# Question Video: Find the Maclaurin Series of a Composition between the Exponential Function and the Square Function Mathematics • Higher Education

Use the Maclaurin series of π^π₯ to express β« π^(π₯Β²) dπ₯ as an infinite series.

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### Video Transcript

Use the Maclaurin series of π to the power of π₯ to express the integral of π to the power of π₯ squared with respect to π₯ as an infinite series.

The question wants us to use the Maclaurin series of π to the power of π₯ to represent this integral as an infinite series. We recall the Maclaurin series of π to the power of π₯ tells us that π to the power of π₯ is equal to the sum from π equals zero to β of π₯ to the πth power divided by π factorial. And this is valid for all real numbers π₯. We want to use this Maclaurin series for π to the power of π₯ to represent the integral of π to the power of π₯ squared with respect to π₯.

To do this, weβre going to use the fact that if we have a Maclaurin series for a function π of π₯, which is valid on the radius of convergence π. Then we can generate a new power series using this Maclaurin series, which is valid on the absolute value of π of π₯ is less than π. Since the question involves π to the power of π₯ squared, weβll set our function π of π₯ to be equal to π₯ squared.

So weβll start with our Maclaurin series for π to the power of π₯, which we know is true for all real numbers π, which is equivalent to saying the radius of convergence is β. And weβre going to change all of our values of π₯ to π₯ squared. This gives us that π to the power of π₯ squared is equal to the sum from π equals zero to β of π₯ squared raised to the πth power divided by π factorial when the absolute value of π₯ squared is less than the radius of convergence for our Maclaurin series of π to the π₯.

However, weβve already shown that our Maclaurin series for π to the power of π₯ has an infinite radius of convergence. So our power series will be true for all real numbers π₯. We can simplify the numerator of our summand by noticing that π₯ squared all raised to the πth power is just equal to π₯ to the power of two π. Next, the question wants us to find an expression for the integral of π to the power of π₯ squared with respect to π₯. So weβre going to integrate both sides of our equation with respect to π₯. This gives us the integral of π to the power of π₯ squared with respect to π₯ is equal to the integral of the sum from π equals zero to β of π₯ to the power of two π divided by π factorial with respect to π₯.

And since weβre integrating the infinite sum of a power series, we can switch the order of our integral and our sum. This gives us the sum from π equals zero to β of the integral of π₯ to the power of two π divided by π factorial with respect to π₯. Next, weβre going to use the fact that if π is not equal to negative one and π is a constant, then to integrate π multiplied by π₯ to the πth power with respect to π₯, we add one to our exponent of π₯ and then divide by this new exponent. Then, we add our constant of integration π. Since our sum starts from π equals zero, two π is never equal to negative one. So our exponent of π₯ is never equal to negative one. And π is independent of our variable π₯. This means we can consider it a constant with respect to π₯.

So we can now use our integral rule to integrate each term of our sum. We add one to our exponents giving us π₯ to the power of two π plus one and then we divide by two π plus one. This gives us the sum from π equals zero to β of π₯ to the power of two π plus one divided by π factorial multiplied by two π plus one. And instead of adding a constant of integration for all of our terms, weβll just add one constant of integration outside of our sum, which we will call π.

Therefore, weβve shown that we can represent the integral of π to the power of π₯ squared with respect to π₯ as the sum from π equals zero to β of π₯ to the power of two π plus one divided by π factorial multiplied by two π plus one plus a constant of integration π.