Improper Integrals (AQA A-Level Further Maths): Revision Notes

Worked Example: Calculate $\int_2^{\infty} \frac{2}{x^3} \, dx$

Solution:

This integral is improper because the upper limit is infinite. Replace ∞ with $t$ and evaluate the limit as $t \to \infty$:

$\int_2^{\infty} \frac{2}{x^3} \, dx = \lim_{t \to \infty} \int_2^{t} 2x^{-3} \, dx$

Integrate using the power rule:

$= \lim_{t \to \infty} \left[-x^{-2}\right]_2^{t}$

$= \lim_{t \to \infty} \left(-\frac{1}{t^2} - \left(-\frac{1}{4}\right)\right)$

$= \lim_{t \to \infty} \left(-\frac{1}{t^2} + \frac{1}{4}\right)$

As $t \to \infty$, we have $\frac{1}{t^2} \to 0$, therefore:

$= \frac{1}{4}$

The integral converges to $\frac{1}{4}$.

Worked Example: Evaluating integrals with two infinite limits

a) Evaluate $\int_{-\infty}^{\infty} \frac{x}{e^{x^2}} \, dx$

b) Show that the improper integral $\int_{-\infty}^{\infty} e^x \, dx$ is divergent.

Solution:

Part a:

Since both limits are infinite, split the integral at $x = 0$:

$\int_{-\infty}^{\infty} \frac{x}{e^{x^2}} \, dx = \int_{-\infty}^{0} \frac{x}{e^{x^2}} \, dx + \int_{0}^{\infty} \frac{x}{e^{x^2}} \, dx$

Use different variables for the two infinite limits:

$= \lim_{a \to -\infty} \int_{a}^{0} xe^{-x^2} \, dx + \lim_{b \to \infty} \int_{0}^{b} xe^{-x^2} \, dx$

To integrate, use the chain rule in reverse. Since $\frac{d}{dx}(e^{-x^2}) = -2xe^{-x^2}$, we have:

$= \lim_{a \to -\infty} \left[-\frac{1}{2}e^{-x^2}\right]_a^{0} + \lim_{b \to \infty} \left[-\frac{1}{2}e^{-x^2}\right]_0^{b}$

$= \lim_{a \to -\infty} \left(-\frac{1}{2}e^{0} - \left(-\frac{1}{2}e^{-a^2}\right)\right) + \lim_{b \to \infty} \left(-\frac{1}{2}e^{-b^2} - \left(-\frac{1}{2}e^{0}\right)\right)$

Since $e^0 = 1$:

$= \lim_{a \to -\infty} \left(-\frac{1}{2} + \frac{1}{2}e^{-a^2}\right) + \lim_{b \to \infty} \left(-\frac{1}{2}e^{-b^2} + \frac{1}{2}\right)$

$= \left(-\frac{1}{2}\right) + \left(\frac{1}{2}\right) = 0$

Both limits exist (as $a^2 \to \infty$ and $b^2 \to \infty$, the exponentials $e^{-a^2} \to 0$ and $e^{-b^2} \to 0$), so the improper integral is convergent and equals 0.

Part b:

Split the integral at $x = 0$:

$\int_{-\infty}^{\infty} e^x \, dx = \lim_{a \to -\infty} \int_{a}^{0} e^x \, dx + \lim_{b \to \infty} \int_{0}^{b} e^x \, dx$

$= \lim_{a \to -\infty} [e^x]_a^{0} + \lim_{b \to \infty} [e^x]_0^{b}$

$= \lim_{a \to -\infty} (e^0 - e^a) + \lim_{b \to \infty} (e^b - e^0)$

$= \lim_{a \to -\infty} (1 - e^a) + \lim_{b \to \infty} (e^b - 1)$

As $a \to -\infty$, $e^a \to 0$, so the first limit is $1 - 0 = 1$.

However, as $b \to \infty$, $e^b \to \infty$, so the second limit does not exist.

Since the second limit does not exist, the improper integral $\int_{-\infty}^{\infty} e^x \, dx$ is divergent. Both parts must converge for the original integral to be convergent.

Worked Example: Evaluate $\int_0^{4} \frac{2}{\sqrt{x}} \, dx$

Solution:

The function $\frac{2}{\sqrt{x}} = \frac{2}{\sqrt{x}}$ is undefined at $x = 0$ (the lower limit), making this an improper integral. Replace 0 with $t$ and take the limit as $t \to 0$ from above:

$\int_0^{4} \frac{2}{\sqrt{x}} \, dx = \lim_{t \to 0^+} \int_t^{4} 2x^{-1/2} \, dx$

Integrate using the power rule:

$= \lim_{t \to 0^+} \left[2 \cdot \frac{x^{1/2}}{1/2}\right]_t^{4}$

$= \lim_{t \to 0^+} \left[4\sqrt{x}\right]_t^{4}$

$= \lim_{t \to 0^+} (4\sqrt{4} - 4\sqrt{t})$

$= \lim_{t \to 0^+} (8 - 4\sqrt{t})$

As $t \to 0^+$, we have $\sqrt{t} \to 0$, therefore:

$= 8 - 0 = 8$

The integral converges to 8.

Worked Example: Calculate $\int_0^{1} (2 - \ln x) \, dx$

Solution:

This integral requires integration by parts because of the logarithm term. Since $\ln x$ is undefined at $x = 0$, replace 0 with $t$ and evaluate the limit:

$\int_0^{1} (2 - \ln x) \, dx = \lim_{t \to 0^+} \int_t^{1} (2 - \ln x) \, dx$

$= \lim_{t \to 0^+} \left[2x - x\ln x + x\right]_t^{1}$

For the $x \ln x$ term, use integration by parts with $u = \ln x$ and $\frac{dv}{dx} = 1$. This gives $\int \ln x \, dx = x \ln x - x$.

$= \lim_{t \to 0^+} [(2 \cdot 1 - 1 \cdot \ln 1 + 1) - (2t - t\ln t + t)]$

Since $\ln 1 = 0$:

$= \lim_{t \to 0^+} [3 - 3t + t\ln t]$

As $t \to 0^+$, we have $3t \to 0$.

For the $t \ln t$ term, use the limit result $x^k \ln x \to 0$ as $x \to 0^+$ with $k = 1$.

Therefore $t \ln t \to 0$.

$= 3 - 0 + 0 = 3$

The integral converges to 3.

Worked Example: Evaluate the improper integral $\int_{-1}^{e} x\ln|x| \, dx$

Solution:

The function $x\ln|x|$ is undefined at $x = 0$, which lies between the limits $-1$ and $e$. Split the integral at this point of discontinuity:

$\int_{-1}^{e} x\ln|x| \, dx = \int_{-1}^{0} x\ln|x| \, dx + \int_{0}^{e} x\ln|x| \, dx$

Evaluate the first integral by replacing 0 with $a$ and taking the limit as $a \to 0^-$:

$\int_{-1}^{0} x\ln|x| \, dx = \lim_{a \to 0^-} \int_{-1}^{a} x\ln|x| \, dx$

Use integration by parts with $u = \ln|x|$ and $\frac{dv}{dx} = x$:

$= \lim_{a \to 0^-} \left[\frac{1}{2}x^2\ln|x| - \frac{1}{4}x^2\right]_{-1}^{a}$

$= \lim_{a \to 0^-} \left[\left(\frac{1}{2}a^2\ln|a| - \frac{1}{4}a^2\right) - \left(\frac{1}{2}(-1)^2\ln|-1| - \frac{1}{4}(-1)^2\right)\right]$

$= \lim_{a \to 0^-} \left[\frac{1}{2}a^2\ln|a| - \frac{1}{4}a^2 + \frac{1}{4}\right]$

Using the limit result $a^k\ln|a| \to 0$ as $a \to 0$ with $k = 2$:

$= 0 - 0 + \frac{1}{4} = \frac{1}{4}$

Now evaluate the second integral by replacing 0 with $b$ and taking the limit as $b \to 0^+$:

$\int_{0}^{e} x\ln|x| \, dx = \lim_{b \to 0^+} \int_{b}^{e} x\ln|x| \, dx$

$= \lim_{b \to 0^+} \left[\frac{1}{2}x^2\ln|x| - \frac{1}{4}x^2\right]_{b}^{e}$

$= \lim_{b \to 0^+} \left[\left(\frac{1}{2}e^2\ln e - \frac{1}{4}e^2\right) - \left(\frac{1}{2}b^2\ln|b| - \frac{1}{4}b^2\right)\right]$

Since $\ln e = 1$:

$= \lim_{b \to 0^+} \left[\frac{1}{2}e^2 - \frac{1}{4}e^2 - \frac{1}{2}b^2\ln|b| + \frac{1}{4}b^2\right]$

$= \frac{1}{4}e^2 - 0 + 0 = \frac{1}{4}e^2$

Since both limits exist, the improper integral converges. The final answer is:

$\int_{-1}^{e} x\ln|x| \, dx = \frac{1}{4} + \frac{1}{4}e^2 = \frac{1}{4}(1 + e^2)$

Worked Example: Determine whether the improper integral $\int_1^{\infty} \left(\frac{2x}{x^2+1} - \frac{4x-1}{2x^2-x}\right) dx$ is convergent or divergent.

Solution:

Replace ∞ with $a$ and evaluate as $a \to \infty$:

$\int_1^{\infty} \left(\frac{2x}{x^2+1} - \frac{4x-1}{2x^2-x}\right) dx = \lim_{a \to \infty} \int_1^{a} \left(\frac{2x}{x^2+1} - \frac{4x-1}{2x^2-x}\right) dx$

Both terms have the form $\frac{f'(x)}{f(x)}$, which integrates to $\ln|f(x)|$. Recognise that:

• For $\frac{2x}{x^2+1}$: the derivative of $x^2+1$ is $2x$
• For $\frac{4x-1}{2x^2-x}$: the derivative of $2x^2-x$ is $4x-1$

$= \lim_{a \to \infty} \left[\ln(x^2+1) - \ln(2x^2-x)\right]_1^{a}$

Combine the logarithms using the law $\ln A - \ln B = \ln\left(\frac{A}{B}\right)$:

$= \lim_{a \to \infty} \left[\ln\left(\frac{x^2+1}{2x^2-x}\right)\right]_1^{a}$

$= \lim_{a \to \infty} \ln\left(\frac{a^2+1}{2a^2-a}\right) - \ln\left(\frac{1+1}{2-1}\right)$

$= \lim_{a \to \infty} \ln\left(\frac{a^2+1}{2a^2-a}\right) - \ln 2$

To find the limit of the rational expression, divide numerator and denominator by $a^2$:

$\frac{a^2+1}{2a^2-a} = \frac{1 + \frac{1}{a^2}}{2 - \frac{1}{a}}$

As $a \to \infty$, we have $\frac{1}{a^2} \to 0$ and $\frac{1}{a} \to 0$, so:

$\frac{a^2+1}{2a^2-a} \to \frac{1}{2}$

Therefore:

$\lim_{a \to \infty} \ln\left(\frac{a^2+1}{2a^2-a}\right) = \ln\left(\frac{1}{2}\right)$

The final result is:

$\ln\left(\frac{1}{2}\right) - \ln 2 = -\ln 4$

Since the limit exists and equals a finite value, the integral converges to $-\ln 4$.