What is the best approach for solving the given exponential integral?

[Hip2dagame]

Homework Statement

∫(0 to 2) ∫(2x to 4) [e^(-y^2)] dydx

This was a question given for a multivariable class exam a couple years ago, and we don't ever learn about the exponential integral in that class.

Homework Equations

We know how to go about solving this, BUT...

The Attempt at a Solution

So everyone knows you can't integrate e^(-y^2) dy directly. So i changed the bounds of integration to get ∫(0 to 4) ∫(y/2 to 2) [e^(-y^2)] dxdy. Doing that, the first integral became

xe^(-y^2) from y/2 to 2, which gave me

∫(0 to 4) [2e^(-y^2) - ye^(-y^2)/2] dy

So now I'm stuck with the same problem with that first part of the integral. Now what do I do?! Thanks.

 

[SammyS]

Homework Statement

∫(0 to 2) ∫(2x to 4) [e^(-y^2)] dydx

This was a question given for a multivariable class exam a couple years ago, and we don't ever learn about the exponential integral in that class.

Homework Equations

We know how to go about solving this, BUT...

The Attempt at a Solution

So everyone knows you can't integrate e^(-y^2) dy directly. So i changed the bounds of integration to get ∫(0 to 4) ∫(y/2 to 2) [e^(-y^2)] dxdy. Doing that, the first integral became

xe^(-y^2) from y/2 to 2, which gave me

∫(0 to 4) [2e^(-y^2) - ye^(-y^2)/2] dy

So now I'm stuck with the same problem with that first part of the integral. Now what do I do?! Thanks.

Sketch the region of integration.

You integrated over a different region. In the initial integration, y goes from 2x to 4, so in that region, y > 2x. If y > 2x, then x < y/2 .

 

[Hip2dagame]

ok, done, fixed integral is now

∫(0 to 4) ∫(0 to y/2) [(-e^(-y^2))/4] dy. Same problem with the e^(-y^2) not being integratable.

 

[Ray Vickson]

ok, done, fixed integral is now

∫(0 to 4) ∫(0 to y/2) [(-e^(-y^2))/4] dy. Same problem with the e^(-y^2) not being integratable.

Your original integral was: first integrate y, then integrate x. You want to swap x and y, so you should first integrate x, then integrate y. If you sketch the integration region and take care you should get a simple final answer. If you don't take care you can certainly get into difficulties.

 

[lurflurf]

You do not need a lesson for each integral. The e^(-y^2) is integrable. You might have meant the primitive function of e^(-y^2) is not an elementary function, which is unimportant. What is needed is to integrate by parts or interchange the integrals.

by interchange the integrals
\int_0^2 \int_{2x}^4 e^{-y^2} \mathop{\text{dy dx}}=\int_0^4 \int_0^{y/2}e^{-y^2} \mathop{\text{dx dy}}

by integration by parts
\int_0^2 \int_{2x}^4 e^{-y^2} \mathop{\text{dy dx}}=-\int_0^2 x \dfrac{d}{dx}\int_{2x}^4 e^{-y^2} \mathop{\text{dy dx}}

 

[SammyS]

You do not need a lesson for each integral. The e^(-y^2) is integrable. You might have meant the primitive function of e^(-y^2) is not an elementary function, which is unimportant. What is needed is to integrate by parts or interchange the integrals.

by interchange the integrals
\int_0^2 \int_{2x}^4 e^{-y^2} \mathop{\text{dy dx}}=\int_0^4 \int_0^{y/2}e^{-y^2} \mathop{\text{dx dy}}

by integration by parts
\int_0^2 \int_{2x}^4 e^{-y^2} \mathop{\text{dy dx}}=-\int_0^2 x \dfrac{d}{dx}\int_{2x}^4 e^{-y^2} \mathop{\text{dy dx}}

@ lurflurf

The anti-derivative of \displaystyle e^{-y^2} is the error function. That's a special function. Therefore we often say that \displaystyle e^{-y^2} is not integrable. Integration by parts does work for this function.


@Hip2dagame:

Integrating over x, \displaystyle \ \int_0^{y/2}e^{-y^2}\,dx\,,\ treat the exponential, which is a function of y, as a constant. The result then allows integration by substitution, for the integration over y.

 

[lurflurf]

That is a most unusual definition of integrable function, can you provide a reference to a book that uses that definition? The usual definition of integrable is that the integral should converge, you are referring to primitive functions that are not expressible in terms of elementary function. In this problem it is quite possible to avoid using the error function as I have above, it is also possible to use the error function when computing an elementary integral.

\int -2x e^{-x^2} \mathop{\text{dx}}=-x \sqrt{\pi} \text{erf}(x)+\int \sqrt{\pi} \text{erf}(x) \mathop{\text{dx}}=\text{Constant}+e^{-x^2}

 

[SammyS]

That is a most unusual definition of integrable function, can you provide a reference to a book that uses that definition? The usual definition of integrable is that the integral should converge, you are referring to primitive functions that are not expressible in terms of elementary function. In this problem it is quite possible to avoid using the error function as I have above, it is also possible to use the error function when computing an elementary integral.

You are correct regarding the term "integrable". I should have said the indefinite integral of $\ \displaystyle e^{-y^2}\$ cannot be expressed in closed form as a ordinary function -- and then mentioned the error function, erf(y).

Regarding

\int -2x e^{-x^2} \mathop{\text{dx}}=-x \sqrt{\pi} \text{erf}(x)+\int \sqrt{\pi} \text{erf}(x) \mathop{\text{dx}}=\text{Constant}+e^{-x^2}

I appears that you are trying to use integration by parts to obtain the middle expression, the one involving the error function. That expression looks wrong to me. What steps are involved in obtaining that ?

 

[lurflurf]

\int f(x)g ^\prime (x) \mathop{\text{dx}}=f(x)g(x)-\int f ^\prime (x)g(x)\mathop{\text{dx}}
where in this case
f(x)=x
g(x)=sqrt(pi)erf(x)
and
\text{erf}(x)=\dfrac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} \mathop{\text{dx}}

\int -2x e^{-x^2} \mathop{\text{dx}}=-x \sqrt{\pi} \text{erf}(x)+\int \sqrt{\pi} \text{erf}(x) \mathop{\text{dx}}=\text{Constant}+e^{-x^2}
can be written
\int -x (\sqrt{\pi} \text{erf}(x))^\prime \mathop{\text{dx}}=-x \sqrt{\pi} \text{erf}(x)+\int x^\prime \sqrt{\pi} \text{erf}(x) \mathop{\text{dx}}=\text{Constant}+e^{-x^2}

 

[SammyS]

\int f(x)g ^\prime (x) \mathop{\text{dx}}=f(x)g(x)-\int f ^\prime (x)g(x)\mathop{\text{dx}}
where in this case
f(x)=x
g(x)=sqrt(pi)erf(x)
and
\text{erf}(x)=\dfrac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} \mathop{\text{dx}}

\int -2x e^{-x^2} \mathop{\text{dx}}=-x \sqrt{\pi} \text{erf}(x)+\int \sqrt{\pi} \text{erf}(x) \mathop{\text{dx}}=\text{Constant}+e^{-x^2}
can be written
\int -x (\sqrt{\pi} \text{erf}(x))^\prime \mathop{\text{dx}}=-x \sqrt{\pi} \text{erf}(x)+\int x^\prime \sqrt{\pi} \text{erf}(x) \mathop{\text{dx}}=\text{Constant}+e^{-x^2}

It's certainly true that \displaystyle \int -2x e^{-x^2}\,dx=\text{Constant}+e^{-x^2}.\ That result can arrived at through the substitution, u = -x² .

 

[Ray Vickson]

You do not need a lesson for each integral. The e^(-y^2) is integrable. You might have meant the primitive function of e^(-y^2) is not an elementary function, which is unimportant. What is needed is to integrate by parts or interchange the integrals.

by interchange the integrals
\int_0^2 \int_{2x}^4 e^{-y^2} \mathop{\text{dy dx}}=\int_0^4 \int_0^{y/2}e^{-y^2} \mathop{\text{dx dy}}

by integration by parts
\int_0^2 \int_{2x}^4 e^{-y^2} \mathop{\text{dy dx}}=-\int_0^2 x \dfrac{d}{dx}\int_{2x}^4 e^{-y^2} \mathop{\text{dy dx}}

There is no need to integrate by parts.
\int_{y=0}^4 \int_{x=0}^{y/2} e^{-y^2} \; dx \; dy<br /> = \int_{y=0}^4 \frac{y}{2} e^{-y^2} dy = -\frac{1}{4} \int_{y=0}^4 d(e^{-y^2}).