How Do You Integrate (x^3)(e^2x)?

[shinkansenfan]

How to integrate (x^3)(e^2x) ?

Thanks.

 

[Dickfore]

by parts.

 

[Count Iblis]

Or by integrating exp(p x) and differentiating the result 3 times w.r.t. the parameter p.

Or by integrating exp(p x) and comptuting the third order Taylor expansion around p = 2.

The last option requires the least number of computations, because Taylor expansions can be obtained by using the standard Taylor expansions for the standard functions, rather than using the general Taylor expansion formula in which you have to repeatedly differentiate the function.

 

[shinkansenfan]

Thanks very much.

 

[g_edgar]

Or: get a computer to do it for you. Maple, Mathematica, etc,

(Is that a lot different that getting Physics Forums to do it for you?)

 

[Count Iblis]

Or: get a computer to do it for you. Maple, Mathematica, etc,

(Is that a lot different that getting Physics Forums to do it for you?)

Yes it is, because the whole point of this exercise is to let the student learn how to compute such integrals, not merely to know the answer to this problem. Using Mathematica is ok. for people like me who know all about computing integrals but who need to go on with their work and not waste a lot of time doing integrals themselves.

When I use Mathematica for complicated problems I often have to write small programs or give commands that tells Mathematica what to do. Those steps that I put in myself require far more thinking than the steps needed to solve simple integrals by substitution. All that Mathematica is doing for me is to mechanically use certain rules to produce hundreds of pages full of formula, something that would take me more than a lifetime to do by hand. I then use that output to extract the information I want.

 

[ibmichuco]

by parts.

So I did this three times and I got something like e^2x[x^3/2-3x^2/4+6x/8-16/6] + C
which sounds reasonable, but then when I tried

\int_{-\infty}^{\infty} x^2 e^{-2 x^2} dx

with integration by parts, I ended up with something like infinity ..?

Michuco

 

[Cyosis]

The integral

<br /> \int_{-\infty}^{\infty} x^2 e^{-2 x^2} dx<br />

does converge.

Show the steps you've made so we can see where it went wrong.

 

[ibmichuco]

The integral

<br /> \int_{-\infty}^{\infty} x^2 e^{-2 x^2} dx<br />

does converge.

Show the steps you've made so we can see where it went wrong.

u = x^2 -> du = 2xdx
dv = e^{2x^2} -> v = \sqrt{\pi/2}

\int ... = x^2\sqrt{\pi/2}|_{-\infty}^\infty -\int_{-\infty}^\infty \sqrt{\pi/2} 2 x dx

The first term looks like zero, and the second doesn't looks like it would converge.
I also tried u=e^{-2x^2} and dv = x^2dx, but then I ran into problem with v.

 

[Cyosis]

You need to put [.tex][./tex] tags around your latex code (without the dots).

Fixing your tex so I can read it.
<br /> u = x^2 -&gt; du = 2xdx<br /> \;<br /> dv = e^{2x^2} -&gt; v = \sqrt{\pi/2}<br /> <br /> \int ... = x^2\sqrt{\pi/2}|_{-\infty}^\infty -\int_{-\infty}^\infty \sqrt{\pi/2} 2 x dx

I am not sure what you're doing here, but it is definitely wrong.

<br /> \begin{align*}<br /> \int_{-\infty}^{\infty} x^2 e^{-2 x^2} dx&amp; =\int_{-\infty}^{\infty} x \left(x e^{-2 x^2}\right) dx<br /> \\<br /> &amp;= -\frac{1}{4}x e^{-2 x^2}|_{-\infty}^\infty+\frac{1}{4} \int_{-\infty}^\infty e^{-2x^2}dx<br /> \end{align*}<br />

 

[ibmichuco]

You need to put [.tex][./tex] tags around your latex code (without the dots).

Fixing your tex so I can read it.
<br /> u = x^2 -&gt; du = 2xdx<br /> \;<br /> dv = e^{2x^2} -&gt; v = \sqrt{\pi/2}<br /> <br /> \int ... = x^2\sqrt{\pi/2}|_{-\infty}^\infty -\int_{-\infty}^\infty \sqrt{\pi/2} 2 x dx

I am not sure what you're doing here, but it is definitely wrong.

<br /> \begin{align*}<br /> \int_{-\infty}^{\infty} x^2 e^{-2 x^2} dx&amp; =\int_{-\infty}^{\infty} x \left(x e^{-2 x^2}\right) dx<br /> \\<br /> &amp;= -\frac{1}{4}x e^{-2 x^2}|_{-\infty}^\infty+\frac{1}{4} \int_{-\infty}^\infty e^{-2x^2}dx<br /> \end{align*}<br />

I take that by the parenthesis, you define u=x and dv=xe^{-2x^2}? But wouldn't this give v=0 with xe^{-2x^2} being an odd function?

Sorry, I am particularly densed today

 

[Cyosis]

Why would v be zero? The primitive of an odd function is not zero. Example: according to you \int xdx=0, yet \int x dx= 1/2 x^2, which I am sure you know.

 

[ibmichuco]

Why would v be zero? The primitive of an odd function is not zero. Example: according to you \int xdx=0, yet \int x dx= 1/2 x^2, which I am sure you know.

I am thinking of
<br /> \begin{equation*}<br /> v = \int_{-\infty}^{\infty} x e^{-2 x^2} d x = 0<br /> \end{equation*}<br />

Don't I need the definite integral in this step also?

 

[Dickfore]

So I did this three times and I got something like e^2x[x^3/2-3x^2/4+6x/8-16/6] + C
which sounds reasonable, but then when I tried

\int_{-\infty}^{\infty} x^2 e^{-2 x^2} dx

with integration by parts, I ended up with something like infinity ..?

Michuco

That's because you took the lower bound to be -\infty and the exponential function e^{2x} is infinitely large in that limit.

 

[Cyosis]

I am thinking of
v = \int_{-\infty}^{\infty} x e^{-2 x^2} d x = 0

Don't I need the definite integral in this step also?

Sure that integral is zero. However that integral never enters in the solution so I am not sure why you're trying to calculate. It has become clear to me now what you did wrong in post #9 as well. You're using the boundary values to determine v. This is wrong, do not plug in the boundary values until you're done integrating. I really suggest you review integration by parts since it seems you don't understand how it works when limits are involved.

 

[ibmichuco]

Sure that integral is zero. However that integral never enters in the solution so I am not sure why you're trying to calculate. It has become clear to me now what you did wrong in post #9 as well. You're using the boundary values to determine v. This is wrong, do not plug in the boundary values until you're done integrating. I really suggest you review integration by parts since it seems you don't understand how it works when limits are involved.

My bad. Most of what I have read indicate that you use the boundary in the integration and I thought that also includes the evaluation of v. Thanks for clearing that out. Can you suggest any particular book that helps in evaluating tough(er) integrals. I am trying eventually integrate

<br /> \int_{-\infty}^\infty(x-a)^n(x-b)^m e^{\alpha(x-c)^2} dx<br />

where n and m are integers. I know the result from Mathematica but would like how to work it out. I tried the simplest one and encountered gamma function.

Again, thanks for you patient explanations

 

[cstoos]

I got e^2x(\frac{1}{2}x³-\frac{3}{2}x²-3x-3)
Of course, I did it by hand and have not checked for mistakes.

 

[Cyosis]

My bad. Most of what I have read indicate that you use the boundary in the integration and I thought that also includes the evaluation of v. Thanks for clearing that out. Can you suggest any particular book that helps in evaluating tough(er) integrals. I am trying eventually integrate

LaTeX Code: <BR>\\int_{-\\infty}^\\infty(x-a)^n(x-b)^m e^{\\alpha(x-c)^2} dx<BR>

where LaTeX Code: n and LaTeX Code: m are integers. I know the result from Mathematica but would like how to work it out. I tried the simplest one and encountered gamma function.

Again, thanks for you patient explanations

Have you solved the previous integrals yet? Also I suggest you check what kind of restrictions you need on all the constants in order to have the integral converge.