MHB How can I simplify this integral using integration by parts?

[ra_forever8]

Consider the integral
\begin{equation}
I(x)=\int^{2}_{0} (1+t) e^{xcos[\pi (t-1)/2]} dt
\end{equation}
show that
\begin{equation}
I(x)= 4+ \frac{8}{\pi}x +O(x^{2})
\end{equation}
as $x\rightarrow0$.=> Using integration by parts, but its too complicated for me because of huge exponential term.
please help me.

 

[Evgeny.Makarov]

First represent the integrand as $f(t)+xg(t)+O(x^2)$.

 

[chisigma]

Consider the integral
\begin{equation}
I(x)=\int^{2}_{0} (1+t) e^{xcos[\pi (t-1)/2]} dt
\end{equation}
show that
\begin{equation}
I(x)= 4+ \frac{8}{\pi}x +O(x^{2})
\end{equation}
as $x\rightarrow0$.=> Using integration by parts, but its too complicated for me because of huge exponential term.
please help me.

If You expand I(x) in McLaurin series You have...

$\displaystyle I(x) = I(0) + I^{\ '} (0)\ x + \mathcal{O} (x^{2})\ (1)$

The first term is...

$\displaystyle I(0) = \int_{0}^{2} (1 + t)\ d t = 4\ (2)$

The $\displaystyle I^{\ '} (x)$ can be obtained deriving into the integral and is...

$\displaystyle I^{\ '} (0) = \int_{0}^{2} \cos [\frac{\pi}{2}\ (t-1)]\ (1 + t)\ d t = - \frac{2}{\pi^{2}}\ | \pi\ (1+ t)\ \cos ( \frac{\pi}{2}\ t )- 2\ \sin ( \frac{\pi}{2}\ t) |_{0}^{2} = \frac{8}{\pi}\ (3) $

Kind regards

$\chi$ $\sigma$

 

[ra_forever8]

That was so clear. you made the solution so easy to understand. Thank you very much sir.

 

[ra_forever8]

The $\displaystyle I^{\ '} (x)$ can be obtained deriving into the integral and is...

$\displaystyle I^{\ '} (0) = \int_{0}^{2} \cos [\frac{\pi}{2}\ (t-1)]\ (1 + t)\ d t $

how did you get the above term or $I'(x)$ term?
please clarify me. other than that everything is perfect.

 

[Saitama]

The $\displaystyle I^{\ '} (x)$ can be obtained deriving into the integral and is...

$\displaystyle I^{\ '} (0) = \int_{0}^{2} \cos [\frac{\pi}{2}\ (t-1)]\ (1 + t)\ d t $

how did you get the above term or $I'(x)$ term?
please clarify me. other than that everything is perfect.

chisigma used this: Differentiation under the integral sign - Wikipedia, the free encyclopedia

He differentiated wrt $x$.