficult to me. If you are OK, please help me.(heart)

1.2.
55.jpg

I have solutions about above problems.
However, I don't know how I approach and find the way for solving them.
 

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chisigma

Well-known member
Feb 13, 2012
1,704
Re: Help me~♡ Difficult improper integrals in real analysis.

Hello. I'm studying improper integrals in real analysis. However, two problems are very difficult to me. If you are OK, please help me.(heart)
Let's consider the second integral. A primitive of the function is...

$\displaystyle \int \frac{d x}{x\ \ln^{\alpha} x} = \frac{ \ln^{1 - \alpha} x} {1 - \alpha} + c\ (1)$

... but the fact that the integral is from 0 to infinity introduces several difficulties. In particular for $\alpha> 1$ the (1) has a singularity in x= 1 and that means that the integral diverges and for $\alpha< 1$ the (1) tends to infinity if x tends to infinity and also in this case the integral diverges...

Kind regards

$\chi$ $\sigma$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: Help me~♡ Difficult improper integrals in real analysis.

We have to consider \(\displaystyle \alpha>-1 \) to solve the first integral by Laplace transform .
 

chisigma

Well-known member
Feb 13, 2012
1,704

chisigma

Well-known member
Feb 13, 2012
1,704
Re: Help me~♡ Difficult improper integrals in real analysis.

We have to consider \(\displaystyle \alpha>-1 \) to solve the first integral by Laplace transform .
May be that the following 'forgotten formula' ...

http://mathhelpboards.com/analysis-50/laplace-transform-7057.html#post32097

$\displaystyle \mathcal {L} \{f_{1}(t)\ f_{2}(t)\} = \frac{1}{2\ \pi\ i}\ \int_{\gamma - i\ \infty}^{\gamma + i\ \infty} F_{1} (z)\ F_{2}(s - z)\ dz\ (1)$

... is useful in this case. Here is $\displaystyle f_{1} (t) = t^{\alpha} \implies F_{1} (s) = \frac{\Gamma (1 + \alpha)}{s^{1 + \alpha}}$ and $\displaystyle f_{2} (t) = \sin t \implies F_{2}(s)= \frac{1}{1 + s^{2}}$. In any case that is not a trivial task...

Kind regards

$\chi$ $\sigma$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Last edited:

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Re: Help me~♡ Difficult improper integrals in real analysis.

We may be considering finding the explicit values for the improper integrals, but is that really the question? Bw0young0math, are you interested in results for the integrals or just in the analysis of their convergence? :)
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: Help me~♡ Difficult improper integrals in real analysis.

May be that the following 'forgotten formula' ...

http://mathhelpboards.com/analysis-50/laplace-transform-7057.html#post32097

$\displaystyle \mathcal {L} \{f_{1}(t)\ f_{2}(t)\} = \frac{1}{2\ \pi\ i}\ \int_{\gamma - i\ \infty}^{\gamma + i\ \infty} F_{1} (z)\ F_{2}(s - z)\ dz\ (1)$

... is useful in this case. Here is $\displaystyle f_{1} (t) = t^{\alpha} \implies F_{1} (s) = \frac{\Gamma (1 + \alpha)}{s^{1 + \alpha}}$ and $\displaystyle f_{2} (t) = \sin t \implies F_{2}(s)= \frac{1}{1 + s^{2}}$. In any case that is not a trivial task...
An Italian adage says 'The Devil is not so ugly as You paint him' so that we try to compute the integral...

$\displaystyle I(s) = \int_{c - i\ \infty}^{c + i\ \infty} \frac{d z}{z^{1 + \alpha}\ \{1 + (s - z)^{2}\}}\ (1)$

... and we do that integrating along the path illustrated in the figure...



If T tends to infinity the integral along the 'big circle' tends to 0 so that integral is given by...

$\displaystyle I(s)= - \lim_{z \rightarrow s + i} \frac{1}{z^{1 + \alpha}\ (s - z - i)} - \lim_{z \rightarrow s - i} \frac{1}{z^{1 + \alpha}\ (s - z + i)}= \frac{1}{2\ i\ (s+i)^{1 + \alpha}} + \frac{1}{2\ i\ (s-i)^{1 + \alpha}}\ (2)$

... so that the final result should be...

$\displaystyle \int_{0}^{\infty} t^{\alpha}\ \sin t\ e^{-t}\ d t = \frac{\Gamma(1 + \alpha)}{2\ \pi\ i} I(1) = \frac{\Gamma (1 + \alpha)}{4\ \pi} \frac{(1 + i)^{1 + \alpha} + (1-i)^{1 + \alpha}}{2^{1 + \alpha}}\ (3)$

Since in pure calculus I am very poor, it should be useful that some 'young mind' verifies all that...

Kind regards

$\chi$ $\sigma$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: Help me~♡ Difficult improper integrals in real analysis.

I did not see the answer but I think it should be

\(\displaystyle \Im \left (\frac{\Gamma (\alpha +1)}{(1-i)^{\alpha +1}}\right)\)
\(\displaystyle (1-i) = \sqrt{2} e^{-i \frac{\pi }{4}}\)

\(\displaystyle \frac{1}{(1-i)^{\alpha +1}} = 2^{-\frac{1+\alpha }{2}} e^{i \frac{\pi (\alpha+1) }{4}}\)

\(\displaystyle \Im \left( \frac{1}{(1-i)^{\alpha +1}} \right) = 2^{-\frac{1+\alpha }{2}} \sin \left(\frac{ \pi (\alpha+1) }{4} \right)\)

\(\displaystyle I = \Gamma \left(\alpha +1 \right)\Im \left( \frac{1}{(1-i)^{\alpha +1}} \right) = 2^{-\frac{1+\alpha }{2}} \sin \left(\frac{ \pi (\alpha+1) }{4} \right) \Gamma (\alpha+1)\)

However , W|A says that this result is for all \(\displaystyle \alpha >-2\) , so we have to extend it for the singularity at \(\displaystyle \alpha =- 1\).
 

bw0young0math

New member
Jun 14, 2013
27
Hello.
At first, I want to tell you I'm so sorry because I wrote problems in detail.

The problem is judging when the improper integrals are convergent.
(i.e., the problem is finding the ranges for the improperintegrals: convergent.)

I wrote the problems' solutions I have in my blog.
(The file is so large that I uploaded in my blog. Please click here. Thank you.)(heart)
I have solutions someone solved. However, i don't know how I have to solve them.

p.s. ZaidAlyafey sovled by using Laplace but I don't know Laplace.
Are my problems so difficult??
 
Last edited:

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
That result could of course be generalized.


$ \displaystyle \int_{0}^{\infty} {e^{-ax}} \sin (bx) x^{s-1} \ dx \ (a,s >0)$

$ \displaystyle = -\text{Im} \int_{0}^{\infty} e^{-(a+ib)x} x^{s-1} \ dx $

$ \displaystyle = -\text{Im} \ \mathcal{L}\{x^{s-1} \}(a+ib) $

$ \displaystyle = -\text{Im} \ \frac{\Gamma(s)}{(a+ib)^{s}}$

$ \displaystyle = - \text{Im} \ \frac{\Gamma(s)}{\left( \sqrt{a^{2}+b^{2}} \right)^{s}} \ e^{-is \arctan \frac{b}{a}}$

$ \displaystyle = \frac{\Gamma(s)}{\left( \sqrt{a^{2}+b^{2}} \right)^{s}} \ \sin \left( s \arctan \frac{b}{a} \right)$


And if $-1 < s < 0$,

$ \displaystyle \int_{0}^{\infty} {e^{-ax}} \sin (bx) x^{s-1} \ dx = \frac{\pi \csc (\pi s) }{\Gamma(1-s) \left( \sqrt{a^{2}+b^{2}} \right)^{s}} \ \sin \left( s \arctan \frac{b}{a} \right)$
 
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