Question about Changing Limits of Integration

[anoegenetic]

If the limits of integration of my integral are from -Infinity to zero, can I change those limits such that they're from zero to +Infinity? If so, how?

Thanks!

 

[micromass]

Not sure what you mean, but a substitution u=-x should do the trick (if x is your integration variable).

 

[dreit]

You would not be able to change those limits, because that would be integrating on the complete opposite side of the graph

 

[Mark44]

You would not be able to change those limits, because that would be integrating on the complete opposite side of the graph

What's the problem with "integrating on the complete opposite side of the graph"?

Following micromass's lead, suppose we have
\int_{-\infty}^0 x dx

Making the substitution u = -x, (hence du = -dx) and changing the limits of integration, we have
\int_{u = \infty}^0 -u (-du) = -\int_{u = \infty}^0 -u du
= \int_{u = 0}^{\infty} -u du

Geometrically, the first integral represents the infinitely large triangle bounded by the negative x-axis and the line y = x. The last integral represents the infinitely large triangle bounded by the positive u-axis and the line y = -u. Geometrically the two regions are the same size: any point in one region can be shown to be in the other region, and vice versa.

 

[dreit]

Oh that's right I didnt realize that you would be using -U, sorry for any misleading

 

[Prove It]

What's the problem with "integrating on the complete opposite side of the graph"?

Following micromass's lead, suppose we have
\int_{-\infty}^0 x dx

Making the substitution u = -x, (hence du = -dx) and changing the limits of integration, we have
\int_{u = \infty}^0 -u (-du) = -\int_{u = \infty}^0 -u du
= \int_{u = 0}^{\infty} -u du

Geometrically, the first integral represents the infinitely large triangle bounded by the negative x-axis and the line y = x. The last integral represents the infinitely large triangle bounded by the positive u-axis and the line y = -u. Geometrically the two regions are the same size: any point in one region can be shown to be in the other region, and vice versa.

Of course, \displaystyle \int_{-\infty}^0{x\,dx} is divergent, so I don't see any reason to evaluate it anyway...

 

[Mark44]

Of course, \displaystyle \int_{-\infty}^0{x\,dx} is divergent, so I don't see any reason to evaluate it anyway...

This was just an example that showed how you could change the limits of integration.

 

[SammyS]

This was just an example that showed how you could change the limits of integration.

So, let's generalize Mark44's reply

<br /> \displaystyle \int_{-\infty}^0{f(x)\,dx}<br />

Making the substitution u = -x, (hence du = -dx) and changing the limits of integration, we have

<br /> \int_{\infty}^0 f(-u) (-du) = -\int_{\infty}^0 f(-u) du <br />

<br /> =\int_{0}^\infty f(-u) \,du

Since u is a "dummy" variable, we can use x (or most any variable) instead, so we have

<br /> \displaystyle \int_{-\infty}^0{f(x)\,dx}=\int_{0}^\infty{f(-x)\,dx}<br />