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How would I go about to trying to evaluate this Triple Intergral?

$$\int_{a}^{\infty} \int_{4+a}^{2a+1} \int_{4a+b}^{2b+1} \frac{{2a+1}}{ {2a}(b^2+c)^{1/2}} \, {dc}\,{db}\,da=$$

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    What have you tried ? Also are you sure you typed in the integral bounds for $b$ correctly? The $b$ bounds should depend on either $a,c$ or be constants, but should not have $b$ in them . – 2017-02-22
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    your internal integral is $db$ but limits also depend on $b$. Did you mean to integrate $dcdbda$? – 2017-02-22
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    @gt6989b And, the outer integral depends on $a$. You can't have any variables on the limits of integration for the outermost integral. – 2017-02-22
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    Please make our life easier: what is the original problem that turns into this mess after an incorrect change of variables? – 2017-02-22
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    Yeah that was my mistake. I put the order wrong initially but this is the correct order. So I am just trying to figure out first if there is a solution based on the order of which I am trying to integrate? – 2017-02-22
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    Could I factor out the constant and some how use a trig identity? – 2017-02-22

1 Answers 1

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Assuming the lower bound on the outermost integral is supposed to be $0$, then we have

\begin{align} \int_{0}^{\infty} \int_{4+a}^{2a+1} \int_{4a+b}^{2b+1} \frac{2a+1}{2a\sqrt{b^2+c}} dc \, db \, da &= \int_{0}^{\infty} \int_{4+a}^{2a+1} \frac{2a+1}{2a} \left[\int_{4a+b}^{2b+1} \frac{1}{\sqrt{b^2+c}} dc \right]\, db \, da \nonumber \\ &= \int_{0}^{\infty} \int_{4+a}^{2a+1} \frac{2a+1}{2a} \left[2\sqrt{c+b^2} \right]_{c=4a+b}^{2b+1}\, db \, da \nonumber \\ &= \int_{0}^{\infty} \int_{4+a}^{2a+1} \frac{2a+1}{2a} \left[2\sqrt{b^2+2b+1} - 2\sqrt{b^2+b+4a} \right]\, db \, da \nonumber \\ &= \int_{0}^{\infty} \int_{4+a}^{2a+1} \frac{2a+1}{2a} \left[2\sqrt{(b+1)^2} - 2\sqrt{\left(b + \frac{1}{2} \right)^2+ \left(4a-\frac{1}{4} \right)} \right]\, db \, da \nonumber \\ &= \int_{0}^{\infty} \frac{2a+1}{2a} \left[ \int_{4+a}^{2a+1} \left(2b+2 - 2\sqrt{\left(b + \frac{1}{2} \right)^2+ \left(4a-\frac{1}{4} \right)} \right) db \right]\, da \end{align}

To evaluate the inner integral, we find the following indefinite integral using integration by parts and the substitution $x = \sqrt{c} \sinh u$ (implying $dx = \sqrt{c} \cosh u \, dx$): \begin{align*} \int \sqrt{x^2+c} \, dx &= x\sqrt{x^2+c} - \int \frac{x^2}{\sqrt{x^2+c}} dx \\ &= x\sqrt{x^2+c} - \int \left(\frac{x^2+c}{\sqrt{x^2+c}} - \frac{c}{\sqrt{x^2+c}} \right) dx \\ &= x\sqrt{x^2+c} - \int \frac{x^2+c}{\sqrt{x^2+c}} dx + \int\frac{c}{\sqrt{x^2+c}} dx \\ &= x\sqrt{x^2+c} - \int \sqrt{x^2+c}\, dx + \int\frac{c}{\sqrt{x^2+c}} dx \\ 2\int \sqrt{x^2+c} \, dx &= x\sqrt{x^2+c} + \int\frac{c}{\sqrt{x^2+c}} dx \\ &= x\sqrt{x^2+c} + \int\frac{c}{\sqrt{c \sinh^2 u +c}} (\sqrt{c} \cosh u) du \\ &= x\sqrt{x^2+c} + \int\frac{c}{\sqrt{c \cosh^2 u}} (\sqrt{c} \cosh u) du \\ &= x\sqrt{x^2+c} + \int\frac{c}{\sqrt{c} \cosh u} (\sqrt{c} \cosh u) du \\ &= x\sqrt{x^2+c} + \int c du \\ &= x\sqrt{x^2+c} + cu + C \\ &= x\sqrt{x^2+c} + c \sinh^{-1} \left(\frac{x}{\sqrt{c}} \right) + C \end{align*} Therefore: \begin{align*} &\int_{4+a}^{2a+1} \left(2b+2 - 2\sqrt{\left(b + \frac{1}{2} \right)^2+ \left(4a-\frac{1}{4} \right)} \right) db \\ &= \left[b^2 + 2b - \left(b + \frac{1}{2} \right)\sqrt{\left(b + \frac{1}{2} \right)^2+\left(4a-\frac{1}{4} \right)} - \left(4a-\frac{1}{4} \right) \sinh^{-1} \left(\frac{b + \frac{1}{2}}{\sqrt{4a-\frac{1}{4}}} \right) \right]_{b = 4+a}^{2a+1} \\ &= \left[b^2 + 2b - \frac{1}{4}(2b + 1)\sqrt{(2b + 1)^2+(16a-1)} - \frac{1}{4}(16a-1) \sinh^{-1} \left(\frac{2b + 1}{\sqrt{16a-1}} \right) \right]_{b = 4+a}^{2a+1} \end{align*}

I stop here because it is getting quite messy. I hope this is of help to you. Note the above is just the integrand for one more integral in $a$.