Do these improper integrals converge or not? If they do, which are their values?

Question

abb0t · Answer

Can you give an example? There are tests which you can use to determine whether the integral diverges or converges.

anonymous · Answer

Here are the integrals: 

$$\int\limits_{2}^{5}\frac{ 1 }{ \sqrt{x-2} }dx$$

$$\int\limits_{-\infty}^{\infty}x.e ^{\frac{ x^{-2} }{ 2 }}dx$$

anonymous · Answer

$$\int\limits_{0}^{3}\frac{ 1 }{ x-3 }dx$$

anonymous · Answer

$$\begin{align*}\int_2^5\frac{dx}{\sqrt{x-2}}&=\lim_{t\to2^+}\int_t^5(x-2)^{-1/2}~dx\
&=\lim_{t\to2^+}\left[2\sqrt{x-2}\right]_t^5\
&=2\sqrt{5-2}-\lim_{t\to2^+}(2\sqrt{t-2})
\end{align*}$$

anonymous · Answer

$$\int_{-\infty}^\infty x~e^{-\frac{x^2}{2}}~dx$$
Let $u=-\dfrac{x^2}{2}$, so that $-du=x~dx$. Keep in mind the change in the limits: as $x\to\infty$, you have $u\to-\infty$, and as $x\to-\infty$, you have $u\to-\infty$.

So you have
$$\int_{-\infty}^{-\infty}e^u~(-du)$$
What do you know about definite integrals with upper/lower limits that are the same?

anonymous · Answer

$$\begin{align*}\int_0^3\frac{dx}{x-3}&=\lim_{t\to3^-}\int_0^3\frac{dx}{x-3}\
&=\lim_{t\to3^-}\left[\ln|x-3|\right]_0^t\
&=\lim_{t\to3^-}\ln|t-3|-\ln3
\end{align*}$$

anonymous · Answer

@JGMach, do you follow?

anonymous · Answer

Sorry for the stupid question but, why is it required to "put" a limit in the first one you solved?

anonymous · Answer

The function is undefined at $x=2$. The area under the curve from some interval near 2 appears to be approaching $\infty$; we don't know for sure. Using the limit allows to find the area under the curve as our interval slowly approaches this asymptote.