Solve Integral by Parts: exsqrt(x)

[leprofece]

integral is e^xsqrt(x)
ok here u = sqrt(x) du = 1/(2sqrt(x))
dv e^x= v= e^x
so e^xsqrt(x) - integral( 1/2sqrt(x)e^x)
And I can't continue because i can not get rid of e^x??
How must I proceed??

 

[Prove It]

It doesn't have an antiderivative in terms of the elementary functions. Apparently you need to use the imaginary error function.

 

[soroban]

Hello, leprofece!

$\int e^{\sqrt{x}}dx$

Let $w \,=\,\sqrt{x} \quad\Rightarrow\quad x \,=\,w^2 \quad\Rightarrow\quad dx \,=\,2w\,dw$

Substitute: $\displaystyle\;\;\int e^w(2w\,dw) \;=\;2\int we^w\,dw$

By parts: $\;\begin{Bmatrix}u &=& w && dv &=& e^wdw \\ du &=& dw && v &=& e^w\end{Bmatrix}$

We have: $\displaystyle\;2\left(we^w - \int e^wdx\right)$

. . . . . $=\;2(we^w - e^w) + C \;=\;2e^w(w-1) + C$Back-substitute: $\;2e^{\sqrt{x}}\left(\sqrt{x}-1\right)+C$

 

[Prove It]

Hello, leprofece!


Let $w \,=\,\sqrt{x} \quad\Rightarrow\quad x \,=\,w^2 \quad\Rightarrow\quad dx \,=\,2w\,dw$

Substitute: $\displaystyle\;\;\int e^w(2w\,dw) \;=\;2\int we^w\,dw$

By parts: $\;\begin{Bmatrix}u &=& w && dv &=& e^wdw \\ du &=& dw && v &=& e^w\end{Bmatrix}$

We have: $\displaystyle\;2\left(we^w - \int e^wdx\right)$

. . . . . $=\;2(we^w - e^w) + C \;=\;2e^w(w-1) + C$Back-substitute: $\;2e^{\sqrt{x}}\left(\sqrt{x}-1\right)+C$

None of this post is relevant, as it is NOT the function to be integrated, nor is this integral a result of the integration by parts.

The integral to be computed is actually $\displaystyle \begin{align*} \int{ \sqrt{x}\,\mathrm{e}^x\,\mathrm{d}x } \end{align*}$

 

[chisigma]

integral is e^xsqrt(x)
ok here u = sqrt(x) du = 1/(2sqrt(x))
dv e^x= v= e^x
so e^xsqrt(x) - integral( 1/2sqrt(x)e^x)
And I can't continue because i can not get rid of e^x??
How must I proceed??

You can proceed by parts...

$\displaystyle \int \sqrt{x}\ e^{x}\ dx = \frac{2}{3}\ x^{\frac{2}{3}}\ e^{x} - \frac{4}{15}\ x^{\frac{5}{2}}\ e^{x} + \frac{8}{105}\ x^{\frac{7}{2}}\ e^{x} - ... + (-1)^{n+1} \sqrt{x}\ e^{x}\ \frac{(2\ x)^{n}}{(2\ n + 1)!}\ + ...\ (1) $

Now You can use the nice series expansion...

$\displaystyle \sum_{n=0}^{\infty} \frac{t^{n}}{(2\ n +1)!} = \sqrt{\frac{\pi}{2\ t}} \ e^{\frac{t}{2}}\ \text{erf}\ (\sqrt{\frac{t}{2}})\ (2) $

... to arrive at...

$\displaystyle \int \sqrt{x}\ e^{x}\ dx = \sqrt{x}\ e^{x} - \frac{1}{2\ i}\ \sqrt{\frac{\pi}{x}}\ \text{erf}\ (i\ \sqrt{x}) + c\ (3)$

Kind regards

$\chi$ $\sigma$

 

[DreamWeaver]

Alternatively, you could apply the Exponential Integral function \(\displaystyle \text{Ei}(x)\) defined below:
\(\displaystyle \text{Ei}(x) = \int \frac{e^x}{x}\, dx = \log|x|+\sum_{k=1}^{\infty}\frac{x^k}{k\cdot k!}\)

\(\displaystyle \text{Ei}'(x) = \frac{d}{dx}\text{Ei}(x) = \frac{d}{dx}\, \int \frac{e^x}{x}\, dx = \frac{e^x}{x} \)Hence\(\displaystyle \int e^x\sqrt{x}\, dx = \int x^{3/2} \frac{e^x}{x}\, dx = \int x^{3/2} \left[ \frac{d}{dx} \text{Ei}(x) \right]\, dx = \)\(\displaystyle x\sqrt{x}\,\text{Ei}(x) - \frac{3}{2}\, \int \sqrt{x}\, \text{Ei}(x)\, dx = \)\(\displaystyle x\sqrt{x}\,\text{Ei}(x) - \frac{3}{2}\, \int \sqrt{x}\, \Bigg\{ \log|x|+\sum_{k=1}^{\infty}\frac{x^k}{k\cdot k!} \Bigg\} \, dx = \)\(\displaystyle x\sqrt{x}\,\text{Ei}(x) - \frac{3}{2}\, \int \sqrt{x} \log|x|\, dx + \frac{3}{2}\, \sum_{k=1}^{\infty}\frac{1}{k\cdot k!} \int x^{k+1/2} \, dx = \)\(\displaystyle x\sqrt{x}\,\text{Ei}(x) - \frac{3}{2}\, \int \sqrt{x} \log|x|\, dx + 3x\sqrt{x}\, \sum_{k=1}^{\infty}\frac{x^k}{k(2k+3)\cdot k!} \)Finally, assuming that \(\displaystyle x > 0\), we can drop the absolute value sign in the logarithm.

\(\displaystyle \int \sqrt{x}\log x\, dx = \frac{2 x\sqrt{x}}{3}\, \log x - \frac{4 x\sqrt{x}}{9}\)Hence\(\displaystyle \int e^x\sqrt{x}\, dx =\)\(\displaystyle x\sqrt{x}\,\text{Ei}(x) - x\sqrt{x}\log x + \frac{2x\sqrt{x}}{3} + 3x\sqrt{x}\, \sum_{k=1}^{\infty}\frac{x^k}{k(2k+3)\cdot k!} =
\)\(\displaystyle x\sqrt{x}\, \left[ \text{Ei}(x) - \log x + \frac{2}{3} + 3\, \sum_{k=1}^{\infty}\frac{x^k}{k(2k+3)\cdot k!} \right]
\)

 

[leprofece]

Alternatively, you could apply the Exponential Integral function \(\displaystyle \text{Ei}(x)\) defined below:
\(\displaystyle \text{Ei}(x) = \int \frac{e^x}{x}\, dx = \log|x|+\sum_{k=1}^{\infty}\frac{x^k}{k\cdot k!}\)

\(\displaystyle \text{Ei}'(x) = \frac{d}{dx}\text{Ei}(x) = \frac{d}{dx}\, \int \frac{e^x}{x}\, dx = \frac{e^x}{x} \)Hence\(\displaystyle \int e^x\sqrt{x}\, dx = \int x^{3/2} \frac{e^x}{x}\, dx = \int x^{3/2} \left[ \frac{d}{dx} \text{Ei}(x) \right]\, dx = \)\(\displaystyle x\sqrt{x}\,\text{Ei}(x) - \frac{3}{2}\, \int \sqrt{x}\, \text{Ei}(x)\, dx = \)\(\displaystyle x\sqrt{x}\,\text{Ei}(x) - \frac{3}{2}\, \int \sqrt{x}\, \Bigg\{ \log|x|+\sum_{k=1}^{\infty}\frac{x^k}{k\cdot k!} \Bigg\} \, dx = \)\(\displaystyle x\sqrt{x}\,\text{Ei}(x) - \frac{3}{2}\, \int \sqrt{x} \log|x|\, dx + \frac{3}{2}\, \sum_{k=1}^{\infty}\frac{1}{k\cdot k!} \int x^{k+1/2} \, dx = \)\(\displaystyle x\sqrt{x}\,\text{Ei}(x) - \frac{3}{2}\, \int \sqrt{x} \log|x|\, dx + 3x\sqrt{x}\, \sum_{k=1}^{\infty}\frac{x^k}{k(2k+3)\cdot k!} \)Finally, assuming that \(\displaystyle x > 0\), we can drop the absolute value sign in the logarithm.

\(\displaystyle \int \sqrt{x}\log x\, dx = \frac{2 x\sqrt{x}}{3}\, \log x - \frac{4 x\sqrt{x}}{9}\)Hence\(\displaystyle \int e^x\sqrt{x}\, dx =\)\(\displaystyle x\sqrt{x}\,\text{Ei}(x) - x\sqrt{x}\log x + \frac{2x\sqrt{x}}{3} + 3x\sqrt{x}\, \sum_{k=1}^{\infty}\frac{x^k}{k(2k+3)\cdot k!} =
\)\(\displaystyle x\sqrt{x}\, \left[ \text{Ei}(x) - \log x + \frac{2}{3} + 3\, \sum_{k=1}^{\infty}\frac{x^k}{k(2k+3)\cdot k!} \right]
\)

Thank for your answer but you must solve it by normal or traditional methods students have not studied series yet
Could anybody do that way??

 

[DreamWeaver]

Thank for your answer but you must solve it by normal or traditional methods students have not studied series yet
Could anybody do that way??

Hello, Leprofece! (Sun)

It sounds to me like they've asked you to solve a problem that isn't solvable (with what they've taught you so far). A possible case of "Bad teacher!", I think...