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Question Number 167526 by Gbenga last updated on 18/Mar/22

$${\int }_0^{\frac{\mathit{\pi }}{2}}\underset{\mathbf{n}=1}{\mathbf{\sum }^{\mathrm{\infty }}}\frac{1}{{\mathbf{n}}^2+1}\mathbf{dn}=???$$

Commented by aleks041103 last updated on 18/Mar/22

$$howdoyousumandintegrateover$$$$thesamething-n???$$

Answered by MJS_new last updated on 18/Mar/22

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+1}=\frac{\pi }{2\tanh \pi }-\frac{1}{2}=C$$$$\underset{0}{\stackrel{\pi /2}{\int }}Cdn=\frac{C\pi }{2}=\frac{{\pi }^2}{4\tanh \pi }-\frac{\pi }{4}$$

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