# Improper Integral Calculator

inf = ∞ , pi = π and -inf = -∞
This will be calculated

Mobile Apps

## Improper Integral calculator

Improper integral calculator is used to integrate the definite integrals with one or both boundaries at infinity. This integral convergence calculator measures the convergence or divergence of the functions with defined limits.

## How to use this improper integral calculator?

Follow the below steps to check the convergence or divergence of the functions.

• Input the function.
• Use the keypad icon to write the mathematical symbols.
• Write the upper and the lower limits of the function.
• Choose the integrating variable.
• Click the load example button to use the sample examples.
• Hit the calculate button.
• Press the clear button to enter another function.

## What is the improper integral?

Improper integral is the definite integral with one or both limits are infinity and covers the unbounded area. The function can also be unbounded if it provides the undefined result by applying the upper and lower limits.

There are four types of improper integral.

1. The lower limit at minus infinity

$\int _{-\infty }^vf\left(x\right)dx=\lim _{u\to -\infty \:}\left(\int _u^{v\:}\:f\left(x\right)dx\right)$

1. The upper limit is at infinity

$\int _u^{\infty }f\left(x\right)dx=\lim _{v\to \infty }\left(\int _u^{v\:}\:f\left(x\right)dx\right)$

1. Both the upper and lower limits are at infinity.

$\int _{-\infty }^∞\:f\left(x\right)dx=\int _{-∞}^a\:f\left(x\right)dx+\int _a^{∞\:}f\left(x\right)dx$

1. The points are finite but the functions are unbounded such as:

$\int _0^1\frac{1}{\sqrt{2x}}dx=\lim _{b\to 0}\left(\int _b^1\frac{1}{\sqrt{2x}}dx\:\right)$

How to calculate the improper integral?

Following is a solved example of the improper integral.

Example

Find the convergence of the given improper integral $\frac{1}{\:2x^2}$ from -infinity to -1.

Solution

Step 1: Write the given improper integral along with the integral notation.

$\int _{-\infty }^{-1}\left(\frac{1}{\:2x^2}\right)dx$

Step 2: Write the general formula of improper integral for minus infinity.

$\int _{-\infty }^vf\left(x\right)dx=\lim _{u\to -\infty \:}\left(\int _u^{v\:}\:f\left(x\right)dx\right)$

Step 3: Write the given improper integral according to the above type.

$\int _{-\infty }^{-1}\left(\frac{1}{2x^2}\right)dx=\lim _{u\to -\infty }\left(\int _u^{-1}\left(\frac{1}{2x^2}\right)dx\right)$

Step 4: Integrate the improper integral with respect to x.

$\int _{-\infty }^{-1}\left(\frac{1}{2x^2}\right)dx=\lim _{u\to -\infty }\left(\int _u^{-1}\left(\frac{x^{-2}}{2}\right)dx\right)$

$\int _{-\infty }^{-1}\left(\frac{1}{2x^2}\right)dx=\lim _{u\to -\infty }\frac{1}{2}\int _u^{-1}x^{-2}dx$

$\int _{-\infty }^{-1}\left(\frac{1}{2x^2}\right)dx=\lim _{u\to -\infty \:}\frac{1}{2}\left[\frac{x^{-2+1}}{-2+1}\right]^{-1}_u$

$\int _{-\infty }^{-1}\left(\frac{1}{2x^2}\right)dx=\lim _{u\to -\infty \:}\frac{1}{2}\left[\frac{x^{-1}}{-1}\right]^{-1}_u$

Step 5: Use the fundamental theorem of calculus and apply the upper and lower limits.

$\int _{-\infty }^{-1}\left(\frac{1}{2x^2}\right)dx=\lim _{u\to -\infty \:}-\frac{1}{2}\left[\frac{1}{x}\right]^{-1}_u$

$\int _{-\infty }^{-1}\left(\frac{1}{2x^2}\right)dx=\lim _{u\to -\infty \:}-\frac{1}{2}\left[-\frac{1}{1}-\frac{1}{u}\right]$

$\int _{-\infty }^{-1}\left(\frac{1}{2x^2}\right)dx=\lim _{u\to -\infty \:}-\frac{1}{2}\left[-1-\frac{1}{u}\right]$

Step 6: Apply the limit value.

$\int _{-\infty }^{-1}\left(\frac{1}{2x^2}\right)dx=-\frac{1}{2}\left[-1-\frac{1}{-\infty \:\:}\right]$

$\int _{-\infty }^{-1}\left(\frac{1}{2x^2}\right)dx=-\frac{1}{2}\left[-1+0\right]$

$\int _{-\infty }^{-1}\left(\frac{1}{2x^2}\right)dx=\frac{1}{2}$

Hence, the given function converges.

### References

What is the improper integral? | Khan Academy (n.d.)

Example of Improper integrals. Brilliant Math & Science Wiki. (n.d.)