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#### bw0young0math

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- Jun 14, 2013

- 27

- Thread starter bw0young0math
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- Jun 14, 2013

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- Feb 13, 2012

- 1,704

Let's consider the second integral. A primitive of the function is...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)

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

... 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

$\chi$ $\sigma$

- Jan 17, 2013

- 1,667

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

- Feb 13, 2012

- 1,704

Regarding to the first integral 'Monster Wolfram' supplies...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)

int e^(- x) sin x x^a from 0 to infinity - Wolfram|Alpha

How to obtain that result however is a difficult task...

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

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

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

$\chi$ $\sigma$

- Jan 17, 2013

- 1,667

I did not see the answer but I think it should beRegarding to the first integral 'Monster Wolfram' supplies...

int e^(- x) sin x x^a from 0 to infinity - Wolfram|Alpha

How to obtain that result however is a difficult task...

Kind regards

$\chi$ $\sigma$

\(\displaystyle \Im \left (\frac{\Gamma (\alpha +1)}{(1-i)^{\alpha +1}}\right)\)

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- Feb 29, 2012

- 342

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?

- Feb 13, 2012

- 1,704

An Italian adage says 'The Devil is not so ugly as You paint him' so that we try to compute the integral...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...

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

... 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

$\chi$ $\sigma$

- Jan 17, 2013

- 1,667

\(\displaystyle (1-i) = \sqrt{2} e^{-i \frac{\pi }{4}}\)I did not see the answer but I think it should be

\(\displaystyle \Im \left (\frac{\Gamma (\alpha +1)}{(1-i)^{\alpha +1}}\right)\)

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

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

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

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

- Thread starter
- #10

- 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??

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??

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- 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)$

$ \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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