# Difficult Improper Integrals in Real Analysis.

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#### chisigma

##### Well-known member
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
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
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)
Regarding 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

$\chi$ $\sigma$

#### chisigma

##### Well-known member
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
Re: Help me~♡ Difficult improper integrals in real analysis.

Regarding 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

$\chi$ $\sigma$
I did not see the answer but I think it should be

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

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#### Fantini

MHB Math Helper
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
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
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
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.
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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#### Random Variable

##### Well-known member
MHB Math Helper
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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