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I'm having a really hard time figuring out how to solve the following integrals: $$ \int_0^1 \frac{e^{-t} }{1+t} dt\, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \int_0^1 \frac{e^{-t} }{(1+t)^2} dt\, $$ According to my professor, the m-th Exponential Integral is suppose to be used. I'm not that familiar with the Exponential Integral and am really unsure how to get the above integrals in the right format, particularly for the bounds (Does a transformation need to be done?), or solve it. The equation we were given for the Exponential Integral is $E_m(x)=\int_1^\infty \frac{e^{-xt} }{t^m} dt\,$. Any insight would be incredibly helpful (especially steps!!). Thank you!

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    Do I not need to worry about the bounds matching the exponential integral? – 2017-02-28
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    What about the Taylor series for $e^{-t} $ and for $\frac{1}{1+t}$ and then multiply? Interesting pattern to product and you can get pattern. – 2017-02-28

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Note that we can write

$$\begin{align} \int_0^1 \frac{e^{-t}}{t+1}\,dt&=e\int_1^2 \frac{e^{-t}}{t}\,dt\\\\ &=e\left(\int_1^\infty \frac{e^{-t}}{t}\,dt-\int_2^\infty \frac{e^{-t}}{t}\,dt\right)\\\\ &=e\left(\int_1^\infty \frac{e^{-t}}{t}\,dt-\int_1^\infty \frac{e^{-2t}}{t}\,dt\right)\\\\ &=e\left(E_1(1)-E_1(2)\right) \end{align}$$

Therefore, we can write

$$\bbox[5px,border:2px solid #C0A000]{\int_0^1 \frac{e^{-t}}{t+1}\,dt=e\left(E_1(1)-E_1(2)\right)}$$

Note that we can write

$$\begin{align} \int_0^1 \frac{e^{-t}}{(t+1)^2}\,dt&=e\int_1^2 \frac{e^{-t}}{t^2}\,dt\\\\ &=e\left(\int_1^\infty \frac{e^{-t}}{t^2}\,dt-\int_2^\infty \frac{e^{-t}}{t^2}\,dt\right)\\\\ &=e\left(\int_1^\infty \frac{e^{-t}}{t}\,dt-\frac12 \int_1^\infty \frac{e^{-2t}}{t}\,dt\right)\\\\ &=e\left(E_2(1)-\frac12 E_2(2)\right) \end{align}$$

Therefore, we can write

$$\bbox[5px,border:2px solid #C0A000]{\int_0^1 \frac{e^{-t}}{(t+1)^2}\,dt=e\left(E_1(1)-E_1(2)\right)}$$

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    Once you split the integrals up, why does the second one have an e in front of it? – 2017-03-01
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    @Sherlock Those were typos. I've edited. And nice catch! -Mark – 2017-03-01
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How about using the geometric series? That is

$$\frac{1}{1+t}=\sum_{n=0}^\infty (-t)^n$$

Then we have

$$ \int^1_0\frac{e^{-t}}{1+t}\,dt = \sum_{n=0}^\infty \int^1_0 e^{-t}(-t)^n\,dt. $$

This isn't quite the exponential integral, since the power of $t$ is in the numerator instead of the denominator, and the bounds are $0\leq t\leq1$ instead of $1\leq t\leq\infty$. So let's invert $t$ to transform that. Let $u=\frac{1}{t}$ and we have:

$$ \int^1_0e^{-t}t^n\,dt = -\int^1_\infty \frac{e^{-1/u}}{u^n}\frac{1}{u^2}\,du = \int^\infty_1 \frac{e^{-1/u}}{u^{n+2}}\,du $$

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    that geometric series should alternate – 2017-02-28
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    @spaceisdarkgreen thanks for the correction. – 2017-02-28
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    That's not in the form of the exponential integrals. – 2017-02-28