How Can Double Integration Solve This Tricky Infinite Integral?

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SUMMARY

The discussion focuses on solving the infinite integral ∫ 0 to ∞ [(e^-x)-(e^-3x)]/x dx using double integration techniques. Participants suggest introducing a second variable and reversing the boundaries to facilitate the integration process. The hints provided by the professor include integrating with respect to y first and manipulating the integral to achieve a form that cancels the denominator. Despite challenges, the consensus is that a solution is attainable through persistence and exploration of various mathematical strategies.

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Homework Statement



“Fun Homework Problem” assigned for extra credit...
∫ 0 to ∞ [(e^-x)-(e^-3x)]/x dx



2. Homework Equations (supplied hints)

I can get an abstract answer from wolfram but it’s not how the professor wants us to do it.
He gave us a couple “hints”; try to introduce a second variable and reverse the boundries
for example: ∫∫ f(y) f(x) dy dx? I’m assuming he means integrate with respect to y first.
A second possible hint was to try to get to ∫ 1/x dx by getting -[(e^-x)-(e^-3x)] on top to cancel the other.



The Attempt at a Solution



The biggest problem I’m running into is the x on the bottom.
Can I set new boundries by introducing 1 to y+1? I feel like the only way to do this problem is by getting +n in the denominator in order to prevent division by 0 when evaluating 0 to infinity.
I’ve also tried multipling by x/x in order to get an x^2 and move it to the top but the following integration by parts seems to get me no where.

Is there a trig sub I’m not seeing?

Even my professor admits he’s having trouble with it, but “knows” it’s possible.


Thanks for any help...
 
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Scrutinize the integral [itex]\int_{a}^{b}e^{-yx}dy[/itex]
 
KevinMWHM said:

The biggest problem I’m running into is the x on the bottom


Hi,

Yeah, that x on the bottom. How about this: I haven't worked this problem yet ok but that's not the point I wish to make. Rather, the essential quality in successful mathematics of trying things and cultivating a tolerance of not getting discouraged when they don't work but rather, motivating yourself into finding other things to try.

Ok, back to that x on the bottom. Well, that reminds me of another problem recently in here:

https://www.physicsforums.com/showthread.php?t=717360

Now, I'm not sure that's even relevant to solving this problem but again, that's not the point I'm trying to make. :)
 

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