calculusimproper-integrals
I do not understand improper integrals.
Is
$$ \int_1^e \frac{ \mathrm{dx}}{x(\ln x)^{1/2}}$$ an
improper integral? Is
$$ \int_0^2 \frac{\mathrm{dx}}{x^2+6x+8}$$
an improper integral?
For both I need to evaluate the integral or show that it's divergent. I have no idea what to do and need serious help for math homework since we never covered this in our Calc 1 class.
The integrand is a well-behaved function except perhaps at 0. Thus, if you replace
$\int_0^1{\frac{1}{x^p}}\mathrm{d} x$
with
$\int_{\epsilon}^{1}{\frac{1}{x^p}}\mathrm{d} x$,
for some small $\epsilon >0$, then you can safely integrate:
$\int{\frac{1}{x^p}}\mathrm{d} x = \int x^{-p} \mathrm{d}x = \frac{1}{-p+1}x^{-p+1} + C$, except for the special cases $p=0,1$ with which you can deal separately. Now substitute the limits of integration and check when letting $\epsilon \to 0$ results in a blow-up. To be precise, you might want to consider $\epsilon = \frac{1}{n}$ and send $n \to \infty$.
Let me also address your concern about $\int_0^1{\frac{1}{x^0}}\mathrm{d} x$. Recall that $x^0 = 1$ for all $x$ (including $x=0$), so in fact $\frac{1}{x^0} = 1$ and hence there is no discontinuity.
Finally, for a general $f$, there is probably no one recipe, but usually one has to be careful with end-points of integration (e.g. replace $0$ by $\epsilon$ and $1$ by $1-\delta$) and also identify for what $x$ the function $f$ blows up (if any) and whether this translates to the blow-up for the value of the integral.
I believe these integrals have a simple analytical form. I will demonstrate for $I_1$ and I hope you can see how to do the others similarly.
I write $I_1$ out as originally stated:
$$I_1 = \int_0^{\infty} d\lambda \, \lambda^{1/2} \, e^{-\lambda} \, J_1(\lambda r) \, J_{1/2}(\lambda t)$$
Note that
$$J_{1/2}(\lambda t) = \sqrt{\frac{2}{\pi \lambda t}} \sin{(\lambda t)}$$ $$J_1(\lambda r) = \frac1{i \pi} \int_0^{\pi} d\theta \, \cos{\theta} \, e^{i \lambda r \cos{\theta}} $$
Plugging back into the integral definition of $I_1$ and changing the order of integration, we get
$$I_1 = \frac1{i \pi} \sqrt{\frac{2}{\pi t}} \int_0^{\pi} d\theta \, \cos{\theta} \, \int_0^{\infty} d\lambda \, e^{-\lambda} \, \sin{(\lambda t)} \, e^{i \lambda r \cos{\theta}} $$
Rewriting the sine in exponential form, the integral over $\lambda$ is simple, and we are left with the integral over $\theta$:
$$I_1 = -\frac1{2 \pi} \sqrt{\frac{2}{\pi t}} \int_0^{\pi} d\theta \, \cos{\theta} \, \left [\frac1{1-i t - i r \cos{\theta}} - \frac1{1+i t - i r \cos{\theta}} \right ] $$
Now let's consider the integral
$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} $$
where $a$ and $b$ may be complex; in our case $a=1\pm i t$ and $b=-i r$. While there are at least a couple of ways of evaluating this integral, I will demonstrate how it is done using contour integration.
Consider the contour integral
$$-i \oint_C \frac{dz}{z} \, \frac{z^2+1}{b z^2+2 a z+b} $$
where $C$ is the following contour:
The semicircle has unit radius. Note that, because the real integral is only over a half-cycle rather than a full cycle, the contour $C$ includes a traversal along the real axis. Nevertheless, because of the pole at the origin, there needs to be a small detour of radius $\epsilon$ around the origin as shown.
The contour integral is then equal to
$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} - i \, PV \int_{-1}^1 \frac{dx}{x} \, \frac{x^2+1}{b x^2+2 a x+b} - i (i \epsilon) \int_{\pi}^0 d\phi \, e^{i \phi} \, \frac1{\epsilon \, e^{i \phi}} \frac{\epsilon^2 e^{i 2 \phi}+1}{b \epsilon^2 e^{i 2 \phi}+ 2 a \epsilon \, e^{i \phi} + b} $$
The first integral is what we seek (for now). The third integral is, in the limit as $\epsilon \to 0$, $-\pi/b$. The second integral, the principal value integral, may be evaluated as follows:
$$\begin{align} PV \int_{-1}^1 \frac{dx}{x} \, \frac{x^2+1}{b x^2+2 a x+b} &= \frac1{b} \, PV \int_{-1}^1 \frac{dx}{x}\, \left (1 - \frac{2 a x}{b x^2+2 a x+b} \right ) \\ &= \frac1{b} \, PV \int_{-1}^1 \frac{dx}{x} - \frac{2 a}{b} \int_{-1}^1 \frac{dx}{b x^2+2 a x+b}\end{align}$$
Note that the first principal value integral on the RHS vanishes by symmetry. The second integral on the right needs not be expressed using principal value notation because the pole at the origin is removed. Accordingly,
$$\begin{align} PV \int_{-1}^1 \frac{dx}{x} \, \frac{x^2+1}{b x^2+2 a x+b} &= - \frac{2 a}{b} \int_{-1}^1 \frac{dx}{b x^2+2 a x+b} \\ &= -\frac{2 a}{b^2} \int_{-1}^1 \frac{dx}{\left ( x+\frac{a}{b} \right )^2 + 1-\frac{a^2}{b^2}}\\ &= -\frac{2 a}{b^2} \frac1{\sqrt{1-\frac{a^2}{b^2}}} \left [ \arctan{\left ( \frac{x+\frac{a}{b}}{\sqrt{1-\frac{a^2}{b^2}}} \right )} \right ]_{-1}^1 \\ &= \frac{\pi}{b} \frac{a}{\sqrt{b^2-a^2}} \end{align}$$
For convenience later on, we may write
$$ PV \int_{-1}^1 \frac{dx}{x} \, \frac{x^2+1}{b x^2+2 a x+b} = -i \frac{\pi}{b} \frac{a}{\sqrt{a^2-b^2}} $$
This way, we may write that the contour integral is equal to
$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} - \frac{\pi}{b} \frac{a}{\sqrt{a^2-b^2}} - \frac{\pi}{b} $$
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand of the contour integral inside the contour $C$. In this case, the only pole inside the contour is at $z_+ = -\frac{a}{b} + \sqrt{\frac{a^2}{b^2}-1}$. Computing the residue at this pole, the resulting equation for the integral we seek for now is
$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} - \frac{\pi}{b} \frac{a}{\sqrt{a^2-b^2}} - \frac{\pi}{b} = -\frac{2 \pi}{b} \frac{a}{\sqrt{a^2-b^2}}$$
or
$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} = -\frac{\pi}{b} \left ( \frac{a}{\sqrt{a^2-b^2}} - 1 \right ) $$
Now we may use this result to determine $I_1$. Again, subbing $a=1 \pm i t$ and $b=-i r$, we get that
$$\begin{align} I_1 &= -\frac1{2 \pi} \sqrt{\frac{2}{\pi t}} \frac{\pi}{i r} \left ( \frac{1-i t}{\sqrt{(1-i t)^2+r^2}} - \frac{1+i t}{\sqrt{(1+i t)^2+r^2}} \right ) \\ &= \sqrt{\frac{2}{\pi t}} \frac1{r} \operatorname{Im}{\left (\frac{1+i t}{\sqrt{(1+i t)^2+r^2}} \right )}\end{align}$$
And with that, we are technically finished. But as someone who likes explicit results, I will take this a bit further and express the result as follows:
$$I_1 = \int_0^{\infty} d\lambda \, \lambda^{1/2} \, e^{-\lambda} \, J_1(\lambda r) \, J_{1/2}(\lambda t) = \\ \frac1{\sqrt{\pi t r^2}} \frac{t \sqrt{\sqrt{(1+r^2-t^2)^2+4 t^2}+(1+r^2-t^2)} - \sqrt{\sqrt{(1+r^2-t^2)^2+4 t^2}-(1+r^2-t^2)}}{\sqrt{(1+r^2-t^2)^2+4 t^2}}$$
I have verified this in Mathematica numerically.
Best Answer
Improper integrals occur in primarily two ways: an bound that goes off to infinity or a bound where the function goes off to infinity (infinitely wide vs. infinitely tall). Clearly anything of the form$$\int_0^\infty f(x)\mathrm{d}x$$ is improper and we have to really consider what happens as we approach infinity. Which should scream limit! So we set it up as follows:
$$\lim_{b\to\infty}\int_0^b f(x)\mathrm{d}x$$ and perform the integration as normal, and then take the limit.
The other possibility is when the function in the integral does funky things on our bounds (such as going off to infinity or dividing by zero, etc.) One example is this simple integral:
$$\int_0^1 \frac{\mathrm{d}x}{x-1}$$ This could potentially have an infinity area (look at the graph) because as we get close to $x=1$ we are getting close to dividing by $0$. So we pull the same trick and we look at the integral $$\int_0^{0.9} \frac{\mathrm{d}x}{x-1}$$ and then $$\int_0^{0.99} \frac{\mathrm{d}x}{x-1}$$ and it doesn't take long to realize you're taking a limit. $$\lim_{b\to1}\int_0^b \frac{\mathrm{d}x}{x-1}$$
where this time $b$ approaches a finite value.
Now, to visit your examples: neither of them have an infinite bound and so our first type of improper integral can be ruled out. Now we see if we divide by $0$ or take the $\log$ of $0$ or other math no-no's. (As Andre corrected me) The first one has the $\ln{1}=0$ in a denominator which is the same as trying to divide by $0$ and therefore, we have an improper integral. And we would want to solve:
$$\lim_{b\to 1^+}\int_b^e \frac{\mathrm{d}x}{x\left(\ln{x}\right)^{1/2}}$$ and solve the integral first (in terms of a new variable $b$) and then take the limit, as if it had nothing to do with an integral.
In the current state, it seems that the second integral has no such "funk" and therefore is not improper.
I hope this helps clear up what an improper integral is!
EDIT: I do want to point out that if the bounds on the second integral were this:
$$\int_{-2}^0\frac{\mathrm{d}x}{x^2+6x+8}$$
then we would have an improper integral because if we look at $x=-2$ we end up with $1/0$ and that's a no-no. So we, instead, consider:
$$\lim_{b\to -2^+}\int_b^0\frac{\mathrm{d}x}{x^2+6x+8}$$
and solve the integral first (in terms of a new variable $b$) and then take the limit, as if it had nothing to do with an integral.