Evaluate the following integrals or show that the integral diverges. \int_{-\infty}^{\infty}\frac{x^{3}}{1+x^{8}}dx

David Young 2021-12-16 Answered
Evaluate the following integrals or show that the integral diverges.
\(\displaystyle{\int_{{-\infty}}^{{\infty}}}{\frac{{{x}^{{{3}}}}}{{{1}+{x}^{{{8}}}}}}{\left.{d}{x}\right.}\)

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Expert Answer

nghodlokl
Answered 2021-12-17 Author has 2431 answers

Step 1
Given- \(\displaystyle{\int_{{-\infty}}^{{\infty}}}{\frac{{{x}^{{{3}}}}}{{{1}+{x}^{{{8}}}}}}{\left.{d}{x}\right.}\)
To evaluate- The above improper integral.
Concept Used- The above integral can be solved by using the substitution method.
Step 2
Explanation- Rewrite the given integral as,
\(\displaystyle{I}={\int_{{-\infty}}^{{\infty}}}{\frac{{{x}^{{{3}}}}}{{{1}+{x}^{{{8}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{-\infty}}^{{\infty}}}{\frac{{{x}^{{{3}}}}}{{{1}+{\left({x}^{{{4}}}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}\)
Now, substituting \(x^{4}=t\) and differentiating w.r.t. x, we get,
\(\displaystyle{4}{x}^{{{3}}}{\left.{d}{x}\right.}={\left.{d}{t}\right.}\)
\(\displaystyle{x}^{{{3}}}{\left.{d}{x}\right.}={\frac{{{\left.{d}{t}\right.}}}{{{4}}}}\)
At \(\displaystyle{x}=-\infty,{t}=-\infty\) and at \(\displaystyle{x}=\infty,{t}=\infty\).
Now, from the above integral,we can write as,
Step 3
\(\displaystyle={\infty_{{-\infty}}^{{\infty}}}{\frac{{{\frac{{{\left.{d}{t}\right.}}}{{{4}}}}}}{{{1}+{t}^{{{2}}}}}}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}{\int_{{-\infty}}^{{\infty}}}{\frac{{{1}}}{{{1}+{t}^{{{2}}}}}}{\left.{d}{t}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}{{\left[{{\tan}^{{-{1}}}{\left({t}\right)}}\right]}_{{-\infty}}^{{+\infty}}}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}{\left[{{\tan}^{{-{1}}}{\left(\infty\right)}}-{{\tan}^{{-{1}}}{\left(-\infty\right)}}\right]}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}{\left[{\frac{{\pi}}{{{2}}}}-{\left({\frac{{-\pi}}{{{2}}}}\right)}\right]}\)
\(\displaystyle={\frac{{\pi}}{{{4}}}}\)
Answer- Hence, the value of the integral \(\displaystyle{\int_{{-\infty}}^{{\infty}}}{\frac{{{x}^{{{3}}}}}{{{1}+{x}^{{{8}}}}}}{\left.{d}{x}\right.}\) is \(\displaystyle{\frac{{\pi}}{{{4}}}}\).

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Marcus Herman
Answered 2021-12-18 Author has 2249 answers
It is required to calculate:
\(\displaystyle\int{\frac{{{x}^{{{3}}}}}{{{x}^{{{8}}}+{1}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}\int{\frac{{{1}}}{{{u}^{{{2}}}+{1}}}}{d}{u}\)
Now we calculate:
\(\displaystyle\int{\frac{{{1}}}{{{u}^{{{2}}}+{1}}}}{d}{u}\)
This is the well-known tabular integral:
\(\displaystyle={\arctan{{\left({u}\right)}}}\)
We substitute the already calculated integrals:
\(\displaystyle{\frac{{{1}}}{{{4}}}}\int{\frac{{{1}}}{{{u}^{{{2}}}+{1}}}}{d}{u}\)
\(\displaystyle={\frac{{{\arctan{{\left({u}\right)}}}}}{{{4}}}}\)
Reverse replacement \(\displaystyle{u}={x}^{{{4}}}:\)
\(\displaystyle={\frac{{{\arctan{{\left({x}^{{{4}}}\right)}}}}}{{{4}}}}\)
\(\displaystyle\int{\frac{{{x}^{{{3}}}}}{{{x}^{{{8}}}+{1}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{\arctan{{\left({x}^{{{4}}}\right)}}}}}{{{4}}}}+{C}\)
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nick1337
Answered 2021-12-28 Author has 10701 answers

We put the expression \(4*x^{3}\) under the differential sign, i.e.:
\(4x^{3}dx=d(x^{4}), t=x^{4}\)
Then the original integral can be written as follows:
\(\int \frac{1}{4*(t^{2}+1)}*dt\)
This is a table integral:
\(\int \frac{1}{4*(t^{2}+1)}*dt=\frac{\arctan (t)}{4}+C\)
To write down the final answer, it remains replace t with \(x^{4}\)
\(\frac{\arctan (x^{4})}{4}+C\)

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