Solve Exponential Integral with Polynomial | Math Homework Help

[abotaha]

Dear all,
I have exponential integral with polynomial. I tried to solve it but I could not.
the integral is :

Homework Statement

The integral equation is:
\int_{-\infty}^{\infty} (ax^2+bx+c) e^{ax^2+bx+c} dx$

Homework Equations

The Attempt at a Solution

Can anyone help me please.

Thanks in advance.

 

[Dickfore]

Hint: Complete the square and make a substitution. Then use the following results:

<br /> \int_{-\infty}^{\infty}{e^{-x^{2}} \, dx} = \sqrt{\pi}<br />

<br /> \int_{-\infty}^{\infty}{x^{2} \, e^{-x^{2}} \, dx} = \frac{\sqrt{\pi}}{2}<br />

 

[abotaha]

Thanks for the replay.
unfortunately, I had mistake in writing with the previous integral.
the correct one is:


\int_{-\infty}^{\infty} (-ax^2+bx+c) e^{-dx^2+bx+f} dx

and i complete the square as you mentioned where the integral equation becomes:

\int_{-\infty}^{\infty} -(\sqrt{a}x-\frac{b}{2\sqrt{a}})^2+(c+\frac{b^2}{4a}) \times e^{-(\sqrt{d}x-\frac{b}{2\sqrt{d}})^2+(f+\frac{b^2}{4d})} dx

However, the integration is still not easy for me to solve it.
could you please tell me how i can solve this one.

your help would appreciate.

 

[Dick]

You don't need to complete both squares, just the one in the exponential. Then substitute u=(sqrt(d)*x-b/(2sqrt(d)). Now express the quadratic multiplying the exponential in terms of u also.

 

[abotaha]

You don't need to complete both squares, just the one in the exponential. Then substitute u=(sqrt(d)*x-b/(2sqrt(d)). Now express the quadratic multiplying the exponential in terms of u also.

Thanks for the replay. It is still not clear for me. Could you please explain more.

 

[Dick]

Thanks for the replay. It is still not clear for me. Could you please explain more.

Do a u substitution where u is the linear expression in the exponential. That's really the whole suggestion. If you multiply the result out you can express it completely in terms of the integrals Dickfore gave you. It's just messy and complicated. It's not hard.

 

[abotaha]

Thanks a lot guys,
I learned a lot from you, but to be honest with you I have not solved the problem yet.
I cannot figure it out, but i am still looking around.

a great appreciation for all of you.

Cheers.

 

[HallsofIvy]

After completing the square in the exponential, you will have something like
\int (ay^2+ by+ c)e^{my^2+ b) dy= e^by\int (ay^2+ by+ c)e^{my^2}dy/tex]<br /> <br /> Now separate that into three integrals:<br /> e^by (a\int y^2 e^{my^2}dy+ b\int ye^{my^2}dy+ c\int e^{my^2}dy[/math]&lt;br /&gt; &lt;br /&gt; The middle of those can be done by letting z= y^2 but the other two cannot be done in terms of elementary functions.

 

[abotaha]

After completing the square in the exponential, you will have something like
\int (ay^2+ by+ c)e^{my^2+ b} dy= e^b\int (ay^2+ by+ c)e^{my^2}dy

Now separate that into three integrals:
e^b (a\int y^2 e^{my^2}dy+ b\int ye^{my^2}dy+ c\int e^{my^2}dy

The middle of those can be done by letting z= y^2 but the other two cannot be done in terms of elementary functions.

Thanks too much. It helps a lot.
The integral now is simplified a lot but i wonder how the integration would be if we define it, especially in case:

e^b \left (a\int_0^{\infty} y^2 e^{my^2}dy+ b\int_0^{\infty} ye^{my^2}dy+ c\int_0^{\infty} e^{my^2}dy\right)

any suggestion please.

 

[Dickfore]

why is the lower bound on your integrals 0?

 

[abotaha]

why is the lower bound on your integrals 0?

It is zero because the variable that I am going to integrate is never be less than zero (That is what I recognized later).

 

[Dickfore]

What variable is that?

 

[abotaha]

The variable is y as shown in the integration equation. This variable is called Leaf Area Index (LAI) which represents the area of the leaves of the trees.

 

[Dickfore]

Oh, so the boundaries on your initial integrals were incorrect.

Nevertheless, these integrals can be expressed in terms of the Euler gamma function:

<br /> I_{k}(m) = \int_{0}^{\infty}{x^{k} \, e^{m \, x^{2}} \, dx}, \; m &lt; 0<br />

make the subst:

<br /> t = |m| \, x^{2}<br />

<br /> x = \sqrt{\frac{t}{|m|}} \Rightarrow dx = \frac{dt}{2 \, \sqrt{|m| \, t}}<br />

<br /> I_{k}(m) = \frac{1}{2 |m|^{(k + 1)/2}} \, \int_{0}^{\infty}{t^{\frac{k + 1}{2} - 1} \, e^{-t} \, dt} = \frac{\Gamma(\frac{k + 1}{2})}{2 |m|^{(k + 1)/2}}<br />

Then, use the rules:
<br /> \Gamma(n + \frac{1}{2}) = \frac{(2n - 1)! \, \sqrt{\pi}}{2^{n}}, \; k = 2 n, \; n \ge 0<br />

<br /> \Gamma(n + 1) = n!, \; k = 2 n + 1, \; n \ge 0<br />

 

[abotaha]

Thanks for this nice explanation.
I have not had any knowledge about the Gamma function.
However if you are suggesting me to use it and you are sure about this method then I will used.

thanks again, you are very kind and generous.

 

[Dickfore]

Notice that m has to be negative in order that the integral converges.

 

[abotaha]

Thank Dickfore very much,
It helps me a lot and the whole problem is solved.

High appreciation to you.

one last thing please
can i apply Gamma function (the method above that you mentioned) when the integral boundaries from constant to infinity, like

<br /> <br /> I_{k}(m) = \int_{c}^{\infty}{x^{k} \, e^{m \, x^{2}} \, dx}<br /> <br />

or is there any transformation to make the low boundary zero in order to apply Gamma function formula.

 

[Dickfore]

that integral is expressible in terms of the upper incomplete gamma function. There are a lot of resources discussing it.

 

[abotaha]

well you are right.
i am trying to figure out this thing.

cheers,

 

[Dick]

well you are right.
i am trying to figure out this thing.

cheers,

The integral from 0 to infinity of x^k*exp(-m*x^2) is sqrt(pi/m)*(1/2) if k=0, 1/(2m) if k=1 and sqrt(pi/m)/(4m) if k=2. k=1 is elementary and you can get the others by substituting x=sqrt(m)*u into the forms Dickfore gave you and cutting them in half.