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Integration Under the Integral Sign

April 12th, 2009 |

Author: Merrin

Integration under the integral sign is nothing more than Fubini’s theorem artfully applied. Suppose we start with a definite integral that involves a parameter t.

$\displaystyle \int _{a}^{b}f(x,t)dx$

The next step is to build up a more complicated multiple integral by integrating over some range with respect to the parameter t

$\displaystyle \int _{\alpha }^{\beta }\left[\int _{a}^{b}f(x,t)dx \right] dt$

Now since the limits of integration are non variable we can use Fubini’s theorem and reverse the order of integration.

$\displaystyle \int _{a}^{b}\left[\int _{\alpha }^{\beta }f(x,t)dt \right] dx$

These two integrals are equal by Fubini’s theorem. If we can solve the double integral one way but not with the order of integration interchanged, then it is said that we have integrated under the integral sign and solved the more difficult integral. To put this technique into practice build up your target integral by integrating a parameter.

Example 1. We can find by the universal trigonometric substitution that

$\displaystyle \int _{0}^{\pi }\cfrac{dx}{1+a\sec x} =\cfrac{a\pi }{\sqrt{a^{2} -1} } \quad \, a>1$

Integrate under the integral sign with respect to the parameter \textit{a}.

Solution 1. The result of integration under the integral sign is to prove a fairly difficult definite integral. We will rewrite the integral in our $(x,t)$ form so we have

$\displaystyle {\int _{t=\alpha }^{\beta }\left[\int _{x=0}^{\pi }\cfrac{dx}{1+t\sec x} \right] dt=\int _{\alpha }^{\beta }\cfrac{\pi tdt}{\sqrt{t^{2} -1} } \qquad t>1} \\ {\int _{\alpha }^{\beta }\cfrac{\pi tdt}{\sqrt{t^{2} -1} } =\cfrac{1}{2} \int _{\alpha^2-1 }^{\beta^2-1 }\cfrac{\pi du}{\sqrt{u} } =\cfrac{\pi }{2} \left. \cfrac{u^{1/2} }{1/2} \right|_{\alpha ^{2} -1}^{\beta ^{2} -1} =\pi (\sqrt{\beta ^{2} -1} -\sqrt{\alpha ^{2} -1} )} \\ {\int _{t=\alpha }^{\beta }\left[\int _{x=0}^{\pi }\cfrac{dx}{1+t\sec x} \right] dt=\int _{0}^{\pi }\int _{\alpha }^{\beta }\cfrac{dt}{1+t\sec x} dx=\int _{0}^{\pi }dx\cos x\ln \left|\cfrac{1+\beta \sec x}{1+\alpha \sec x} \right| }$

Combining the two orders of integration we have found a nontrivial definite integral.

$\displaystyle {\int _{0}^{\pi }\cos x\ln \left|\cfrac{1+\beta \sec x}{1+\alpha \sec x} \right| dx=\pi (\sqrt{\beta ^{2} -1} -\sqrt{\alpha ^{2} -1} )}$

Example 2. Show that

$\displaystyle \int _{0}^{1}dx\cfrac{(x^{\beta } -x^{\alpha } )}{\ln x} =\ln \left|\cfrac{\beta +1}{\alpha +1} \right|$

Solution 2. By equate the two orders of integration, we have solved an unusual integral by integration under the integral sign.

$\displaystyle {\int _{\alpha }^{\beta }\left[\int _{0}^{1}x^{t} dx \right] dt=\int _{\alpha }^{\beta }\cfrac{dt}{t+1} =\ln \left|\cfrac{\beta +1}{\alpha +1} \right| } \\ {\int _{x=0}^{1}\left[\int _{t=\alpha }^{\beta }x^{t} dt \right]dx =\int _{0}^{1}dx\cfrac{(x^{\beta } -x^{\alpha } )}{\ln x} }$

$\displaystyle \int _{0}^{1}dx\cfrac{(x^{\beta } -x^{\alpha } )}{\ln x} =\ln \left|\cfrac{\beta +1}{\alpha +1} \right|$

This result is related to one of the previous problems.

$\displaystyle \int _{0}^{\infty }\cfrac{e^{-\mu x} -e^{-\nu x} }{x} dx=\ln \left(\nu /\mu \right)$

Example 3. Show how integration under the integral sign works for the following generalization
of the previous example.

$\displaystyle \int _{a}^{b}\cfrac{dx}{f(x)} \ln \left|\cfrac{1+\alpha f(x)}{1+\beta f(x)} \right|$

Solution 3. We may be able to solve an integral of the above form if we can integrate under the integral sign.

$\displaystyle \int _{a}^{b}\left[\int _{\alpha }^{\beta }\cfrac{dt}{1+tf(x)} \right] dx=\int _{a}^{b}\cfrac{dx}{f(x)} \ln \left|\cfrac{1+\alpha f(x)}{1+\beta f(x)} \right|$

This means we have to be able evaluate the integral

$\displaystyle \int _{\alpha }^{\beta }\left[\int _{a}^{b}\cfrac{dx}{1+tf(x)} \right]dt$

This integral is the same as the previous integral except the order of integration has been reversed.

Example 4. Integrating a known result by integration under the integral sign. Use the result

$\displaystyle \int _{0}^{\pi /2}\cfrac{dx}{(\alpha -\cos x)^{2} } =\cfrac{2\pi \alpha }{(\alpha ^{2} -1)^{3/2} }$

to find

$\displaystyle \int _{0}^{\pi /2}\cfrac{dx}{(\alpha -\cos x)} =?$

Solution 4. In this problem we undo the result that we found by differentiation under the integral sign in the previous section. Integrate the left and right hand sides with respect to \textit{t}.

$\displaystyle {\int _{0}^{\pi /2}dx\int _{-\infty }^{\alpha }\cfrac{dt}{(t-\cos x)^{2} } =\int _{0}^{\pi /2}\cfrac{-1}{\alpha -\cos x} dx } \\ {\int _{-\infty }^{\alpha }\cfrac{2\pi tdt}{(t^{2} -1)^{3/2} } =\left. -\cfrac{\pi }{\sqrt{t^{2} -1} } \right|_{-\infty }^{\alpha } =-\cfrac{\pi }{\sqrt{\alpha ^{2} -1} } } \\ {\int _{0}^{\pi /2}\cfrac{1}{\alpha -\cos x} dx=\cfrac{\pi }{\sqrt{\alpha ^{2} -1} } }$

Notice how we took the limit of integration to be $latex \beta \to -\infty &s=1$ and we have the result that

$\displaystyle \int _{0}^{\pi /2}\cfrac{dx}{\alpha -\cos x} =\cfrac{\pi }{\sqrt{\alpha ^{2} -1} }$

Example 5. Apply Fubini’s theorem over the rectangle $x\in [a,b],t[\alpha ,\beta ]$ to show that

$\displaystyle \int _{a}^{b}[(\beta +\ln x)^{2} -(\alpha +\ln x)^{2} ]dx=(b-a)(\beta -\alpha )^{2} +2(b\ln b-a\ln a)(\beta -\alpha )$

Solution 5. Fubini’s theorem gives us another result by integration under the integral sign, however this integral is not too impressive because it can be solved in either order using standard techniques.

$\displaystyle {\int _{a}^{b}\int _{\alpha }^{\beta }(t+\ln x)dtdx } = {\int _{a}^{b}\int _{\alpha +\ln x}^{\beta +\ln x}ududx =\cfrac{1}{2} \int _{a}^{b}[(\beta +\ln x)^{2} -(\alpha +\ln x)^{2} ]dx} \\ {\int _{\alpha }^{\beta }\int _{a}^{b}(t+\ln x)dx dt} = {\int _{\alpha }^{\beta }[(b-a)t+(b\ln b-a\ln a) ]dt}$

Equating the two we have

$\displaystyle \int _{a}^{b}\cfrac{1}{2} [(\beta +\ln x)^{2} -(\alpha +\ln x)^{2} ]dx =\cfrac{1}{2} (b-a)(\beta -\alpha )^{2} +(b\ln b-a\ln a)(\beta -\alpha )$

Cancel the factors of two and get

$\displaystyle \int _{a}^{b}[(\beta +\ln x)^{2} -(\alpha +\ln x)^{2} ]dx =(b-a)(\beta -\alpha )^{2} +2(b\ln b-a\ln a)(\beta -\alpha )$

A few things can go wrong when trying to integrate under the integral sign. When attempting integration under the integral sign if the integrals are too difficult we may will end up with a relationship between integrals, rather than a result. If the function is too simple, perhaps we will just integrate through both ways.

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