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A 'too difficult' integral...

chisigma

Well-known member
Feb 13, 2012
1,704
As You can see in…

Integrale di analisi complessa • Forum Scienze Matematiche

…an Italian student failed to solve the following integral…

$\displaystyle\int_{0}^{\infty} \frac{\sin 2 t}{1+t^{3}}\ dt$

After several 'attempts' [that pratically meansthe use of Mathematica,WolframAlpha or other computer tools...] the student has been answered that the integral is 'too difficult'. Effectively if You instruct WolframAlpha to solve the integral...

$\displaystyle \int_{0}^{\infty} \frac{\sin 2t}{1+t^{n}}\ dt$ (2)

... the solution is possible for n=1 and n=2 but not for n>2. Because I think that is not a good idea to replace the brain with a computer, let me try to find a solution to integral (1).


The starting point is the following Laplace Transform relation...

$\displaystyle \mathcal{L} \{\frac{1}{t+a}\}=e^{a s}\ \text{Ei}\ (a s)$ (3)

... where Ei(*) if the Exponential Integral Function. The (3) permits us to arrive to a first nice result...

$\displaystyle \int_{0}^{\infty} \frac{e^{i\ \omega\ t}}{t+a}\ dt = e^{-i\ a\ \omega}\ \text{Ei} (-i\ a\ \omega)$ (4)

... that, remembering the identity involving Sine Integral, Cosine Integral and Exponential Integral functions...

$\displaystyle \text{Ei}\ (i x)= - \text{Ci} (x) + i\ [\frac{\pi}{2} + \text{Si} (x) ] $ (5)

… is equivalent to write...


$\displaystyle \int_{0}^{\infty} \frac{e^{i \omega\ t}}{t+a}\ dt = - \cos (a \omega)\ \text{Ci}(a \omega) + \sin (a \omega)\ [\frac{\pi}{2} - \text{Si} (a \omega)] +i\ \{\cos (a \omega)\ [\frac{\pi}{2} - \text{Si} (a \omega)] + \sin (a \omega)\ \text{Ci} (a \omega) \}$ (6)

… or alternatively…


$\displaystyle\int_{0}^{\infty} \frac{\cos \omega\ t}{t+a}\ dt = - \cos (a \omega)\ \text{Ci}(a\omega) + \sin (a \omega)\ [\frac{\pi}{2} - \text{Si} (a \omega)]$ (7)

$\displaystyle \int_{0}^{\infty} \frac{\sin\omega\ t}{t+a}\ dt = \cos (a \omega)\ [\frac{\pi}{2} - \text{Si} (a \omega)] + \sin (a \omega)\ \text{Ci} (a \omega)$ (8)

In a successive post we will see how to use these result in computation of (1)...

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Using what obtained in the previous post, now we compute the integral...

$\displaystyle \int_{0}^{\infty}\frac{\sin\omega t}{1+t^{n}}\ dt$ (1)

... for several values of n [of course the integral including $\cos \omega t$is completely similar...]. It is well known the partial fraction expansion...


$\displaystyle\frac{1}{1+t^{n}}=\frac{1}{n}\sum_{k=0}^{n-1} \frac{w_{n,k}^{1-n}}{t -w_{n,k}}\ ,\ w_{n,k}=e^{i\ \frac{2k+1}{n}\ \pi}\ $(2)

... so that, using the resultof the previous post, we obtain the final result...


$\displaystyle \int_{0}^{\infty}\frac{\sin \omega t}{1+t^{n}}\ dt =\frac{1}{n}\ \sum_{k=0}^{n-1} w_{n.k}^{1-n}\ \{\cos (w_{n,k} \omega)\ [\frac{\pi}{2}+\text{Si} ( w_{n,k} \omega)] -\sin(w_{n,k} \omega)\ \text{Ci}(w_{n,k}\omega)\}$ (3)


Let's compute the (1)for n=1....

$\displaystyle n=1 \implies w_{1,0}= -1\implies \int_{0}^{\infty}\frac{\sin\omega t}{1+t}\ dt = \cos (\omega)\ [\frac{\pi}{2} - \text{Si}(\omega)] +\sin(\omega)\ \text{Ci} (\omega)$


… and the task is relatively comfortable. For n=2 is…

$\displaystyle n=2\implies w_{2,0}=i\ ,\ w_{2,1}=-i \implies \int_{0}^{\infty} \frac{\sin\omega t}{1+t^{2}}\ dt =$


$\displaystyle = -\frac{i}{2}\ \{ \cos (i\omega)\ [\frac{\pi}{2} + \text{Si}\ (i \omega)] + \sin (i \omega)\ \text{Ci}(i \omega) + \frac{i}{2} \{ \cos(i\omega)\ [\frac{\pi}{2} - \text{Si}\ (i \omega)] - \sin (i \omega)\ \text{Ci}(i \omega)\}=$

$\displaystyle = -i\ \cos (i \omega)\ \text{Si} (i\omega) - i\ \sin (i \omega)\ \text{Ci} (i \omega)$

… and the computational effort greatly increases. For n=3 or even more hand computationis hard and a computer approach tom the problem based on (3) is strongly recommended…

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
As You can see here…

AoPS Forum - Integral • Art of Problem Solving


… in the Aps site it has been proposed the‘difficult integral’ …


$\displaystyle I= \int_{0}^{\infty} \frac{\sin t}{1+t^{2}}\ dt$ (1)


First an user has considered the integral'impossible' because the standard complex path integration approach fails butimmediately after another 'very clever' user has found on 'Monster Wolfram' thesolution...


$\displaystyle I= \frac { \text{Ei}\ (1) -e^{2}\ \text{Ei}\ (-1)}{2\ e} \ $ (2)


In my previous post I’m arrived to the relation…


$\displaystyle \int_{0}^{\infty}\frac{\sin\omega t}{1+t^{2}}\ dt = -i\ \cos (i \omega)\ \text{Si} (i \omega) - i\ \sin (i\omega)\ \text{Ci} (i \omega)$ (3)


Now considering the relations...

$\displaystyle \cos (i x)= \cosh (x)$

$\displaystyle \sin (i x)= i\ \sinh (x)$

$\displaystyle \text{Ci} (i x)= \frac{ \text{Ei} (x) + \text{Ei} (-x)}{2}$


$\displaystyle\text{Si} (i x)=\frac{\text{Ei}(-x)-\text{Ei}(x)}{2 i}$ (4)


... the (3) becomes...


$\displaystyle \int_{0}^{\infty}\frac{\sin \omega t}{1+t^{2}}\ dt = \cosh \omega\ \frac{\text{Ei}(\omega) - \text{Ei}(- \omega)}{2}\ + \sinh \omega\ \frac{\text{Ei} (\omega) + \text{Ei}(- \omega)}{2} \ $ (5)

... according with WolframAlpha...

Kind regards

$\chi$ $\sigma$
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
In www.artofproblemsolving ...AoPS Forum - Definite integral for (sin x)/(1+x) • Art of Problem Solving

... the following definite integral has recently been proposed...

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{1+x}\ dx$ (1)

In the post #1 we found a general expression for an integral like (1) so that we now can write...

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{1+x}\ dx = \cos (1)\ \{\frac{\pi}{2} - \text{Si}\ (1)\ \} + \sin (1)\ \text{Ci} (1)$ (2)

Scope of this post however is not to show again this result, but to ask to the numerous 'experts' of MHB to solve one personal doubt. It is well known that the integral (1) converges if it is considered as a Riemann integral and diverges if it is considered as a Lebesgue integral. Well!... the [very obvious...] question is : if in a pratical situation I meet the (1) what I have to do? (Nerd)...

Kind regards

$\chi$ $\sigma$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
In www.artofproblemsolving ...AoPS Forum - Definite integral for (sin x)/(1+x) • Art of Problem Solving

... the following definite integral has recently been proposed...

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{1+x}\ dx$ (1)

In the post #1 we found a general expression for an integral like (1) so that we now can write...

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{1+x}\ dx = \cos (1)\ \{\frac{\pi}{2} - \text{Si}\ (1)\ \} + \sin (1)\ \text{Ci} (1)$ (2)

Scope of this post however is not to show again this result, but to ask to the numerous 'experts' of MHB to solve one personal doubt. It is well known that the integral (1) converges if it is considered as a Riemann integral and diverges if it is considered as a Lebesgue integral. Well!... the [very obvious...] question is : if in a pratical situation I meet the (1) what I have to do? (Nerd)...

Kind regards

$\chi$ $\sigma$
I think that this is analogous to the situation with infinite series, where a series can be conditionally convergent without being absolutely convergent. It is essential to the Lebesgue integral that it is an absolute integral. In other words, a function is only defined to be integrable if its absolute value is integrable. In the case of the function $\frac{\sin x}{1+x}$, its Lebesgue integral over $[0,\infty)$ would be defined as the integral of its positive part $\Bigl[\frac{\sin x}{1+x}\Bigl]_+$ minus the integral of its negative part $\Bigl[\frac{\sin x}{1+x}\Bigl]_-.$ Since neither of these integrals is finite, the integral is not defined. However, if you define $\displaystyle \int_{0}^{\infty} \frac{\sin x}{1+x}\, dx$ to mean $\displaystyle \lim_{X\to\infty}\int_{0}^X \frac{\sin x}{1+x}\, dx$, then that limit does exist and so the integral can be legitimately defined.
 

chisigma

Well-known member
Feb 13, 2012
1,704
I think that this is analogous to the situation with infinite series, where a series can be conditionally convergent without being absolutely convergent. It is essential to the Lebesgue integral that it is an absolute integral. In other words, a function is only defined to be integrable if its absolute value is integrable. In the case of the function $\frac{\sin x}{1+x}$, its Lebesgue integral over $[0,\infty)$ would be defined as the integral of its positive part $\Bigl[\frac{\sin x}{1+x}\Bigl]_+$ minus the integral of its negative part $\Bigl[\frac{\sin x}{1+x}\Bigl]_-.$ Since neither of these integrals is finite, the integral is not defined. However, if you define $\displaystyle \int_{0}^{\infty} \frac{\sin x}{1+x}\, dx$ to mean $\displaystyle \lim_{X\to\infty}\int_{0}^X \frac{\sin x}{1+x}\, dx$, then that limit does exist and so the integral can be legitimately defined.
Of course I expected an answer like that... in other words Riemann's and Lebesgue's definitions are simply two fully equivalent definitions of integral and we are 'free' to choose whichever we like... course that is 'all right'!... anyway all that reinforces my firm belief, allready expressed in the past, that in Math there are 'good definitions', which leads to valid solutions [in this case the Riemann's definition] and 'wrong definitions', which don't lead anywhere [in this case Lebesgue's definition]...

Kind regards

$\chi$ $\sigma$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Of course I expected an answer like that... in other words Riemann's and Lebesgue's definitions are simply two fully equivalent definitions of integral and we are 'free' to choose whichever we like... course that is 'all right'!... anyway all that reinforces my firm belief, allready expressed in the past, that in Math there are 'good definitions', which leads to valid solutions [in this case the Riemann's definition] and 'wrong definitions', which don't lead anywhere [in this case Lebesgue's definition]...

Kind regards

$\chi$ $\sigma$
Yes, there are good definitions and bad definitions, and they are subject to the process of natural selection. Good definitions survive, bad definitions get forgotten. The fact that Lebesgue integration has flourished for more than a century, and has supplanted the Riemann integral in many contexts, indicates that it is not such a bad definition.

Coming back to the integral $\displaystyle\int_0^\infty\frac{\sin x}{1+x}\,dx$, strictly speaking this does not exist either as a Lebesgue integral or as a Riemann integral. The Riemann integral is only defined over bounded intervals. In order to deal with cases such as this one, where the interval is unbounded, it is necessary to extend the theory to include what in English are known as "improper integrals" (maybe Italian has a less pejorative name for this construction?). When defining $\displaystyle\int_0^\infty\frac{\sin x}{1+x}\,dx$ as a limit of integrals over bounded intervals $[0,X]$, one can take the integrals over the bounded intervals either in the Riemann or in the Lebesgue sense. In that way, you can have "improper Lebesgue integrals" as well as "improper Riemann integrals".

The big advantages of Lebesgue integration are first, that it integrates a much wider class of functions than the Riemann integral, and second, that it has much neater and more easily applicable convergence theorems.