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For example, $$I=\int xe^{x}dx$$ By taking the derivative of x, and then repeating integration by parts once, the integral can be evaluated trivially. However, when taking the derivative of $e^{x}$ and integrating $x^2$, the process goes on forever. Another example would be $$J=\int \sin{x}\cos{x}dx$$ By repeating integration by parts, $$J= sin^2x + cos^2x + sin^2x + cos^2x ....$$ Does this have any use/any interesting links? e.g. could be used to evaluate integrals where other "combination" of choosing which term is differentiated/integrated is not plausible.

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  • $\begingroup$ Usually this technique is used to derive a series, especially if the variable we are doing the series expansion in is in the limits of the integral. $\endgroup$ – Ninad Munshi Feb 25 at 20:35
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On example I could see is the following :

Imagine you want to compute $$\int_{0}^{1}\frac{\text{d}x}{\left(1+x^2\right)^2}$$

You can use that $$ \int_{0}^{1}\frac{\text{d}x}{1+x^2}=\frac{\pi}{4} $$ and then you integrate by part "in the other way" you usually do $$ \int_{0}^{1}\frac{\text{d}x}{1+x^2}=\int_{0}^{1}\frac{1}{1+x^2}\times 1\text{ d}x=\left[\frac{x}{\left(1+x^2\right)^2}\right]^{1}_0+\int_{0}^{1}\frac{2x^2}{\left(1+x^2\right)^2}\text{d}x $$ Hence $$ \frac{\pi}{4}=\frac{1}{4}+2\int_{0}^{1}\left(\frac{1}{1+x^2}-\frac{1}{\left(1+x^2\right)^2}\right)\text{d}x$$and we find

$$\int_{0}^{1}\frac{\text{d}x}{\left(1+x^2\right)^2}=\frac{2+\pi}{8}$$

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To expand on my comment, I will explicitly use your first example and another example I have.

For $I(z) = \int_0^z xe^x\:dx$, going in the "opposite" direction gives us

$$I(z) = \frac{1}{2}z^2e^z - \int_0^z \frac{1}{2}x^2e^x\:dx \implies I(z) = e^z\left(\frac{1}{2}z^2-\frac{1}{6}z^3+\cdots\right)$$

$$=e^z\left(e^{-z}-1+z\right)=1-e^z+ze^z$$

getting the series from repeated integration by parts. Another example would be

$$\int_0^{\frac{1}{2}}\frac{1}{\sqrt{1-x^2}}\:dx = \frac{x}{\sqrt{1-x^2}}\Biggr|_0^{\frac{1}{2}}-\int_0^{\frac{1}{2}}\frac{x^2}{(1-x^2)^{\frac{3}{2}}}\:dx$$

which gives a series for

$$\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} = \frac{1}{\sqrt{3}}-\frac{1}{3}\left(\frac{1}{\sqrt{3}}\right)^3+\frac{1}{5}\left(\frac{1}{\sqrt{3}}\right)^5-\cdots$$

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