Is the double integral convergent?

[Fizic]

Evaluate the integral ∫(2,∞) ∫(2/x,∞) 1/(y^2)*e^(-x/y) dydx by changing the order of integration.

I get ∫(1,∞) ∫(2y,∞) 1/(y^2)*e^(-x/y)dxdy

etc. etc. etc.

I get to ∫(1,∞) (e^(-2)/y) dy

Which is (ln∞-ln1)/e^2 = ∞

Does this thing not converge?

 

[SammyS]

Evaluate the integral ∫(2,∞) ∫(2/x,∞) 1/(y^2)*e^(-x/y) dydx by changing the order of integration.

I get ∫(1,∞) ∫(2y,∞) 1/(y^2)*e^(-x/y)dxdy

etc. etc. etc.

I get to ∫(1,∞) (e^(-2)/y) dy

Which is (ln∞-ln1)/e^2 = ∞

Does this thing not converge?

Sketch the region of integration.

You have errors in your integration limits for your integral, \displaystyle \int\int \frac{1}{y^2}e^{-x/y}\,dx\,dy\ .

It looks to me like you will need to write that as the sum of two integrals.

 

[Fizic]

Sketch the region of integration.

You have errors in your integration limits for your integral, \displaystyle \int\int \frac{1}{y^2}e^{-x/y}\,dx\,dy\ .

It looks to me like you will need to write that as the sum of two integrals.

You're right, it should be ∫(2,2y) instead of ∫(2y,∞)

I'm still getting ∞

 

[SammyS]

You're right, it should be ∫(2,2y) instead of ∫(2y,∞)

I'm still getting ∞

I also get ∞. However, the limits of integration are not what you have.

The original integral has y going from y = 2/x to y = ∞.

If y = 2/x , then x = 2/y not what you have, which is x = 2y.

Furthermore:

When y ≥ 1 , x goes from x = 2 to x = +∞ .

And when 0 ≤ y ≤ 1 , x goes from x = 2/y to x = +∞ .

 

[Fizic]

I also get ∞. However, the limits of integration are not what you have.

The original integral has y going from y = 2/x to y = ∞.

If y = 2/x , then x = 2/y not what you have, which is x = 2y.

Furthermore:

When y ≥ 1 , x goes from x = 2 to x = +∞ .

And when 0 ≤ y ≤ 1 , x goes from x = 2/y to x = +∞ .

My bad, it was supposed to be x/2, not 2/x.

I think it's infinity but I can't integrate e^(-2/y)/y.

 

[SammyS]

My bad, it was supposed to be x/2, not 2/x.

I think it's infinity but I can't integrate e^(-2/y)/y.

O.K.

Then you do get \displaystyle \int_{1}^{\infty}\,\int_{2}^{2y} \frac{1}{y^2}e^{-x/y}\,dx\,dy \ .

I also get that this does not converge.

Yes. That does not integrate to an elementary function.

 

[Michael Redei]

Surely
$$
\int_{2}^{\infty}\int_{x/2}^{\infty}\frac 1{y^2}e^{-x/y} \, \mathrm dy \, \mathrm dx =
\int_{1}^{\infty} \frac{e^{1/y}}{y^2} \int_{2y}^{\infty}e^{-x} \, \mathrm dx \, \mathrm dy
$$
is integrable. I get $e^{-2}$ as its value.

 

[SammyS]

Surely
$$
\int_{2}^{\infty}\int_{x/2}^{\infty}\frac 1{y^2}e^{-x/y} \, \mathrm dy \, \mathrm dx =
\int_{1}^{\infty} \frac{e^{1/y}}{y^2} \int_{2y}^{\infty}e^{-x} \, \mathrm dx \, \mathrm dy
$$
is integrable. I get $e^{-2}$ as its value.

Except that \ \ e^{-x/y}\ne e^{-x}e^{1/y}\ .

 

[Michael Redei]

I'd still try moving $1/y^2$ forwards:
$$
\int_2^\infty\int_{x/2}^\infty\frac1{y^2}e^{-x/y}\,\mathrm dy\,\mathrm dx = \int_2^\infty\frac1{y^2}\int_{2y}^\infty e^{(-1/y)\cdot x}\,\mathrm dx\,\mathrm dy.
$$