Improper integralsAQA A-Level Further Maths: Revision notes

An integral is improper if either

• a limit of integration is infinite, for example $\int_1^\infty x^{-2}\,dx$, or
• the integrand is undefined at a point in the range of integration, for example $\int_0^1\ln x\,dx$ (undefined at $x=0$) or $\int_0^4\frac{1}{\sqrt x}\,dx$.

An improper integral converges if the limit that defines it exists and is finite; otherwise it diverges. You cannot substitute $\infty$ or the undefined value straight into the antiderivative: you must use a limit.

Replace $\infty$ by $t$, integrate, then let $t\to\infty$: $\int_a^\infty\mathrm{f}(x)\,dx=\lim_{t\to\infty}\int_a^t\mathrm{f}(x)\,dx.$ Example: $\int_1^\infty\frac{1}{x^2}\,dx=\lim_{t\to\infty}\left[-\frac1x\right]_1^t=\lim_{t\to\infty}\left(1-\frac1t\right)=1$ (converges).

Example: $\int_1^\infty\frac1x\,dx=\lim_{t\to\infty}\ln t=\infty$ (diverges).

For $\int_1^\infty x^{-p}\,dx$: the integral equals $\frac{1}{p-1}$ if $p>1$ and diverges if $p\le1$. Also $\int_0^\infty\mathrm{e}^{-2x}\,dx=\lim_{t\to\infty}\left[-\frac12\mathrm{e}^{-2x}\right]_0^t=\frac12$.

If the integrand is undefined at an end-point, replace that end-point by $t$ and let $t$ approach it from inside the range: $\int_0^b\mathrm{f}(x)\,dx=\lim_{t\to0^+}\int_t^b\mathrm{f}(x)\,dx.$ Example: $\int_0^4x^{-1/2}\,dx=\lim_{t\to0^+}\left[2\sqrt x\right]_t^4=4$ (converges).

Example: $\int_0^1\frac1x\,dx=\lim_{t\to0^+}\left(-\ln t\right)=\infty$ (diverges).

If the problem point lies inside the range, split the integral at that point and take a limit for each part; the whole integral converges only if both parts do.

You may use these limits for any constant $k>0$:

• $x^k\mathrm{e}^{-x}\to0$ as $x\to\infty$ (the exponential beats any power);
• $x^k\ln x\to0$ as $x\to0^+$ (the power beats the logarithm).

They are what remove the awkward terms in improper integrals, such as $t\mathrm{e}^{-t}$ as $t\to\infty$ and $t\ln t$ or $t^2\ln t$ as $t\to0^+$. Also $\mathrm{e}^{-t}\to0$ as $t\to\infty$ and $\ln t\to-\infty$ as $t\to0^+$.

(1) $\int_0^\infty x\mathrm{e}^{-x}\,dx$. By parts, $\int_0^t x\mathrm{e}^{-x}\,dx=\left[-x\mathrm{e}^{-x}-\mathrm{e}^{-x}\right]_0^t=1-t\mathrm{e}^{-t}-\mathrm{e}^{-t}$. As $t\to\infty$ both $t\mathrm{e}^{-t}$ and $\mathrm{e}^{-t}$ tend to 0, so the integral is $1$.

(2) $\int_0^1\ln x\,dx$. $\int_t^1\ln x\,dx=\left[x\ln x-x\right]_t^1=-1+t-t\ln t\to-1$ as $t\to0^+$, since $t\ln t\to0$.

(3) $\int_0^1x\ln x\,dx$. $\int_t^1x\ln x\,dx=\left[\frac{x^2}{2}\ln x-\frac{x^2}{4}\right]_t^1=-\frac14-\frac{t^2}{2}\ln t+\frac{t^2}{4}\to-\frac14$.

1. Say why the integral is improper.
2. Introduce $t$ and write the limit notation.
3. Integrate (by parts or by a standard form) and substitute $t$ and the finite limit.
4. State the standard limits you use, then the value, or say 'diverges' with the reason ($\ln t\to\infty$, or the limit is infinite).

Compare powers: $\int_1^\infty x^{-p}\,dx$ converges for $p>1$, and $\int_0^1x^{-p}\,dx$ converges for $p<1$.